CONSTRUCTION MATHEMATICS
By Spencer Hinkle
Construction
Mathematics
 Left Coast Publishing
5235 SW 26th Drive
Portland, Oregon. 97201
Phone 503-246-8194 • Email swhinkle@msn.com
Table of Contents
CHAPTER 1
LI NEAR
CHAPTER 4
MEASURING
Linear measuring
1
Imperial System of Measuring
1
Sixteenths
2
Tape Measures
3
Homework
4
CHAPTER 2
UNDERSTANDING FRACTIONS
AREA MEASUREMENTS
The square
26
Rectangles
27
Parallelograms
27
Rhombus
28
Trapezoids
28
Triangles
29
Hero’s Formula
31
The Area of a Circle
32
The Anatomy of a Fraction
6
Complex Area Problems
32
Adding Fractions
7
Square Yards
35
Subtracting Fractions
7
Materials Sold by the Square
35
Expressing Fractions as Decimals
8
Area Formulas
37
Expressing Decimals as Fractions
8
Using the Calculator
9
Decimal Feet VS Feet and Inches
10
Homework
13
CHAPTER 5
CALCULATING THE COST OF BUILDING
MATERIALS
Framing Lumber Terminology
38
CHAPTER 3
Linear Measure of Lumber
39
SURFACE MEASUREMENTS
Board Foot Measure
40
Polygons
18
Pricing by the Linear Foot
44
Circles
20
Pricing by the Square Foot
45
Linear and Perimeter Measurements
21
Pricing Lap Siding
Circumference
22
CHAPTER 6
REPETITIVE FRAMING MEMBERS
On-center Measurements
47
Laying Out Evenly Spaced Balusters
53
CHAPTER 7
VOLUMETRIC MEASUREMENTS
CHAPTER 11
The Cubic Unit
57
TRIGONOMETRY FOR CONSTRUCTION
Cubic Yardage
59
Definition
111
Closed Foundation Walls & Footings
60
Triangles in Construction
111
Air Volume
64
Right Triangles
112
Volume Formulas
65
Labeling the Triangle for Trig
117
Step By Step Trig
118
Trig Functions
121
Using Your Calculator for Trig
123
More on Trig
134
CHAPTER 8
RATIOS AND PROPORTIONS
Ratios and Factors
67
Proportions
70
Indirect Proportions
74
CHAPTER 9
STAIR LAYOUT
Terminology
76
Calculating Stair Rise & Run
77
Sizing Treads & Risers
78
CHAPTER 12
PERCENTAGES, DISCOUNTS & MARKUP
Percentages
137
Discounts
140
Double Discounts
144
Working Backwards
142
Markup
144
THE HP48G CALCULATOR
CHAPTER 10
CONSTRUCTION GEOMETRY
Getting Started with the HP48G
147
Symbolic Mode
149
Definition and History of Geometry
83
Setting Decimal Places
150
Pythagorean Theorem
85
3-4-5 Squaring Method
85
Adding, Subtracting, Multiplying
& Dividing
150
Squaring Building Corners
88
Setting the Clock, Date and Alarms
152
Roof Framing
91
Transferring Data
153
Roof Terminology
92
Calculating Roof Properties
93
Line Length Calculation
94
More Roof Terminology
96
Dropping the Ridge
98
The Hip Roof
101
Calculating the Hip Constant
104
Calculating the Common Difference
107
Roof Overhang
109
1
Chapter
M E A S U R I N G
LINEAR MEASURING
A measure of length
Carpenters are constantly taking linear measurements. They read measurements written
on blueprints and they use tape measures to layout and cut materials to length.
Cabinetmakers measure spaces in which cabinets are to be installed and estimators take
measurements off of blueprints and calculate material and labor costs.
In the United States the Imperial system, which expresses linear distances in feet and
inches is still widely used. Much of the rest of the world uses the metric system. If you
visit a construction site in Canada you will likely see the carpenters carrying metric tape
measures.
The metric system is actually quite easy to learn and use because there are no
conversions. Conversely, the Imperial system, using feet, inches and fractions of an
inch, requires conversions in order to carry out calculations. For example in the metric
system if you wanted to add 793 mm (millimeters) to 620 mm it’s a simple addition
problem. 793 + 620 = 1413 mm. You can probably do that without pencil and paper.
The same measurements in the Imperial system would be approximately - 2-7 1/4 +
2-4 15/16. Add that up in your head! The answer is 5- 0 3/16. The methods for
solving these types of problems may be found in the next chapter.
Note: You will find a dash between feet and inches, as above, in architectural dimensioning.
THE IMPERIAL SYSTEM OF MEASURING
Let’s talk about the Imperial system of measure a little more since we are stuck with it.
Most of us have a pretty good idea of what one-foot looks like. The paper on which
these words are written is one half inch shy of being one foot in height. A foot is
further broken down into smaller units called inches. We all know there are twelve
inches in one foot.
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M E A S U R I N G
Inches are broken down into smaller increments, which we express as fractions, onehalf inch, one-eighth inch, etc. Take a look at your tape measure and count how many
divisions there are in one inch. Did you count sixteen? If you used the first inch on the
tape it may have been broken down into 32 divisions or thirty seconds of an inch.
I once worked with a carpenter who had trouble reading sixteenths on his tape
measure and he would call out a measurement like this - twelve feet, three inches
and one little one past a big one. Hopefully we can improve on his system.
For now, let’s concentrate on the blown up inch illustrated below. Each space between
lines or hash marks is one sixteenth of one inch because there are sixteen spaces. When
a fraction of an inch is expressed it should be reduced to its lowest common
denominator. For example 8/16 is expressed as 1/2 and 4/16 is expressed as 1/4.
Notice the lines representing each sixteenth of an inch are of different lengths. This is
designed to make it easier for you to find and reduce the fractions.
There are five lengths:
1
2
K E Y
1 – Whole inch
2 - Eight inch
3 - Half inch
4 - Quarter inch
5 - Sixteenth inch
3
4
5
Why sixteenths? Sixteenths are normally the smallest unit
carpenters use. Smaller divisions such as thirty seconds or sixty
fourths could be used, but in most cases they are not practical.
There are exceptions however.
In cabinetmaking and finish work, carpenters use thirty
seconds in an unusual and very practical way. If a measurement of 3 3/32 were
taken, it would be written down or called out as 3 1/16+.
Explanation: One-sixteenth equal’s 2/32 and the plus sign indicates one additional
thirty-second. A little more sophisticated than one little one past a big one.
With practice you will be able to quickly measure any length. To help shorten the
learning curve remember that 1/2 is 8/16 so the hash mark before the one half inch
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M E A S U R I N G
mark is 7/16 and one past the one half inch mark is 9/16. Three quarters of an inch
is 12/16 so the hash mark before it is 11/16 and the one after is 13/16.
TAPE MEASURES
Lengths – Tape measures come in a variety of lengths: One hundred footers, fifty
footers, thirty footers, twenty-five footers, twenty footers, sixteen footers, twelve
footers and ten footers.
Blades – Most of the commonly used retractable tape measures have a Mylar blade
that is flexible but rigid in its length. The blades of some of the longer tapes are
constructed of cloth or steel.
Units – Most common are the Imperial, metric and engineering where the foot is
divided into 100 equal units.
Retractablity – Blades on standard tapes used by carpenters retract into the case
automatically. This is made possible by a coil-spring; where as longer tapes are retracted
by a hand crank.
THINGS TO CONSIDER WHEN SELECTING A TAPE MEASURE
Next, a few words about the myriad of tape measures available in the market today.
The first thing you should think about is the length. If you are a framer, twenty-five
and thirty-foot tapes are for you because framers often layout walls requiring the longer
tapes. On the other hand if you are a cabinetmaker you will want a twenty-five footer
for taking site measurements and a ten to twelve footer for building cabinets.
A second consideration to make when purchasing a tape measure is what system of
measure is wanted? A cabinetmaker may want a tape with both Imperial and metric
scales because much of the latest hardware used in cabinetmaking requires metric
dimensioning.
THINGS TO WATCH OUT FOR
Look at the hook on the end of your tape. With a little help it should move slightly
back and forth. That movement is by design and should not be restricted. The reason
it moves is to facilitate inside and outside measurements.
An example of an inside measurement would be measuring between two walls. The
outside edge of the hook is butted against one wall and the measurement is taken at the
other wall. When taking an outside measurement the inside edge of the hook is placed
over an object, pulled tight, and the measurement is read on the other end of the tape.
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In each case you are measuring on opposite sides of the hook. Because the hook has
thickness it moves back and forth to insure the measurement always starts at zero
inches. I mention this because I encountered a student in one of my classes who had
wrapped duct tape around his tape measure where the hook attaches. When I inquired
as to the reason for the duct tape, he told me the hook was loose and he had taped it
down to increase the accuracy of his measurements. Good intentions don’t always
bring good results!
One additional point about the hook of your tape measure: Make sure that it remains
straight, flat and oriented ninety degrees to the Mylar blade. Often times when a tape is
dropped, it lands on the hook and bends affecting the tape’s accuracy. Look at the
hook on your tape periodically to make sure it is straight. If it is bent, place it on a flat
surface and carefully flatten it out with a hammer.
TRY THESE:
1. Write the appropriate fractions in sixteenths i.e. 1/16, 2/16, 3/16 above the hash marks on the
illustration below. Write them again under the hash marks in their reduced form i.e. 1/16, 1/8,
3/16
Practice using you tape measure by marking the following measurements on any piece
of material at least two feet in length:
2.
3.
4.
5.
6.
1-3 1/4
1-8 3/8
0-7-3/4
0-4 1/2
1-5 7/8
7. 1-9 11/16
8. 0-10 5/16+
9. 1-2 5/8
10. 0-11 15/16
11. 1-1 7/16
12.
13.
14.
15.
16.
4
0-8 13/16+
1-6 7/16
0-9/16+
0-4 3/16
1-1 1/16
U N D E R S T A N D I N G
F R A C T I O N S
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D E C I M A L S
Measure the lengths of the lines below and record the answers on top of
the line. If necessary, round up to the nearest sixteenth.
Example:
3/4"
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
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2
Chapter
UNDERSTANDING
FRACTIONS AND
DECIMALS
Anyone performing mathematical calculations in the building trades must sooner or late
convert fractions to decimals, and or decimals to fractions. The architectural convention for
dimensioning is feet and inches, which is no problem when simply measuring with a tape
measure, but because most calculators only work with decimals, knowing how to convert back
and forth is essential.
You could simply avoid such calculators and use a HP 48G or a Construction Master, which
both allow the entry and manipulation of decimals, fractions and feet and inches but because
you deciding to attend a construction math class, I am sure you want a deeper understanding of
the subject.
Learning how to convert fractions to decimals and decimals to fractions lies in the heart of the
craft you have chosen to pursue, so let’s take a closer look at fractions and decimals.
The Anatomy of a Fraction
Numerator
3
16
Denominator
FIGURE 2-1
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F R A C T I O N S
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D E C I M A L S
The denominator indicates the number of equal parts something is divided into. For
example one foot can be divided into 12 inches and one inch can be divided into
sixteenths of an inch. In Figure 2-1, 1-inch (the whole unit) is divided into sixteenths but
it could be divided into 2 equal parts with a denominator of 2, or into 4 equal parts with a
denominator of 4. In any case the number expressed in the denominator tells how many
parts the whole is divided into.
The numerator indicates how many of the total parts, expressed in the denominator, are
actually present. In Figure 2-1 the fraction’s numerator is three, meaning there are three
of sixteen equal divisions represented in the measurement.
ADDING AND SUBTRACTING FRACTIONS WITHOUT A
CALCULATOR
One of the easiest ways to add and subtract whole numbers and fractions is to simply use
your tape measure and your thumbnail. This method of adding and subtracting mixed
numbers is a very visual way of carrying out addition and subtraction. It may seem crude,
but when your calculator fails or you cannot remember your decimal equivalencies it
works great, plus it’s a great way to get comfortable reading a tape measure.
ADDING
Example 2-A: You want to add 13 3/16 + 4 3/8 + 22 1/2 on your tape.
STEP 1: Add the whole numbers in your head, on paper or right on the tape.
13 + 4 + 22 = 39
STEP 2: Extend the tape and add the fractions right on the tape. Starting with the largest
fraction add up the fractions using your thumbnail as a guide.
Pull the tape out to 39. Next using your thumbnail add the fractions on the tape starting
with the largest fraction first, then the next and so on. Using this method for this example
you would end up at 1 1/16, in which case you would add 1 1/16 to the whole number
39. You can add your fractions starting at 10 so that you don’t have to deal with the first
inch where the tape hook is in the way.
1/2 + 3/8 + 3/16 = 1 1/16
STEP 3: Put it all together 39 + 1 1/16 = 40 1/16
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F R A C T I O N S
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SUBTRACTING
Example 2-B:
23 5/16 – 14 7/8 = ?
STEP 1: Subtract the whole numbers. 23 – 14 = 9
STEP 2: Subtract the fractions. Find 9 on your tape and slide your thumbnail up to 9 5/16.
Next count back 7/8 and you end up at 8 7/16.
Here is another subtraction problem that is a little different.
Example 2-C: 46 – 22 7/16 =?
STEP 1: 46 – 22 = 24
STEP 2: Find 24 on your tape and subtract 7/16 by counting off with your thumbnail.
24 – 7/16 = 23 9/16
EXPRESSING FRACTIONS AS DECIMALS
When adding, subtracting, multiplying and dividing on most calculators, fractions must
be converted to decimals. This is done by simply divide the denominator into the
numerator. For example 1/2 expressed as a decimal is .5.
NOTE:
Think of a fraction as a division problem that hasn’t yet happened.
Every fraction has a decimal equivalent, which can be obtained by dividing the numerator
(the number above the line), by the denominator (the number below the line). i.e. 1/2 =
12 = .5
Convert the following fractions to decimals:
21. 3/8 = _______
22. 3/16= _______
23. 8. 5/8 = _______
24. 5/16= _______
25. 15/16 = _______
Easy!
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F R A C T I O N S
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EXPRESSING DECIMALS AS FRACTIONS
When a carpenter uses a calculator to obtain a measurement the answer is given in
decimal form. Since the tape measure is in fractional form (sixteenths) decimals must be
converted so the fraction may be found on the tape measure.
Before carrying out this operation the fraction’s denominator must be determined. Nine
times out of ten a carpenter or cabinetmaker will want the denominator to be sixteen.
Typically sixteenths are the smallest unit used by cabinetmakers and carpenters.
Also as stated before, sixteenths are the basic unit used on most tape measures.
Once a fractional unit has been chosen (16ths, 8ths, etc.), multiply it times the decimal.
Example 2-D: You want your fraction expressed in 16ths. So how would you convert
.5 into a fraction? By deciding on 16ths as your fractional unit you already have part of
your fraction ?/16. You know the denominator, because you chose it. What you don't
know is the numerator. So, multiply the decimal (don't forget to include the decimal point
when you multiply) by the denominator you chose.
.5  16 = 8 Now you have your numerator. Put it all together and your fraction is 8/16. This
fraction can of course be reduced to 1/2.
In the example above when the decimal was multiplied by 16 the answers came out as a whole
number. Things get sketchy when the answer comes out as a mixed number.
Example 2-E: Express .3487 in sixteenths
.3487  16 = 5.5792
Can the fraction be written as 5.5792/16? Of course not, but what does this number
(5.5792) tell us? It tells us that the fraction is 5/16" plus a little over half of 1/16. What
is half of 1/16? ______. If you answered 1/32 that is correct. If you weren’t sure how
to find one half of one sixteenth, here’s how. Simply double the denominator and leave
the numerator as is.
In the example above .5792 may be eliminated by rounding the fraction up or down. In
it’s final form the fraction could be written as 5/16 in which case .5792 is simply drop or
round up to 6/16 or 3/8. Rounding up or down depends on the situation. Remember, as
carpenters we work in sixteenths so there is no reason to calculate to greater precision.
USING THE CALCULATOR TO ADD, SUBTRACT, MULTIPLY AND
DIVIDE FRACTIONS
By now, converting fractions to decimals and visa versa should be no problem, but what about
adding, subtracting, multiplying and dividing fractions? Armed with your trusty pocket
calculator and the knowledge of converting fractions to decimals you have no worries.
Most carpenters and cabinetmakers memorize the decimal equivalents of the fractions
with which they work. If you haven't memorized them yet, don't worry, it just means one
added step. In the next example you should have no problem adding the whole numbers.
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1 + 3 + 5 = 9. In order to add the fractions using the calculator, they must be converted
into decimals. No matter what the fraction is simply divide the numerator by the
denominator.
Example 2-F:
1 5/8 + 3 3/16 + 5 7/16 = _____________
1 5/8 = 1.625
3 3/16 = 3.1875
5 7/16 = 5.4375
Now add them up:
1.625
3.1875
+ 5.437
10.25"
Can you find 10.25 on your tape? Probably not, so to be able to find the fraction on your
tape, convert the decimal into a fraction.
How do you convert the decimal (.25) into a fraction? If you want your answer in
16ths, multiply the decimal (.25) times 16. What if you wanted your answer in 32nds? By
what number would you multiply your decimal (.25)? ______
.25  16 = 4 or 4/16 reduced to 1/4
.25  32 = 8 or 8/32 reduced to 1/4
Final answer = 10 1/4
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DECIMAL FEET VS. FEET & INCHES
Architectural drawing are dimensioned in feet and inches so it is quit likely that you will
need to convert feet and inches into decimal form in order to use your calculator.
1 foot = ______. So how hard can it be?
Take a look at this problem:
Express 12.7638 in feet and inches.
First, look at what you already know. The question asks for feet and inches. You know
part of the answer will be 12 feet. It’s .7638' that must be expressed as inches.
What part of a foot is .7638? How many inches are there in one foot? _______ Let’s
apply the same logic we used in our fraction problems to change .7638 into inches. For
example .5 feet is 6 inches (.5  12 = 6). Couldn’t 6 inches be written as 6/12? By
dividing the numerator by the denominator (6 divided by 12) equals .5. Apply the same
principle to this problem. .7638 feet x 12 (why 12? Because there are 12 inches in one foot
just like there are 16, 16ths in one inch.) = 9.1656
Answer: 12- 9.1656
Are you finished? You’re pretty close! All you have to do now is to convert the decimal
inches into a fraction. You are already a pro at this! Remember that the only part of your
answer you can’t read on your tape is the decimal. To get it into 16ths multiply times 16.
.1656  16 = 2.6496
One last detail! The answer as it stands is 12 9 2.6496/16 . Try finding that on your
tape. Round the remaining decimal up or down. Since .6469 is greater than 1/2 of one
sixteenth, round up to 3/16 (2.6496 rounds up to 3)
Final answer: 12- 9 3/16
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What if the measurement is in inches but you need to express it in feet
and inches?
Example 2-G:
246 11/16
Easy! Leave 11/16" as is and divide 246 by 12 (12 inches in one foot) = 20.5
Now .5 equals how many inches? .5  12 = 6
Put it all together
Final answer: 20'-6 11/16
REMEMBER THIS:
Often times when converting decimals to fractions, or fractions to decimals there is
confusion whether to multiply or divide.
Here is a handy
rule to help you
remember:
Think about what you are doing. By converting feet to inches you are increasing the
number of units, right? You are make twelve units out of one unit. So, to increase the
number of units, you multiply times twelve? Conversely when you convert inches to feet
you decrease the number of units. You make one unit from twelve. So you divide by
twelve. The same rule applies to converting decimal inches to sixteenths and visa versa.
Feet to inches
Less units to More units, Multiply by 12
Inches to feet
More units to Less units, Divide by 12
Decimal inches to fraction
Less units to More units, Multiply by 16
Fraction to decimal inches
More units to Less units, Divide by 16
Example 2-H:
Express 24.5634 as feet and inches
24 is a whole number so we can keep it, but .5634 must be converted to inches. We are
increasing the number of units so multiply by 12.
12  .5634 = 6.7608
6 is a whole number so we can use it.
.7608 must be converted to sixteenths. Less to more, so multiply by 16
16  .7608 = 12.288 rounds to 12
Answer = 24-6 12/16 reduced 24-6 3/4
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U N D E R S T A N D I N G
Here is a handy
visual to help you
remember:
F R A C T I O N S
&
D E C I M A L S
Use the triangle compartments to organize the information you know and want to know, as in
Figure 2-2.
The fraction’s numerator is represented in the top compartment, the fractions denominator
and decimal equivalent are represented in the bottom compartments. You should know two of
the three values. For example, in expressing .3125 as a fraction of an inch you know the
decimal because it is given and you know the denominator because you are a carpenter and
carpenters work in sixteenths. So, put what you know and what you need to know in the
triangle see Figure 2-3. Place your finger over the value you want to know. If the remaining
uncovered values are in the bottom compartments, then multiply. If the remaining
uncovered values are in the top compartment and a bottom compartment, divide the
bottom into the top as in Figure 2-4.
In summary, if you are having trouble remembering whether to multiply or divide when
deriving decimals from fractions and visa versa, draw the triangle on a piece of paper and use it
as a tool. Don’t hesitate to use it when taking a test.
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F R A C T I O N S
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D E C I M A L S
Complete the following:
1. How many sixteenths are present in one inch?_____
Express one inch as a fraction with sixteen as the denominator _______
2. How many inches are in one foot? ____ ?
Express one foot as a fraction with twelve as the denominator _______
3. How many 16th’s are present in one half-inch _______?
express as a fraction __/__. Now reduce __/__
Express the following fractions as 16ths:
4.
1/8 = _______
5.
5/8 = _______
6.
3/4 = _______
7.
7/8 = _______
8. 3/8 = _______
Try these using the tape and thumbnail method:
9.
12 5/8 + 16 5/16 + 32 9/16 = ________
10. 20 3/8 – 15 7/16 = ________
11. 18 7/8 + 26 7/16 + 9 13/16 = _________
12. 96 – 4 5/8 = _________
13. 12 7/16 + 22 1/4 + 16 1/16 = _________
14. 46 1/2 + 22 13/16 - 37 9/16 = ____________
15. 25 5/16 - 7 15/16 = _________
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16. 16 3/8 + 12 7/16 + 4 3/4 = ____________
17. 2 1/4 + 7 5/8 - 4 1/2 = _____________
18. 22 1/2 + 13 9/16 = ___________
19. 29 15/16- 2 3/16 + 32 7/8 - 21 3/4 = ____________
20. 5/8 + 14 1/4 - 3 1/2 = ____________
Use your calculator to convert the following fractions to decimals:
21. 3/8 = _______
22. 3/16= _______
23. 5/8 = _______
24. 5/16= _______
25. 15/16 = _______
26. Create a fraction to decimal conversion chart for your reference guide.
Express the following decimals as fractions. Use 16ths to express your
answers. Reduce you answers if possible.
27. .0625
= _______"
28. .125
= _______"
29. .1875
= _______"
30. .25
= _______"
31.
= _______"
.3125
15
U N D E R S T A N D I N G
F R A C T I O N S
&
D E C I M A L S
Try these using your calculator: (round to the nearest 16th.)
32. .9173 ________
33. .2843 ________
34. .4958 ________
35. .5824 ________
36. .7478 ________
37. 15 3/16 + 32 9/16 + 54 13/16+ 23 3/8 = _____________
38. 31 3/32 + 3 11/16 + 4 1/8 = _______________
Express the following values in feet and inches:
39. 13.5268 _______________
40. 28.9736 _______________
41. 2.3857
_______________
42. 15.2948 _______________
43. .3974
_______________
44. 150.34
_______________
45. 12.692
_______________
46. 4.1974
_______________
Convert the following measurements to decimal feet, then add or subtract
as necessary. Express answer in feet and inches:
47. 31-5 3/16 + 27-7 5/8 = ______________
48. 47-9 13/16 - 19-11 11/16 = ___________________
49. 15- 9 3/16 + 22-6 5/8 = __________________
16
U N D E R S T A N D I N G
F R A C T I O N S
&
D E C I M A L S
50. 18-10 9/16 + 43-7 5/16 = ____________________
Express the following problems in feet and inches:
51. 123 5/16" = __________
52.
16 9/16" = __________
53. 143 1/2" = __________
54. 13 3/16 = _________
55. 77 7/16
= __________
56. 29 15/16 = __________
57. 237 7/8 = ___________
58. 478 3/8 = ____________
59. 22 15/16 = _____________
60. 81 11/16 = _____________
17
S U R F A C E
3
Chapter
M E A S U R E M E N T S
SURFACE MEASUREMENTS
Any measurement taken in a two dimensional plane is a surface measurement. There are
many objects evaluated in only two dimensions such as POLYGONS, consisting of
squares, rectangles, parallelograms, triangles, and trapezoids and CIRCLES. Types of
surface measurements taken to evaluate these objects would include linear
measurements to find the length of, or distance around an object, and square
measurements to find the area inside the boundaries of the object.
Taking surface measurements on the job site or off of blueprints is an everyday activity
for the contractor. How many feet of rim joist are needed? How many sheets of plywood
sheathing are needed? How many pieces of bevel siding are needed to cover the walls?
How many yards of carpeting need to be ordered?
Luckily calculating surface measurements is pretty easy. You need to memorize some
formulas, many of which you already know, and carry out some very basic math - adding
subtracting, multiplying, dividing and squaring. Lets look at some of the objects you will
be evaluating.
POLYGONS
Polygons are plane figures bound by three or more line segments, called sides, which are in
turn, joined at their endpoints. The angle formed where two sides meet is called an interior
angle and the point where two sides meet is called a vertex. A polygon with sides of equal
length and equal interior angles is called a regular polygon. Polygons having four sides are
called quadrilaterals. The perimeter of a polygon is simply the sum of the sides. Figure 3-1
illustrates the names and shapes of polygons you will likely encounter.
18
S U R F A C E
M E A S U R E M E N T S
The sum of interior angles of a closed polygon equals (n – 2) 180, where n equals the
number of sides or interior angles of the Polygon.
Example3-A: To calculate the sum of the interior angles of a quadrilateral simply
substitute 4 for the number of sides and do the math.
19
(4 – 2) 180 = 360
S U R F A C E
M E A S U R E M E N T S
Circle – A curved line in a plane that encloses a space. Every point on a circle is the same
distance from the center point.
Circumference: The distance around a circle. C = D
Diameter:
A straight line from one side of a circle to the
other side that passes through the center.
Radius:
A straight line from the center of the circle to a
point on the circle.
Arc:
A curved line whose points are of equal distance from
a single point.
Arc length:
The length of a curved line, symbolized by the letter L
Tangent Line: A straight line which touches only one point on a circle. A tangent line is
also perpendicular to the radius line or diameter line touching the same
point along the circle.
Chord
A straight line from one point on a circle to another point on a circle.
The longest chord of a circle is a diameter.
Pi () = 3.1416 The circumference of a circle with a diameter of one.
More about circle later!
20
S U R F A C E
M E A S U R E M E N T S
LINEAR AND PERIMETER MEASUREMENTS
The simplest surface measurement is a linear measurement, which is a measurement of
length. Measuring the length of one leg of a triangle or the length of a wall would be a
linear measurement. In the course of a day a carpenter takes dozens of linear
measurements.
Another form of linear measurement is a Perimeter (P) measurement. Perimeter
means the distance around an object. If all three legs of a triangle were measured the sum
would represent the perimeter of the triangle. Likewise, if the distance around a circle
were measured that distance would represent its perimeter.
Try these:
1. List the letters corresponding to the quadrilaterals in Figure 3-1
Answer _____________________________________
2. List the letters corresponding to the regular polygons in Figure 3-1. Note: triangle a
in figure 3-1 is a right triangle.
Answer _________________
3. Calculate the sum of interior angles for g, h, i, and j in figure 3-1.
g. ________ h _________ i _________ j __________
4. A triangle has sides measuring 6-5, 3-7 and 5-2. What is the perimeter
measurement of the triangle?
Answer _________
5. What is the perimeter measurement of the foundation illustrated below?
Answer ____________
21
S U R F A C E
M E A S U R E M E N T S
6. What is the perimeter measurement of the foundation illustrated below?
Answer ______________
7. What is the perimeter of the octagon?
Answer _______________
8. A square building has a perimeter measuring 162 feet. What is the length of each
side of the building?
Answer ____________
9. A pentagon has a perimeter measuring 156-7 1/2. What are the lengths of each
side?
Answer __________
22
S U R F A C E
M E A S U R E M E N T S
CIRCLES
Circles deserve some explanation. To calculate the distance around a circle one could
simply roll the circle or cylinder on a flat surface and measure the distance traveled after
one revolution. In fact if you roll a dowel, with a diameter of one inch, one revolution and
measure the distance traveled, it will equal approximately 3 1/8 or 3.1416. Therefore it
follows that if you do the same experiment with a dowel with a 2-inch diameter the
circumference will be 2 times 3.1416. I find this an easy way to remember the formula for
circumference. Diameter times 3.1416….
Circumference = D
 = 3.1416 ……
D = Diameter
Example 3-A: What is the circumference of a circle with a radius of 16-0
Step 1: If radius is given rather than diameter multiply the radius times 2 .
16  2 = 32 = Diameter
Step 2: Solve for circumference
2r
2r
2  3.1416*  16 = 100-6 3/8
*Using the  symbol on your calculator will round pi to 11 decimal places making
your
answer slightly different than if you input 3.1416 into your calculator.
Example 3-B: What is the radius of a circle with a circumference of 46 -0
Step 1: Setup the formula and plug in what you know.
C = D
46 = 3.1416  D
46 = 3.1416  D
46  3.1416 = D
14.6423 or 14 7 11/16
Step 2: We are not finished because the radius is ½ the diameter.
14.6423/2 = 7-3 13/16
Try These:
10. What is the circumference of a circle with a diameter of 12-6?
Answer ___________
11. What is the circumference of a circle with a radius of 12-3?
23
S U R F A C E
M E A S U R E M E N T S
Answer ____________
12. What is the radius of a circle with circumference of 89-11?
Answer ___________
13. What is the arc length of the outer semicircle of the
sidewalk pictured at right?
Answer _____________
What is the arc length of the inner semicircle?
Answer _______
Answer __________________
14. What is the diameter of a circle with a circumference of 48-11?
Answer __________________
15. You are asked to dig a trench for direct burial cable, around the outside perimeter of
the track below.
A. How many linear feet must you dig?
Answer __________________
24
S U R F A C E
M E A S U R E M E N T S
16. What is the perimeter measurement around the inner portion of the track if the track
is 5 wide?
Answer __________________
25
S Q U A R E
4
Chapter
M E A S U R E M E N T S
SQUARE MEASUREMENTS or AREA MEASUREMENTS
As previously stated square measurements tell us how much space is bound by the sides of
a closed polygon or a circle. As a carpenter you will most often measure area in square
feet, square inches or square yards. Plywood is sold by the square-foot, carpeting and vinyl
flooring are sold by the square-yard, and cabinet doors by the square foot. Because circles
and polygons have different area formulas let’s look at them individually.
SQUARE - A square is a four-sided polygon with sides of equal length and equal
interior angles. The square offers an excellent example of how area is calculated
and what happens when numbers are squared.
FIGURE 4-3 measures 6 by 6. The same square in FIGURE 4-4 is divided into six
equal units horizontally and six equal units vertically. When you add up all the
one-unit squares the sum is 36. Therefore 62 = 36
When a number is squared it is simply multiplied times itself.
Area = S2 or S  S
6 units
62 = 36
6 units
Area = S2
6  6 = 36
FIGURE 4-3
FIGURE 4-4
26
S Q U A R E
M E A S U R E M E N T S
RECTANGLE - Opposite sides are parallel and equal and all interior angles are
90.
L = Length
Area = LW
W = Width
8
15
8  15 = 120 Square Feet (SF)
CONVERTING SQUARE INCHES TO SQUARE FEET:
Cabinetmakers typically measure door in inches, but price them by the square foot, so it is
necessary to convert back and forth. One square foot is 12 x 12 so there are 144 square
inches in one square foot. In short, to convert square inches to square feet simply divide
the square inches by 144.
TRY THESE:
1. Calculate the area of a rectangle with length measuring 34 7/8 and width
Measuring 67 11/16. Express answer in square feet.
Answer ___________
2. Calculate the area of a rectangle measuring 15-7 3/16 X 67-11/16. Express
answer in square feet
Answer ___________
3. Calculate the area of a rectangle measuring 18 9/16 X 42 5/8. Express answer in
square feet
Answer ___________
27
S Q U A R E
M E A S U R E M E N T S
PARALLELOGRAM - Opposite sides are parallel and equal in length. Opposite interior
angles are equal and each diagonal cuts the other diagonal into equal parts (they bisect
each other).
Area = B  H
B = Base
H = Height
H
B
RHOMBUS – The rhombus is also a parallelogram except that the diagonals bisect each
other at right angles. Use the parallelogram area formula for the rhombus.
90
TRAPEZOID – The trapezoid has two parallel sides called bases, and two non-parallel sides.
To solve for the area of a trapezoid one must know the perpendicular distance between the two
bases. This distance is called the altitude or height.
AREA
=
B1 +B2  H
2
B2
H
B1
28
S Q U A R E
M E A S U R E M E N T S
Example 4-A:
What is the area of a trapezoid with base1 = 14-0,
base2=8-0 and height = 17- 0
 14  8 

17  187
 2 
Try this:
Express answer in square feet
4. Calculate the area of the trapezoid.
13-5
H = 14-1
8 -2
Answer ______________
TRIANGLE - Even though triangles come in different shapes they all have three things
in common. They have three sides and three angles and the sum of their interior angles is
180. The height or altitude of any triangle is the distance, measured perpendicular to the
base of the triangle to the opposite vertex. Note that the altitude of a scalene triangle is
measured outside of its perimeter.
Area = 1/2BH
TYPES OF TRIANGLES:
Equilateral – All sides and angles are equal.
H or altitude
B
29
S Q U A R E
M E A S U R E M E N T S
Isosceles – Two sides of equal length
H
B
Scalene Triangle– No equal length sides
H
H
B
B
Right Triangle – One interior angle is 90 the other two are acute
angles (less than 90).
H
B
As stated in the formula the area of a triangle is calculated by multiplying the base and altitude
and dividing the product by two.
Here’s why:
Notice in the rectangle shown in Figure 4-5 that a diagonal
has been drawn and half of the rectangle is colored.
We could say that the rectangle has been divided into two equal
triangles.
The area of the rectangle is 30 so what is the area of each triangle?
Because each triangle represents one half of the rectangle the
area of each triangle would be 30  2 = 15. Therefore, the formula
for the area of a triangle is B H
2
30
10
3
Figure 4-5
S Q U A R E
M E A S U R E M E N T S
Try these: Express answer in square feet
5. Calculate the area of the triangle
25- 9
Answer ______________
15- 10
6. Calculate the area of the triangle
Answer _______________
H=29- 6
32- 2
7. Calculate the area of the triangle
28-4 5/8
Answer ____________
67-5
HERO’S FOMULA – In the event that the lengths of the sides of a triangle are known
but the altitude is unknown, the area can be calculated by using Hero’s Formula.
Area of a triangle =
ss  a s  bs  c 
Where s is half the perimeter, or
S
abc
2
Note: The symbol
means square root. A square root equals a number, when multiplied by
itself, equals a given number. For example the square root of 36 is 6, because 6 multiplied
by itself equals 36. Square roots do not always come out even as in this example. The hand
held calculator saves an incredible amount of work solving for square roots.
31
S Q U A R E
M E A S U R E M E N T S
Example 4-B:
Find the area of a triangle with sides measuring 4.6, 7.2, 5.9
Step 1: Solve for s: s = 4.6 + 7.2 +5.9 = 8.85
2
Step 2: Solve for area using Hero’s formula
Area =  8.85( 8.85 – 4.6) (8.85 – 7.2) ( 8.85 – 5.9)
c = 5.9
b = 7.2
a = 4.6
Deal with the brackets first, then multiply the final product by 8.85 then find the square root.
Area =  8.85 (4.25) (1.65) (2.95)
Area =  8.85  20.69
Area =  183.08
Area = 13.53 SF
CIRCLES
– The area of a circle is solved using the following formula:
Area = r2
Example 4-C: What is the area of a circle with a radius of 7-6?
First convert feet and inches to decimal feet 7.5 and set up the formula.
3.1416  7.52 = 176.72 SF
Note:
If you encounter a problem where diameter is given instead of radius, remember that radius is one half the
diameter.
32
S Q U A R E
M E A S U R E M E N T S
SOLVING MORE COMPLEX AREA PROBLEMS
Many times the objects of which you want to know the area are not simple rectangles or
triangles, they are combinations of polygons and sometimes circles. Look at figure 4-6
30
30
20
20
10
10
50
50
Additive Method
Subtractive Method
Figure 4-6
What is the area of the room in Figure 4-6? This problem can be approached in several
ways.
Additive method - Break the object into two rectangles, solve for the area of each and
add together.
30  20 = 600 SF
600 SF
20  10 = 200 SF
+200 SF
800 SF
Subtractive method - Treat the object as if it were one large rectangle and subtract out
what is not included within the boundaries of the object.
50  20 = 1000 SF
20  10  = 200 SF
1000 SF
- 200 SF
800 SF
TRY THESE:
8. You are constructing a dome with a diameter of 24-6. What is the floor area of the
dome?
Answer _________
7-6
9. Calculate the area of the object.
Express answer in square feet.
8-4
Answer __________
2-0
10-5
33
S Q U A R E
M E A S U R E M E N T S
10. Calculate the area of the pictured object. Express
answer in square feet.
12-3
8-0
Answer _____________
11. Calculate the area of the pictured object. Express answer in
square feet.
25-7
16- 0
8- 0
4-0
Answer ______________
8 -0
24- 0
12. Calculate the square footage of the house footprint. Express answer in square feet.
34
S Q U A R E
M E A S U R E M E N T S
Answer __________________
SQUARE YARDS – is a square measure used to quantify building products such as
carpeting and vinyl flooring. We all know that one-yard measures 3 feet in length. A
square yard would be 3  3 which would equal 9 square feet.
1 Square Yard = 9 SF
How many yards of carpeting is needed for a room measuring 12  10 =
120SF 120  9 = 13.33Yards
Example 4-D:
13. How many yards of carpeting would have to be ordered to cover the floor area of
the house in problem 12? (No waste factor to be added.)
Answer ___________
14. If the wall height of the house in problem 12 is 8-0 what is the total square footage of
wall area? Assume the length of the diagonal walls at the bay are 3- 5 9/16.
Answer __________
15. How many sheets of 4 x 8 drywall will it take to cover the walls in problem 12?
Answer _________
16. How many sheets of 4 x 10 drywall will it take to cover the
ceiling in problem 12?
Answer __________
MATERIALS BY THE SQUARE -Roofing and shingle siding are sold by the square.
One square is the area occupied within a 10 X 10  square or 100SF.
1 Square = 100 SF
Example 4-H: A shed roof measures 25-0  15-0. How many
squares of roofing should be ordered?
25 X 15 = 375 SF
35
375  100 = 3.75 Squares
S Q U A R E
M E A S U R E M E N T S
17. Calculate the number of squares of shingle siding necessary to cover the walls of
the house illustrated in problem 16. Make no deductions for door and windows and
use no waste factor.
Answer ____________
18. Calculate the area of the triangle. Give answer in square feet.
a= 25-4
b= 13-5
c= 17-0
c
b
a
Answer ____________
36
S Q U A R E
M E A S U R E M E N T S
AREA FORMULAS
Square:
Area = Side  Side
or
A = S2
Rectangle:
S
Area = Length  Width
or
A=LW
W
L
Triangle: Area = 1/2  Base  Height
or
A = 1/2bh
or
A = BH
2
H
B
Trapezoid:
Area = 1/2 (Base1 + Base2)  Height
or
A= 1/2(B1 + B2)  H
or
A =(B1 + B2)  H
2
B2
H
B1
Circle:
Area =   radius squared
or
A =   radius radius
or
A = r2
r
Square Yards: 1 Square Yard = 3 feet  3 feet = 9 sq. ft.
Y = Area  9 Example: A room measuring 9 feet by 10 feet would require10 yards of
carpeting. 9  10 = 10 SY
9
37
T H E
C O S T
O F
B U I L D I N G
M A T E R I A L S
5
Chapter
CALCULATING THE COST OF BUILDING MATERIALS
The purpose of this section is to help you understand how framing lumber, trim stock and
panel goods are bought and sold. We will explore three methods of costing lumber –
Board footage, linear footage and square footage. Before covering board footage, here are
some of the terms and properties associated with framing lumber.
Framing Lumber – The Western Wood Products Association breaks framing lumber
into three categories. Dimension lumber, structural decking and timber grades. We will
primarily be working with dimension lumber. For the purpose of this book, dimension
lumber will be defined as 2 by and 4 by material.
Common terms used in association with framing lumber:
Nominal Size – The dictionary defines nominal as - existing or being something
in name only. In the context of framing lumber, nominal refers to framing
lumbers width and thickness in name only not actual size. We commonly refer to
framing lumber as 24, 26, 28, 210, 212 etc., however the actual size or
dressed size of the lumber is of lesser dimension. Think of the nominal size as
the size you pay for. Below you will find a chart showing the nominal sizes and
actual sizes of dimensional framing lumber.
Standard Lengths – As noted in the table, standard lengths are offered in
multiples of 2 feet, such as – eight feet, ten feet, twelve feet, etc. There are also
specialty lengths used for wall studs, 88 5/8 and 92 5/8. When studs of these
lengths are used with a single bottom plate and a double top plate, wall heights of
7-8 and 8-0 are attained.
38
T H E
C O S T
O F
B U I L D I N G
M A T E R I A L S
Standard Sizes - Framing Lumber
Dressed Dimensions
Thicknesses & Widths
(inches)
Nominal Size
Product
Description
Thickness
Width
Surfaced
Surfaced
(inches)
2"
Length
(inches)
Dry
Unseasoned
(inches)
2"
1 1/2"
1 9/16"
6' and longer
3"
3"
2 1/2"
2 9/16"
generally
4"
4"
3 1/2"
3 9/16"
shipped in
Dimensional
S4S
5"
4 1/2"
4 5/8"
multipals
Lumber
(Surfaced 4 Sides)
6"
5 1/2"
5 5/8"
of 2'
8"
7 1/4"
7 1/2"
10"
9 1/4"
9 1/2"
12"
11 1/4"
11 1/2"
over 12"
off 3/4"
off 1/2"
FRAMING LUMBER IS BOUGHT AND SOLD THREE WAYS:

By the linear foot

By the piece

By the board foot
.
LINEAR MEASURE
Linear is defined as a measure of length. This type of measurement does not take into
consideration width and thickness, only length.
Example: The linear measurement of an 8-0 212 is 8-0. Likewise, the linear
measurement of an 8-0 22 is also 8-0.
At the retail level, framing lumber is typically sold by the piece or by the linear foot.
Examples:
By the piece - 8-0 2x4s cost $4.80 each
By the linear foot - 2x4s cost 60 cents per linear foot.
39
T H E
C O S T
O F
B U I L D I N G
M A T E R I A L S
BOARD MEASURE
At the wholesale level, framing lumber is sold by the board foot. Unlike linear
measurements, board footage is a measure of a board’s volume. One board foot can be
described as a piece of lumber 12 inches long, 12 inches wide and 1 inch thick. A board
foot measurement is the ratio of a board’s volume compared to the volume of 1 bd.ft.
12
1
12
One Board Foot
Figure. 5-1
The volume of one board foot, measured in cubic inches, can be calculated as follows:
1 board foot =12121= 144 cu. in.
Therefore any piece of lumber equaling 144 cubic inches is one board foot.
As previously stated, board footage is the ratio or relationship between the volume
(LTW) of the board in question and the volume of one board foot. To better
understand this concept, compare the volume of one linear foot of 24, to one board
foot.
Example 5-A: Calculate the number of board feet in one linear foot of 24?
Variables:
L = length expressed in inches
T = thickness expressed in inches
W = width expressed in inches
Equation:
2x4
One board foot
Solution:
LTW = 1224 = 96 cu. in.
LTW 12 12 1 144 cu. in.
96
8

 .667 BF
144 12
40
T H E
C O S T
O F
B U I L D I N G
M A T E R I A L S
Example 5-Bis
offered to help you gain a clear understanding of how board footage
is calculated. In reality the equation may be greatly simplified.
Simplifying The Board Foot Equation:
In order to simplify the equation, express length in feet rather than inches, otherwise
board lengths normally given in feet must be converted to inches. Next, think of length in
units rather than in feet. For example a 12-0 24 is 12 units long. This will allow the
multiplication of feet and inches.
Now the board foot equation looks like this:
Board Feet = L  T  W
12
Example 5-C: Calculate the number of board feet in one linear foot of 24 using feet
for length.
Variables:
L= length expressed in feet (units)
T = Thickness expressed in inches
W = width expressed in inches
Equation:
L  T  W or 1  2  4 Simplified
L  T  W
1  1  12
Note:
1 x 2  4 = .667 bd.ft. (BF)
12
The length of one unit could have been left out of the equation but was left in because lumber most often
comes in lengths greater than one foot. Also, 12-inches could have been used for length but as you can see it
would have cancelled itself out and become one.
Try These - Calculate the board footage for one-foot lengths of the
following dimension lumber sizes:
1. 2 x 4 = ___________BF
2. 2 x 6 = ___________BF
3. 2 x 8 = ___________BF
4. 2 x 10 =___________BF.
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5. 2 x 12 =___________BF.
Try some more, this time with varying lengths:
6. One - 8-0 length of 2 x 8 =
__________ BF
7. One - 12-0 length of 2 x10 = __________ BF
8. One - 14-0 length of 2 x 2 = __________ BF
9. One - 10-0 length of 4 x 10 = __________ BF.
10. One - 8-0 length of 2 x 3 = __________ BF.
INTRODUCING QUANTITY (# OF PIECES) INTO THE BOARD FOOT
EQUATION
Usually lumber orders will include more than one piece of material therefore; a quantity
multiplier must be part of the calculation. To reflect the total number of board feet in
the order, simply place the quantity (#) into the equation as below.
#  L  T  W = BF
12
Example 5-C:
Calculate the board footage of 8 pieces of 10-0 2 x 4
8  10  2  4 = 53.33 BF
12
Notice in the next set of questions, the lumber quantities and sizes are written
differently. This notation is universally accepted, fast and reliable. The first
number indicates the number of pieces; the second indicates the length of the
pieces; and the ___ x ____ indicates the size of the framing lumber. Hence
12/8 412 would indicate 12 pieces of 412, 8 in length.
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T H E
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Try These:
11. 16/10 2 x 4 = __________ bd. ft.
14. 13/8 2x2 = ____________ bd. ft.
12. 10/12 4 x 4 = __________ bd. ft.
15. 28/16 2x14=____________ bd. ft.
13. 15/8 6 x 8 = __________ bd. ft.
16. 56/18 2x3 = ____________ bd. ft.
INTRODUCING COST TO THE BOARD FOOT EQUATION:
Lumberyards selling lumber to homebuilders, contractors, and remodelers typically
quote lumber prices by the thousand. That is, by the cost of one thousand board feet.
For example 12-0, 2 x 4s might cost $575.00 per thousand board feet. You will see
this written as 575M.
Definition: M is the Roman numeral for one thousand, hence $575.00 per thousand board feet.
In order to use this information in the equation we need to know the cost of one board
foot. By simply moving the decimal point three places to the left we have the cost of one
board foot, $00.575 or 57 ½ cents. Finally, multiply the calculated board footage by the
cost per board foot.
#pcs.  L  T W  cost per bd. ft. = Board Foot Cost
12
Example 5-D:
What is the cost of 1/12 24 @575M?
Variables:
# = Number of pieces
L = Length of material
T = Thickness of material
W = Width of material
$ = Cost per board foot
Equation:
#  L  T  W = bd. ft.  $ = Board foot cost
12
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T H E
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Solution:
1 x 12 x 2 x 4
12
= 8 .575 = $4.60
One 12-0 2x4 equals 8 BF and costs $4.60 each
NOTE:
As in the example, you may find it helpful to first calculate the board footage, write it down, then
multiply by the board foot cost otherwise your equation will only reflect the cost.
TRY THESE:
Calculate the board footage and the price of the following lumber quantities:
17.
10/12 22 @ 350M _________ bd. ft. $___________
18.
22/14 28 @ 590M _________ bd. ft.
$___________
19.
5/18 416 @ 730M _________ bd. ft.
$___________
20.
120/8 24 @ 480M _________ bd. ft. $___________
21.
19/10 212 @ 610M _________ bd. ft.
$___________
22.
11/12 44 @ 570M _________ bd. ft.
$___________
23.
15/16 46 @ 590M _________ bd. ft.
$___________
24.
8/8 26 @ 590M
_________ bd. ft.
$___________
25.
4/16 410 @ 790M _________ bd. ft.
$___________
26.
2/16 23 @ 488M _________
$___________
bd. ft.
PRICING BY THE LINEAR FOOT
Finish lumber and trim – Finish lumber is typically used for trim work and is
sold by the linear foot. Working in linear footage makes estimating quantity and
price easy. Add up all of the lengths and number of pieces needed in a particular
size and multiply times the cost per foot.
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T H E
C O S T
O F
B U I L D I N G
M A T E R I A L S
You need 6/7 1x6 and 14/8 1x2 to complete a set of doorjambs. The 1x6 costs
$1.20 per linear foot and the 1x2 costs $00.67 per Ln. Ft. What is the total cost of the trim package?
Example 5-E:
6  7  $1.20= $50.40
14  8  .67 = $75.04
Total = $125.44
Trim, such as base trim, casing, crown molding, and other milled products, are most often
priced by the linear foot.
Try These:
27.
12/16 Colonial Base @ .78 cents/ft. _________
28.
15/8 – 3 ½ crown @ $9.38/piece _________
29.
22/8, 11/7, 14/12 of 2 1/2 colonial casing at .56 cents/L.F. _________
30. 15 pre-hung doors @ $56.00 each ___________
31.
9/14 - 1x4 @ .68/ft, 12/8 - 1x3 @ .59/ft, 23/16- 1x6 @ $24.00 each. _________
PRICING BY THE SQUARE FOOT
Panel Products – Plywood, particleboard, drywall, wafer board, etc.; are sold by the
square foot or by the sheet. Most sheet goods are manufactured in 48, 410 and 412
sizes and come in 1/4, 3/8, 1/2, 5/8”,3/4, and 1 thickness’
CALCULATING THE COST OF PANEL PRODUCTS
By The Sheet - . Multiply the number of sheets by the cost per sheet.
Example 5-F: 20 sheets of drywall costing $4.00 per sheet.
20  $4.00 = $80.00
By The Square Foot – Start with the equation used to find the area of a rectangle.
Area = L W
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Variables:
L = Length
W =Width
# = Number of pieces
$ = Cost per square foot
Equation:
L  W  #  $= Cost
Example 5-G:
Price 40 sheets of 1/2 4 x 8 CDX plywood costing 500M
Note: Prices per thousand in sheet goods do not take thickness into account. Therefore,
move the decimal point three places to the left and multiply by the calculated
square footage.
4  8  40  .50 = $640.00
Price per sheet 4  8  .50 = $16.00
Try These:
32.
20 sheets of 1/2412 drywall costing 200M _______________
33.
15 sheets of 1/448 Masonite costing 320M _______________
34.
45 sheets of 3/448 CDX plywood costing 560M ____________
LAP SIDING
Lap siding is commonly used as an exterior covering on both residential and commercial
structures. There are many types but the common thread is that each successive course of siding
overlaps the course below.
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T H E
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M A T E R I A L S
This lapping of successive courses of siding renders a waterproof shield, provided it is installed
correctly. Some of the types of lap siding commonly used are cedar or redwood bevel siding,
Hardy Plank concrete siding, and Smart Siding (formerly LP Siding).
Lap siding offers the contractor a real challenge when calculating quantity and price and here is
why. Covering a wall with a panel product would entail a fairly simple square footage
calculation, but lap siding is a plank, sold by the board foot based on nominal width, which is
different than the actual width, and to complicate things, the siding is overlapped.
Calculating waste takes experience and varies depending on the structure being sided, so we will
focus on the fundamentals.
STEP BY STEP CALCULATION OF LAP SIDING
Step 1:
Calculate The Square Footage of the wall or walls to be covered by
multiplying the linear footage of wall times the wall height. Generally windows and doors
are not subtracted from the calculation and serve as a waste factor. Let’s say we calculated
2500 SF of wall to be covered by our lap siding.
Step 2:
Determine the Siding Width. Lap siding is sold in a variety of lengths, and
widths such as 4, 6, 8 and 10. The widths are given as nominal sizes not actual sizes,
so make sure you use the actual width for your calculations. For the sake of this exercise
we will use 10 nominal or 9 1/4 actual cedar bevel siding.
Step 3:
Determine The Lap. As the name implies lap siding is over lapped and you
should follow the manufactures recommendations as to what the lap should be. For our
example we will overlap each piece of siding 1.
Step 4:
Determine The Number Of Linear Feet Of Siding Per Square Foot. One
square foot = 12  12 One linear foot of siding would equal 10  12 but the siding
width is actually 8 1/4. Here’s why - The actual width is 9 1/4 and we loose one inch
when we lap one piece over the one below. So one linear foot of siding = 8 1/412. We
can see by the calculations above that one linear foot of siding is less than one square foot,
so we need to know how many linear feet it will take to cover one square foot. Simply
divide 12 by 8.25. 12 8.25= 1.4545. In other words, it takes 1.4545 LF if siding to cover
1SF of wall surface. You have created a factor, which can be used over and over again to
calculate 10-inch bevel siding quantities with a 1 lap.
Note:
Factors are covered in great detail in Chapter 8.
Step 5:
Determine The Total Number Of Linear Feet Of Siding. This is the easy
part, simply multiply your siding factor (1.4545) times the total square footage, in this case
2500 SF. 1.4545 2500 = 3636.25 LF
Step 6:
Calculate The Board Foot Cost Of The Siding. For this calculation we will
use the nominal width of the siding, and a cost of $1400M
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10 x3636.25
 3030.20 BF  $1.40 = $4242.28
12
Example 5-H: You must calculate the cost of 6 inch nominal (5 1/2 actual) beveled siding
applied to 1340 SF of wall surface, use a 1/2 lap and a board foot cost of 1200M.
5.5 - .5 = 5
12
 2 .4
5
2.41340 = 3216 LF
6 x3216
 1608 BF
12
1608 x 1.2 = $1929.60
Try These:
35. Situation
Total square footage to be covered = 4567 SF
Siding 8 Bevel Siding, actual size 7 1/2
Lap = 1
Cost = 980M
Linear feet needed ________________
Board Footage________________
Cost________________
36. You must side a rectangular building with bevel siding. Here is what you know:






The building measures 52-024-0.
The wall height is 9-0
There is a triangular gable at each end of the building with a base of 24-0
and an altitude of 6-0.
The siding is 6 cedar bevel actual size = 5 1/2
The lap is 1
The board foot cost is $1845M
Linear feet needed ________________
Board Footage________________
Cost________________
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49
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6
Chapter
M E M B E R S
CALCULATING REPETITIVE FRAMING MEMBERS
If you could see behind the siding or under the carpeting and sub-flooring of a house, you
would probably see wood studs behind the siding and wood joists under the sub-floor. If you
climbed into the attic you would see wood rafters supporting the roof sheathing and roofing.
Studs, joists and rafters are all examples of repetitive framing members.
When architects draw house plans they seldom show all of the repetitive members but indicate
their presence by writing a note. For example, - 2x6 Studs 16 O.C. Typ. Or 2x12 Flr. Jst. 24
O.C. Typ.
Here is the translation:

2x6, 2x12 = Size of framing members

Stud = Use i.e. wall studs, floor joist

16 on-center, 24 on-center = Spacing (more about this later)

Typ. = Typical, meaning this same assembly will happen everywhere.
ON-CENTER (O.C.)
The term on-center is used to describe the distance from the center of one framing member to
the center of the next. For example the stud wall in Figure 6-1 has studs laid out 12 o.c. Most
repetitive framing members will be 12 inches, 16 inches or 24 inches on-center. The reason is
simple. To efficiently install sheathing (4 x 8) over repetitive members, the sum of the on-center
dimensions must total 48 or 96 inches. For example three 16 o.c. spaces equals 48 inches (the
width of a piece of sheathing), and six spaces equals 96 inches (the length of a piece of
sheathing).
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R E P E T I T I V E
M E M B E R S
Figure 6-1
The wall in Figure 6-1 is eight feet in length and the studs are 12 apart, yet the wall
requires 9 studs. One stud is added because the count began 12 inches from the point
of beginning.
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M E M B E R S
Example 6-A: How many studs are needed for figure 6-1?
The unit length is converted to feet (12  12 = 1)
The total length is divided by the unit length (8  1 = 8 studs)
One stud is added (8 + 1 = 9 studs)
REPETITIVE MEMBER EQUATION
( Total Length  unit length ) + 1 = # of Repetitive Members
NOTE: Framing members are often doubled at openings and at terminal points so don’t
forget to add them to your total.
Example 6-B:
You must frame a wall, which is sixteen feet in length with studs 16
O.C. How many studs will be used?
Use the same steps as above:
16  12 = 1.33 feet
Convert the on-center measurement to feet
Divide the on center measurement into total length
Add one to the calculated number of studs
16  1.33 = 12
12 + 1 = 13 studs
Example 6-C: You are framing a wall that is 22-5 3/8 in length. The studs are to be
laid out 16 O.C. How many studs are needed? This problem is much the same as
Example 5-1 except that it requires the conversion of the total length into decimal feet
in order to carry out the division.
16  12 = 1.33 (memorize this equivalency)
5.375 12= .4479 (conversion of inches to feet)
22.4479  1.33 = 16.878 (round up to seventeen)
17 + 1 = 18 studs
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R E P E T I T I V E
M E M B E R S
Try these:
1. You are to frame a shed roof. The rafters are to be laid out 24 O.C. on a wall plate
21-9 in length. How many rafters are required?
Show work here
Answer_____________
2. You must select materials to construct two interior walls. You have already selected
the plate stock and now must determine the number of studs needed. Both walls total
37 7. How many wall studs are needed at 16 o.c.?
Show work here
Answer ___________
3. How many joists, spaced 16 o.c. are required for a building 59 10 long?
Show work here
Answer_____________
Use the drawing on page 55 to answer questions 4 and 5
4. If the foundation on page 55 has girders spaced 4 O.C., how many 12 48 girders
are needed?
Show work here
Answer _________________
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M E M B E R S
5. How many 44 and 46 posts are needed? Read the note on page 55
Show work here
44 = _____________
46 _____________
6. A gable roof is 46-0 long. Trusses are to be spaced 24 o.c. How many trusses are
Needed?
Show work here
Answer _________________
7. How many joists spaced 16 o.c. are required for a floor 62-6 long?
Show work here
Answer _________________
8. 25 rafters spaced 16 o.c. will require a wall length of ______ _______.
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M E M B E R S
55
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M E M B E R S
LAYING OUT EVENLY SPACED BALUSTERS
When constructing finished stairs and balustrades or exterior decks, it is necessary to
install vertical balusters to protect people from falling through open areas. When I started
out building decks the maximum space allowed between balusters was nine inches, later it
was reduced to six and today the maximum spacing is four inches. Building inspectors
usually scrutinize baluster spacing very carefully. The problem is how do you maintain
equal spacing between the balusters.
If you install your balusters spaced 4 inches apart, everything will be fine until you install
the last one, which unless you are very lucky, will leave a space less than 4 inches. As you
can see the problem with this technique is that there is no guarantee the last space in the
run will be 4 inches. Depending on the length of the run the last space could be just about
anything between 0 and 4. Your customer will not be impressed.
Outlined below is a method of calculating baluster spacing that insures all of the spaces are
uniform.
Layout Procedure For Evenly Spaced Baluster:
Step 1: Determine total length of run and convert to inches.
Step 2: Deduct width of newels at each end of run and add one baluster width.
Step 3: Divide result of step 2 by desired spacing between balusters plus the
width of one baluster. i.e. desired spacing = 4 + 1 1/4”baluster width
4 + 1 1/4 = 5 1/4
Step 4: The answer obtained in step three will typically be an uneven number.
If this is the case round the number up to the next whole number and divide
again.
Step 5: Subtract one baluster width from the answer obtained in step 4 and you have
the correct spacing between balusters.
Step 6: Subtracting 1 from the number of spaces calculated in step 4 gives the correct
number of balusters for the run. Remember you start and end with a space,
therefore one less baluster than spaces.
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Example 6-D: You are constructing a guardrail that will have 3 1/2 newel posts at
each end. The guardrail will be supported in the field by 1 1/2 balusters. The building
code requirement for spacing between balusters is no more than 4. What equal spacing
between balusters would be necessary to maintain the 4code requirement?
Step 1: Convert feet to inches 12-8 3/8 = 152.375
Step 2: 152.375 total
- 7
2 newel widths
145.375
+ 1.5 one baluster width
146.875
Step 3: Code spacing = 4 + 1.5 (baluster width) = 5.5
146.875  5.5 = 26.7045 spaces
Step 4: Round 26.7045 up to 27 spaces (always round up)
146.875  27 = 5.4398 This number represents the spacing plus one baluster
width.
Step 5: Subtract out the baluster width to get the spacing between balusters
5.4398
-1.5
3.9398 or 3 15/16
Step 6: Subtract 1 from the number of spaces calculated in step 4 to determine
the number of balusters. Remember you start with a space and end
with a space. 27 – 1 = 26 Balusters
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M E M B E R S
Try these:
9. Answer the questions below based on the following information:



Balustrade run is 8-2 5/16
4 wide newel posts at each end
1 1/4 balusters and spacing between balusters of
no more than 4
Total run in inches ________
Distance between newels _________
Distance between newels plus one baluster width _________
Maximum code spacing plus one baluster width __________
Actual spacing between balusters __________
Number of balusters __________
10. Answer the questions below based on the following information:
(see illustration on page 83)
 Balustrade run is 6-4 9/16
 4 wide newel posts at one end and rosette at other.
 1 3/4 balusters and spacing between balusters of
no more than 4
Actual spacing of balusters ___________
Number of balusters __________
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R E P E T I T I V E
M E M B E R S
11. Answer the questions below based on the following information :



Balustrade run is 14-5 1/2
3 1/2 newel posts are placed at each end and in the center of the run.
Balusters are 1 3/8 wide and may not be spaced more than 4 apart.
Actual spacing of balusters __________
Number of balusters in each run __________
12. Use information from question 11 with one addition. The newel posts at each end will
have a 1 3/8 baluster attached and the center newel post will have a baluster attached
on both sides.
Actual spacing of balusters __________
Number of balusters in each run __________
59
V O L U M E T R I C
7
Chapter
M E A S U R E M E N T S
VOLUMETRIC MEASUREMENTS AND CUBIC YARDAGE
Now that you understand square footage, volumetric measurements should seem
easy. Volume is a measurement of the space found within three-dimensional
objects. Volume is expressed in cubic units, i.e. Cubic feet, cubic inches, and
cubic yards. As contractors you will use volumetric measurements to determine
quantities of soil, gravel, concrete, water, air and lumber.
1 unit
= One cubic unit
1 unit
1 unit
The cubic unit could be a cubic foot, cubic inch, or a cubic yard.
With a few exceptions, volume may be determined by: Calculating the square footage of
the face of a three-dimensional object and then multiplying its square footage by its
thickness. Look at the example below showing a 10-foot by 10-foot concrete slab with a
thickness of 1 foot.
10
10
concrete slab
FIGURE 7-1
The concrete slab pictured in fig. 7-1 represents
100 square feet.
S2 = Area or 102 = 100 SF
The oblique view of the slab shown in fig. 7-2 shows a
slab thickness or (H)eight of 1-foot.
Volume = S2  H
V = 10  10  1 = 100 ft3
Note: Many texts express cubic footage as ft3 or CF
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V O L U M E T R I C
M E A S U R E M E N T S
10-0
10-0
1-0
Oblique View
FIGURE 7-2
In the example above the cubic footage is the same as the square footage because the
thickness was one-foot, but what if the slab was 6 thick? First you must be careful not to
multiply feet and inches. Six inches must be converted to feet (612 = .5). Having done
this you can now calculate the cubic footage of this new slab.
V = 10  10  .5
V = 50 ft3
Try these:
1. Calculate the volume of a concrete slab measuring 15-6 14-3 4.
Express answer in ft3
Answer __________
2. Calculate the cubic footage of the concrete slab illustrated below.
4 0
15 0
6-0
5 0
9 3
13 6
Concrete Slab, 8 Thick
9 3
Answer _____________
3. You are building four rectangular columns measuring 18  24 120. What is the
total volume of the four columns expressed in cubic feet?
Answer ___________
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V O L U M E T R I C
4.
M E A S U R E M E N T S
You have completed the formwork for 6 cylindrical columns with diameters of 30
and height of 9 feet. What is the combined volume of the six columns expressed in
cubic feet?
Diameter = 30
9-0
Answer _____________
5. Calculate the cubic footage contained within the trapezoidal prism below.
9-0
14-0
23-4
Answer ______________
13-6
CUBIC YARDAGE
Quantities such as concrete, soil and gravel are bought and sold by the cubic yard.
3
= 1 Cubic Yard = 3  3  3 = 27 cubic feet
3
3
As you can see there are 27 cubic feet in one cubic yard. So, to convert cubic feet to cubic
yards simply divide the number of cubic feet by 27.
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Example 8-A: You have set the forms for a concrete slab that will be 25 feet by 37 feet
and 4 inches thick. How many cubic yards of concrete must be ordered?
Volume = 25  37  .33 = 308.03 cu. ft.
Step 1: Calculate the volume in cubic feet.
308.03  27 = 11.41 cubic yards (CY)
Step 2: Convert to cubic yards
Express your answers to problems 1 through 4 in cubic yards (CY)
6. ft3 in problem 1 ____________
7. ft3 in problem 2 ____________
8. ft3 in problem 3 ____________
9. ft3 in problem 4 ____________
10. Calculate the number of yards of concrete needed to build the 8 thick sidewalk
illustrated below.
Large radius = 8 0
4- 0
40- 0
8- 0
30- 0
Cubic footage ________
Cubic Yards
________
4- 0
CLOSED FOUNDATION WALLS AND FOOTINGS
Most foundation walls and footings close on themselves as in Figure 7-3.
What is the volume of the foundation wall in Figure 7-3?
First lets look at the formula for volume:
L  W  H = Volume (use the perimeter length for L)
40  .5  3 = 60 cu. Ft.
Is that an accurate calculation?
10-0
6 wall thickness
Let’s test it. Look
10-0
at Figure 7-4
3-0
6
FIGURE 7-3 (not to scale)
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10-0
12
Centerline
10-0
6
Footing
FIGURE 7-4
Plan View (not to scale)
As you can see from Figure 7-4 the corners overlap. If we use the exterior perimeter
measurement for length and width in the volume formula each corner will be added in
twice. In reality most contractors don’t worry about the over lap and use the extra
concrete as part of their waste factor. However you may run into a situation where exact
volume is necessary, so here is a way to calculate exact volume.
All you need is a perimeter measurement taken at the centerline of the wall. This
represents the average length of the wall. By using the centerline length in the volume
formula, the correct volume for both wall and the footing may be easily obtained. Here is
an easy way to calculate the centerline measurement.
Step 1:
Step 2:
Add up the exterior perimeter of the foundation wall.
10 + 10 + 10 + 10 = 40
Multiply the wall thickness times 4(4 corners)
.5  4 = 2
Step 3:
Subtract the answer from step 2 from the total perimeter measurement.
40 – 2 = 38
Step 4:
Use this number as the perimeter length in the volume formula.
Wall Volume
38  .5  3 = 57cu. Ft.
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38  1  .5 = 19 cu. Ft.
Footing Volume
Note: It doesn’t matter how many zigs or zags there are in the foundation wall as long the wall and footing
closes on itself.
Example 7-B: Calculate the total cubic yards of concrete needed for the foundation
wall below.
45- 6
32- 4
Wall thickness = 8
Wall height = 18
Footing = 15  8
30- 0
10- 0
Note: The answers below were calculated with a calculator set to four decimal places and rounded to
two places.
Step 1: Add up the perimeter measurements
77.8333 + 77.8333 + 40 + 40 = 235.67
4  .6667 = 2.67
Step 2: Multiply 4 times the wall thickness
Step 3: Subtract 2.67 from perimeter measurement
Step 4: Solve for wall volume
Step 5: Solve for footing volume
Step 6: Add wall and footing volumes
Step 7: Calculate cubic yards
235 67 – 2.67 = 233
233  .6667  1.5 = 233 cu. Ft.
233  1.25  .6667 = 194.18 cu. Ft.
233 + 194.18 = 427.17 cu. Ft.
427.17  27 = 16 cubic yards
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Try these:
10- 0
7- 0 typ.
5-0
20 0
22-0
35-0
BUILDING FOOTPRINT
13-0
55-0
Use the illustration above to answer questions 11, 12 and 13.
11. What is the cubic footage of the foundation wall for the structure illustrated
above, if the wall thickness is 8 and the wall height is 2-6?
Answer ___________
12 How many cubic feet are contained within the footing measuring 15 wide by
12 high?
Answer __________
12 How many yards of concrete are necessary for the foundation wall and footing?
Answer __________
13. You are to build a house on a flat lot. The house measures 78-0 by 42-0. The
house is to have a basement requiring you to dig down 8 6. How many 50-yard
dump trucks will it take to remove the soil?
Total # yd 3_______________ Number of dump trucks________________
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AIR VOLUME
Air is most often measured in cubic feet. Typically the reason for measuring air volume is
to determine air exchange. That is, how much air volume is there and how often is it
replaced with fresh air.
Code requires a bathroom, without an operable window, to have a fan that will turn the
air over once every twelve minutes. If you know the air volume of the bathroom you can
determine how powerful a fan is needed. Cubic feet per minute ( cfm ) is the most
common measure of air movement. Fans are rated by cfm.
Example 8-C: You are trying to determine how large a fan is necessary to turn the air
every 12 minutes in a bathroom measuring 8-0  12-0  8-0.
8  12  8 = 768 cu. Ft.
Step 1: Calculate the volume of the room
768 cu. Ft.  12 min. = 64 cfm
Step 2: Divide the room volume by 12 minutes
Therefore a fan with a capacity of moving at least 64 cfm is necessary.
15.
A bathroom with a vaulted ceiling (illustrated and dimensioned below) has no
windows. You must provide a fan that will turn the air over every 12 minutes. What
size fan is needed?
Total cubic feet ____________
Fan size in cfm ______________
8 0
11 3
15 6
8 9
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VOLUME FORMULAS
Square & Rectangles:
Volume = Length  Width  Height
V = LWH
L
Triangular Prisms: Volume = ½  Base  Height Length
V = ½BHL
L
Trapezoidal Prisms: Volume = ½(Base1 + Base2)  H  L
V=
 B1  B 2 

HL
 2 
H
L
Cylinders: Volume =  Radius2 Length
V = r2L
L
Spheres: Volume = 4/3    Radius3
V = 4/3r3
r
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Cones:
V=1/3r2h
S=r r 2  h 2  r 2
h
S=surface area
r
Right Pyrimids:
V=1/3Bh
B=Area of Base
h
Frustum of cone:
r
V= 1/3h(R2+r2+Rr)
H
R
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8
Chapter
P R O P O R T I O N S
RATIOS AND PROPORTIONS
Understanding ratios, and proportions will be one of the most useful math tools you will
encounter in the building trades, in fact man has been fascinated with ratios and proportions
throughout the course of history. Early on it was realized there were proportionalities common
to a living things.
The ancient Greek thinker Vetrueus, called this proportion the Golden Section. You will find
golden section proportions in nature (the Chambered Nautalis), architecture (the Parthenon),
auto design (Volkswagon Beetle), art (
) and much more.
Here is a sample of how the golden section works. Vetruveus found that a rectangle is not a
rectangle, that is to the human eye, some rectangles are more pleasing than others. He found
that a rectangle with a 5 to 8 ratio was the most pleasing. 5 to 8 means that if the short side of
the rectangle is 5 units the long side is 8 units. If 8 is divided by 5 a factor of 1.6 is established.
This factor when multiplied times the short side of any rectangle will yield the length of the long
side.
Example: A golden section rectangle with short side of 25 will have a long side measuring 40.
25  1.6 = 40
One of the most fascinating occurrences of the Golden Section is in the design of the
Chambered Nautilus seashell. As shown in the illustration below the shell of the Nautilus is a
beautiful sweeping curve. So where can you find the golden section?
If you follow theses easy instructions you can draw the curve found in the Nautilus’ shell.
Step 1
RATIOS & FACTORS
Ratio-The relationship in quantity, amount, or size, between two or more things.
Unit Rise = 6
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Unit Run – 12
Figure 8-1
Figure 8-1 illustrates a triangle with a base of 12 and an altitude of 6. The words base and
altitude have been changed to unit run and unit rise respectively. These are terms used in roof
framing to describe the slope of a roof. The word unit refers to the change in rise per foot of
run. For every foot of horizontal run the roof rises 6 vertically.
The relationship between these two numbers can be described as a ratio and may be
written several ways : 6/12, 612, 6:12 or 6to12 It could also be written 12/6, 126, 12:6,
or 12 to 6
Definitions:
The symbol ( : ) is often used when writing a ratio.
Antecedent – The first quantity of the ratio
Consequence – The second quantity in the ratio
Antecedent
6/12
Consequence
If this fraction is reduced, we can say the ratio is 1 to 2.
We can take this a step further and create a FACTOR by carrying out the division.
6  12 = .5
Factor
12  6 = 2
Factor
As illustrated above there are two possible factors. The first factor (.5) tells us that the
antecedent is one half the consequence. The second factor (2) tells us the antecedent is two
times the length of the consequence.
Example 8-A:
Roof slope is 6/12
The run is 15-0
What is the rise?
Factor = 6/12 = .5 (one half the run), therefore:
15  .5 = 7.5 or 7-6.
A 6/12 roof with a run of 15-0 has a rise of 7-6.
What if the problem had been stated this way:
Roof slope is 6/12
Roof rise is 7-6
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What is the run?
This problem asks for the length of the run.
Factor = 12/6 = 2 (twice the rise)
7.5  2 = 15-0
A second approach to solving this problem is to divide the rise by the factor (.5).
7.5  .5 = 15-0
This second approach allows you to work with only one factor. Roof framing slopes are always expressed as rise
over run, so it’s only natural to solve for the factor in this way.
Try these:
Find the factors for the following roof problems and solve for the missing rise or
run.
1. Slope = 8/12
Factor ____
Roof run = 13-6
Roof rise = ___________
2. Slope = 4/12
Factor ____
Roof run = 16-3
Roof rise = ___________
3. Slope = 9/12
Factor ____
Roof run = 13-6
Roof rise = ___________
4. Slope = 14/12
Factor _____
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Roof run = __________
Roof rise = 12-9 1/2
5. Slope = 3/12
Factor ____
Roof run = ___________
Roof rise = 3-11
Because confusion may arise when deciding whether to multiply or
divide, let’s look at a more intuitive way to solve these problems.
PROPORTIONS
There are two types of proportions – direct proportions and indirect proportions. Direct
proportions will be dealt with first followed by indirect proportions.
Definition: PROPORTION - To adjust in size relative to other parts or things. Also, proportion is an
expression of equality between two ratios.
DIRECT PROPORTIONS
Isn’t that what was done in example 8-A? In that example the slope of the roof was given as
6:12. You then calculated the rise of an actual roof where the run was 15-0. You adjusted the
rise to be proportional.
Proportions can be symbolized or stated several ways. ( = and :: are used to symbolize
proportions)
6" 7.5'

12" 15'
6:12::7.5:15
Six inches of rise is to twelve inches of run as seven and one half feet of rise is to fifteen
feet of run.
The first and last terms of a proportion are the extremes and the middle terms are the
means.
means
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extreme
P R O P O R T I O N S
6/12 = 7.5/15
extreme
To test whether you have set-up the proportion correctly, remember this - The product of
the extremes must equal the product of the means. 6  15 = 12  7.5
Another way to insure a proportion is set up properly is to test whether the factors are the
same for both sides of the equation. If we wrote the equation as seen in Figure 8-2, the two
ratios would no longer be proportional nor would their factors be the same (612 = .5
and 15  7.5 = 2) nor would the products of the extremes equal the product of the means
(6  7.5  12  15).
6" 15'

12" 7.5'
WRONG SETUP
Figure 8-2
Rule: When setting up a proportion pay close attention to the units (feet, inches, miles,
pounds, etc.) particularly when the problem uses ratios with different units.
Remember that you can only use two different units in a proportion. For example
you cannot use feet, minutes and hours. You must convert hours to minutes or
minutes to hours.
ORGANIZING UNITS IN THE PROPORTION PROBLEM
You have two choices, to align like units vertically on each side of the equal sign or align
like units horizontally across the equation.
Example 8-B:
6
x

12 15
or
6 12

x 15
SOLVING FOR UNKNOWNS
We have covered the use of factors to solve for unknowns, however setting up a
proportion offers perhaps an easier more intuitive way to arrive at a solution. The key is to
read the problem and then state what you know and what you want to know, written as a
proportion. Look at the word problem in example 8-C
Example 8-C: If it takes 56.2 minutes to load five 12 yard dump trucks, how long does
it take to load one10 yard truck?
Solution:
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Step 1 Look for what you know. It takes 56.2 minutes to load 60 yards (5  12)
Step 2 Put what you know in the form of a ratio.
56.2 min .
60 yards
Step 3 Look for what you want to know. How many minutes does it take to fill 10 yards?.
X min .
10 yards
56.2 X

Step 5 Put both ratios together in a proportion
12
10
Step 4 Put what you want to know in the form of a ratio.
Say it to yourself – 56.2 minutes is to 12 yards as X minutes is to 10 yards.
Step 6 Multiply the means together and the extremes together and place their products on the
opposite sides of an equal sign. Remember their products must be equal.
Extremes
56.2  10 = 562
Means
12  X = 12X
Next isolate x on one side of the equation by dividing both sides by 12
12X = 562
M 
562
 46.83 X = 46.83 minutes
12
Note: Don’t worry about memorizing the words means and extremes. What you just did is also
called cross multiplication, a term we will use throughout the book.
3.5
x

=
13.5 60
Cross Multiplication
The ability to recognize and setup ratios and proportions found in word problems or on the job
site will make you an effective problem solver. Try these.
Try These:
6. Solve example 8-C by factoring. ____________
Show work below
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Setup a proportion formula and solve the following problems. Show work!
7. A carpenter is able to cut 360 pieces of blocking in an eight-hour shift. How many
pieces can she cut in one minute? _________________
8. A journeyman carpenter receives $460.00 for completing a job. This is one and onehalf times what his helper makes. How much pay will the helper receive?
____________
9. You are framing a wall that is 26-6 in length. You are placing your studs 16 o.c.
How many studs would be necessity to complete the job? ___________________
10. A contractor is developing a contour map for a piece of property on which he is to build a
house. He places two stakes in the ground 20 feet apart. The elevation at stake A is 207
feet. The elevation of Stake-B is 210.5 feet. The contractor places Stake-C on line,
between stake A and B and 6 feet from stake A. What is the elevation at stake C?
Hint: Use the sketch to visualize the ratios and proportion. Remember, Stake-C is 6 feet horizontally from Stake-A .
B 210.5
A 207.0
6
C
20
PROFILE
Answer _______________
11. A piece of land is being prepared to receive an asphalt driveway. The driveway is to be 25
feet in width. The driveway is to have a 2% grade (2 feet of drop for every 100 feet of run) from
the centerline to the edge of the driveway. What is the change in elevation from the road’s
centerline to it’s edge? _______________
12. What is the circumference of the circle below? _______________
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13. A roof has a run of 18-6 and a rise of 10-9 1/2. What is the rise per foot?
Answer
____________
14. How many linear feet of 10 nominal (9 1/2 actual) bevel siding will it take to cover 1200
SF if the siding has a 1 overlap?
Answer _________________
INDIRECT PROPORTIONS
Definition: Indirect Proportions – As the terms of one ratio increase or decrease the terms
of the other ratio do the opposite.
Example 7-D: It takes three carpenters 22 days to frame a house. Working at the same
rate how long will it take seven carpenters to frame the same house?
If this problem is setup as a direct proportion (3/21= 7/X) it will take seven carpenters 49
days to do what three carpenters could do in 22. This is illogical. Seven carpenters should
be able to finish the job in less time, right?
To solve this problem an indirect proportion must be used. Set up just like a direct proportion
but switch the denominators.
3carp. 7carp.

21days xdays
Direct proportion
3carp
7

xdays 21days
Indirect proportion
Notice two of the terms are reversed. Once the reversal has been made the problem
can be solved as a direct proportion.
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Solution:
F A C T O R S
63 = 7X
A N D
P R O P O R T I O N S
X = 9 days It will take 7 carpenters 9 day to frame the house.
Try these:
12. Three roofers can install ten squares of roofing in six hours. Working at the same rate
how many hours will it take five roofers to install ten squares?
Answer ______________
13. If the same five roofers worked for six hours, how many squares could they install?
Ask yourself whether the solution to this problem requires a direct or indirect proportion
Just for Fun:
Two people start walking from the same spot, in opposite directions around the three and one
half mile Fairmount Loop. One person is walking at a rate of three and one half miles per hour
and the other person is traveling at a rate of two and one half miles per hour. How long will it
take them to meet and how many miles will the faster walker have walked when they meet?
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Chapter
L A Y O U T
STAIR LAYOUT
The purpose of this chapter is to outline the mathematical calculations necessary to layout
stairs. Many things in construction can be done incorrectly and not detected, however an
improperly cut and installed staircase is not only easily detected but also potentially
dangerous. Have you ever walked up or down a staircase and tripped on the last step? It
probably wasn’t your fault.
There are numerous building codes effecting stair construction which building inspectors
love to enforce and rightfully so. There are codes restricting the height of risers and the
depth of treads. There is a maximum height difference allowed between risers. Handrails
have height requirements and all stairwells have minimum headroom allowances. The list
goes on and on.
Consequently this is an important chapter. Stair building demands attention to detail and
craftsmanship from the carpenter. Good stair layout and construction begins with
accurate mathematical calculations.
TERMINOLOGY. See the illustrations on pages 80 and 81.







Stringers – Sometimes called carriages, are the inclined members of a set of stairs to
which treads and risers are secured.
Total Run – The horizontal distance from the face of the upper stair riser to the face
of the lower stair riser.
Total Rise – The vertical distance from the finished surface of one floor level to the
finished surface of the floor level above or below.
Unit Run – The depth of each tread minus any overhang.
Unit Rise – The height of each riser in a flight of stairs.
Tread – The horizontal member on which one steps
Riser – The vertical height on a stair stringer between two consecutive treads
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STEP-BY-STEP PROCEDURE FOR CALCULATING STAIR RISE AND RUN
The first step (no pun intended) in calculating unit rise and unit run is possibly the most
misunderstood and the most miscalculated part of the process.
Note:
Stair calculations are more easily worked in inches.
Step 1: Determine TOTAL RISE, the distance from finished floor to finished
floor. That is, from the finished surface (carpeting, vinyl, hardwood or tile) of the lower
floor to the finished surface of the floor above. This may seem academic, however the
finished floor surfaces are rarely in place when the calculations are made. If the customer
has not yet decided on floor surfaces you better move on to something else until they do.
Typically, stringers and temporary treads are put in place during the rough framing stage.
At this stage the floor surfaces are usually sub flooring only. This means if carpeting were
to be the finished floor, an additional 1/2 would be added for underlayment and another
1/2 for carpeting. If the same materials are used on both floors then it’s a wash, because
they cancel each other out. In this situation the total rise could be measured from subfloor to sub-floor.
But what if the lower floor receives carpeting and underlayment totaling 1and the upper
floor receives 3/4 hardwood flooring only? Since 3/4 T&G hardwood flooring would
be installed directly over the sub-floor, the gain would be 3/4 on the second story where
as the loss in total rise on the first story (underlayment and carpeting) would be 1. In this
case if the measurement from sub-floor to sub-floor were 101 3/4 the measurement
from finish floor to finish floor would be 101 1/2.
Rule: When measuring total rise from the lower level sub-floor to the second level sub-
floor, subtract what will be added to achieve a finished lower floor from the total
measured distance and add what will be added to achieve a finished floor on the
second level. i.e. 101 3/4 -1 + 3/4 = 101 1/2
Step 2: Divide the total rise by seven. It has been determined that seven inches is the
optimum riser height in terms of human ergonomics. Using the example above
101 1/2  7 = 14 .5 risers. I think you would agree it would not be prudent to have
fourteen and one half rises? Now you know why you sometimes stumble on the first or
last step of a staircase.
Step 3: Calculate the UNIT RISE by rounding the number of risers from step 2
up or down and divide it back into the total rise. Since a partial step is not an option,
make the number of rises a whole number (14) by rounding down.
101.5  14 = 7 1/4. As a rule, try to make your riser height between seven inches and
seven and one half inches.
Step 4: Calculate the number of treads. The number of treads depends on where the
stringer is attached at the upper level. If the first tread is an extension of the upper floor
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then the number of treads is the same as the number of risers. If the first tread starts one
step down from the upper floor level, then there is one less tread than risers.
The second method of placement is the most common because it requires less horizontal
space than the other. The sooner the staircase begins it’s decent the less space it takes to
get to the bottom.
Step 5: Determine the tread depth or UNIT RUN. Walking up stairs with very
shallow treads is much like climbing a ladder. Conversely walking up stairs with very deep
treads may not fit your pace and feel awkward. One of the factors that effects tread depth
is how much space is available for the staircase. If there is plenty of horizontal space for
the stairs then you have more options for tread depth. If the horizontal space is restricted
then tread depth may be determined by the minimum code requirement.
Guidelines for sizing treads and risers
Guideline1 :
The sum of one riser and one tread should equal between 17 and 18.
Guideline2 :
The sum of two risers and one tread should equal 25.
Try These:
1. The measurement from sub-floor to sub-floor is 99 1/2. The lower floor level is to
receive 7/16 “cement-board” with 1/4 tile for the finished floor. The second level is
to receive 1/2 particleboard and 1/2 carpet. What is the finish floor to finish floor
measurement?
Show work here
Answer _____________
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2. The measurement from sub-floor to sub-floor is 106 3/8. The lower floor level is to
receive 3/4 hardwood and second level is to receive 1/2 particleboard and 1/8
vinyl. What is the finish floor to finish floor measurement?
Show work here
Answer__________
3. Fill in the blanks concerning problem one. Assume that the first tread begins below the
second floor level.
# risers __________
unit rise _______
# treads _________
unit run __________ (11)
total run __________
4. Fill in the blanks concerning problem two. Assume that the first tread begins below
the second floor level.
# risers __________
unit rise _______
# treads _________
unit run __________ (your call)
total run __________
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5. You must frame a staircase from an exterior wood deck to the ground level. You have
determined that the total rise is 72 5/8. The stair’s total run must be no more than 711 5/8. Assume the first tread begins below the second floor level.
# risers __________
unit rise _________
# treads __________
unit run __________
total run __________
6. You must construct an L shaped staircase with intermediate landing. The total rise is
101 (finished floor to finished floor). Because of headroom considerations, the
landing must be no more than six risers higher than the first floor elevation. Assume
that both the lower and upper stairs have the same unit rise and unit run.
# risers __________
unit rise _________
# treads __________
finished height of landing ____________
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Chapter
G E O M E T R Y
10
CONSTRUCTION GEOMETRY
What types of construction tasks require an understanding of geometry? The answer is just
about every type. From construction surveying to cabinetmaking, geometry plays an
integral part in defining property boundaries; squaring building corners, cutting roof
rafters, and building cabinets that are square.
What is geometry? Geometry can be defined as a branch of math dealing with the
measurement, relationship, and properties of figures, points, lines, and solids. In the
previous chapter on surface measurements we looked at, defined, and learned area
formulas for polygons and circles. This chapter will take a closer look at the properties of
these objects.
Geometry has been studied for thousands of years. The Egyptians used geometry to build
the great pyramids. An age-old problem, encountered by the builders through out time is
laying out a square corner.
In figure 10-1 an interior wall is to be constructed to intersect with an existing exterior wall
at a right angle (90). Drawing the right angle in Figure 10-1 was easy. I simply pressed the
shift key while I clicked and dragged the mouse. On the construction site it’s a different
story. How did the Egyptians do it?
They used a rope with knots tied in strategic locations. The rope was placed in the corner
and bent into the shape of a right triangle, with the knots at each vertex. Smart, because
one of the three angles of a right triangle is of course 90.
right triangle
interior wall
exterior wall
90
Figure 10-1
Egyptian builders knew from experience where to tie the knots in the rope and how to
bent it into a right triangle but they didn’t know why it worked mathematically. Some of
the pyramids indicate an accurate understanding of Pi, but the mathematical knowledge of
the Egyptians did not include the ability to derive pi by calculation. It is possible that this
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could have been arrived at "accidentally" through a means such as counting the
revolutions of a drum. Their lack of mathematical understanding should take nothing away
from their accomplishments, many of which are not completely understood 4000 years
later.
Around 500 BC the great thinkers of the ancient Greek civilization began to confirm these
field proven techniques through mathematics. One of the greatest Greek philosophermathematicians was Pythagoras who was no doubt inspired after living in Egypt. Today
we particularly remember Pythagoras for his famous geometry theorem. Although the
Babylonians knew the theory 1000 years earlier, Pythagoras may have been the first to
prove it.
The Pythagorean theorem explains why the Egyptian rope method worked. The theorem
of Pythagoras states that for a right-angled triangle the square of the hypotenuse is equal to
the sum of the squares of the other two sides. It should be pointed out that to Pythagoras,
the square on the hypotenuse would certainly not be thought of as a number multiplied by
itself, but rather as a geometrical square constructed on the side of the triangle. To say that
the sum of two squares is equal to a third square meant that the two squares could be cut
up and reassembled to form a square identical to the third square.
Figure 10-3 illustrates how Pythagoras saw the theorem. The triangle in Figure 10-2 is made
up of three sides, one side four feet in length, a second side three feet in length and the
third side five feet in length.
If each side is made into a square, the area of the 4 foot square plus the area of the three
foot square equal the area of the five foot square as seen in figure 10-3
44 = 16
33 = + 9
55 = 25
4
5
3
Figure 10-2
Figure 10-3
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Today we write the Pythagorean theorem in a mathematical formula a2 + b2 = c2
In this formula c represents the side opposite the right angle and is termed the
hypotenuse. See figure 10-4. The other sides of the triangle are represented by a and b. It
is important to note that in the formula above the outcome of adding a2 and b2 is c2.
Lets substitute the values of the triangle in Figure 10-2 into the formula.
42 + 32 = 52
16 + 9 = 25
Our answer is the square of the hypotenuse or C2, however, we want to know C.
Therefore, the formula can be written in a different form, which solves for c as below.
Angle A
Pythagorean theorem
a b  c
2
2
2
The hypotenuse of a right triangle
equals the square root of the sum of
the squares of the other two sides.
Hypotenuse
side c
Angle B
side a
side b
90 Angle C
Figure 10-4
Right Triangle
Example 10-A:
4 2  32  c
Solve for the hypotenuse of figure 10-2
16  9  c
25  5
THE 3-4-5 SQUARING METHOD
Today we still use the same techniques employed by the Egyptians to square a corner,
except we use Mylar tapes and lasers instead of a rope. Carpenters call the method the
three-four-five method. Here is how it works.
As in figure 10-1 we know that a triangle with sides measuring three feet and four feet, and
hypotenuse measuring five feet is a right triangle. Consequently carpenters measure three
feet along an existing wall from the point where the intersecting wall will be constructed
(called the point of beginning) and make a pencil mark. Next from the point of beginning
the tape measure is extended four feet at an approximate right angle to the existing wall.
Another tape is then extended diagonally from the pencil mark to the extended tape. They
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are then aligned so that the five-foot mark on the diagonal tape touches the four-foot
mark on the perpendicular tape. Another pencil mark is made on the floor at this point. A
chalk line is now extended and snapped from the point of beginning to the pencil mark.
Presto, a right angle is formed!
To increase the accuracy of the procedure the lengths of the sides of the triangle can be
doubled. The same procedure would be carried out except that the triangle sides would
measure six feet and eight feet, and the hypotenuse would measure ten feet.
Try these:
Where appropriate express answers in feet and inches
1. Solve for the hypotenuse of the right triangle
12-0
Answer ___________
15-0
2. Solve for the hypotenuse of the right triangle.
10-4
Answer ____________
32-11
3. Solve for the hypotenuse of the right triangle.
25-7 7/8
Answer ___________
10-2 7/16
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4. Solve for the hypotenuse of the right triangle.
3-4 11/16
56-7 3/16
Answer _________
5. You are to construct a tall cabinet measuring 93 in height, 48 in width and 24 in
depth. The cabinet will be assembled in the shop and delivered to the job site. It must be
carried through a doorway, stood up in the room and slid against the wall where it will be
secured. The ceiling height is 96. Will the cabinet clear the ceiling when tilted up?
Answer ______
Example 10-B: How do you solve for one of the sides when you know the length of
the hypotenuse and the other side as in this example? Substitute the values into the
formula and solve.
a 2  9 2  17 2
a 2  81  289
a 2  289  81  208
a  208 (subtract 81 from both sides)
a = 14.42 or 14-5 1/16
a
9
Try these:
6. Solve for side b of the right triangle
30-2
Answer ______________
12-4
b
7. Solve for side a of the right triangle
a
16-5
Answer _________________
13-6
90
17
C O N S T R U C T I O N
G E O M E T R Y
8. Solve for side b of the right triangle
120
33
Answer ________
b
SQUARING BUILDING CORNERS
As a carpenter you will be checking various squares and rectangles for squareness. That is,
checking to make sure that all of the interior angles are 90. For example when foundation
forms are set it is essential to check for square before concrete is poured. Figure 10-4 illustrates a
rectangular building measuring 24-0 by 10-0. If squareness were not checked the building in
Figure 10-5 could easily wind up looking like Figure 10-6
10-0
24-0
FIGURE 10-5
FIGURE 10-6
In Figure 10-6 the linear measurements are still the same, the problem is the interior angles
are no longer right angles. It’s pretty obvious that the corners in Figure 10-6 are not square
but think about what it would actually look like out in the field. Being out of square is not
that easily detected, when you consider the scale.
Luckily we can fall back on the Pythagorean theorem to insure the corners are square.
Drawing an imaginary diagonal across the rectangle makes two identical right triangles. See
Figure 10-7
26- 0
10-0
24-0
FIGURE 10-7
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Example 10-C: Use the Pythagorean theorem, to calculate length of the diagonals.
24 2  10 2  26'0"
Using Diagonals To Square Building Corners: After calculating the length of the
hypotenuse, extend the tape diagonally across the foundation forms. If the diagonal
measures 26-0 your forms are probably square. Measure the diagonal in the opposite
direction and check the length. See Figure 10-8
FIGURE 10-8
What if one diagonal measures exactly 26-0 and the other does not? Is it possible? Yes,
it means that one side of the rectangle is longer or shorter than its counterpart. For
example one side of the rectangle is 24-0 and the opposite side is 24-1.
If both diagonals are of equal length the object is square. Calculating the length of
the hypotenuse is not always necessary. For example if you want to know whether a piece
of plywood is square, measure the diagonals. If they are not equal the plywood is not
square. The same technique may be used when squaring a building or anything for that
matter.
INTERIOR ANGLE REVIEW
You will remember from a previous chapter that the sum of interior angles of any closed
polygon equal the number of sides minus two, times one hundred and eighty.
Sum of interior angles = n  2180
Therefore the sum of interior angles of a triangle is 180. In a right triangle one angle is
90. Consequently the sum of the other two angles is 90.
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TRY THESE:
Note:
 is the symbol for angle
9. What is angle A in the right triangle
B
C = 90
B = 30
A = _______
C
A
10. What is angle B of the triangle
 C = 90
A = 15
 B = _________
B
C
A
11. What is angle C in the non-right angle triangle.
 A = 32
 B = 52
 C = ________
A
C
B
11. What is the sum of the interior angles of the polygon. Remember, interior angles may be
greater than 90.
Answer ___________
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ROOF FRAMING
The ability to calculate, layout, and frame roofs is one of the defining skills of a carpenter. One
needs good visualization skills because roof systems are three-dimensional and a good
understanding of geometry because most roofs are based on the right triangle.
Once the walls are erected and plumbed, the carpenter must begin collecting field
measurements and roof framing information off of the blueprints and such as roof rise and run .
Next, calculations will be carried out to determine the theoretical measurements necessary to
layout the framing members. And finally the carpenter must layout, cut and install the rafters.
Our mission is to learn the vocabulary of roof framing, collect information and calculate the
measurements necessary to layout roof members. Rafter layout requires more skills you will
learn later.
ROOFS
Roofs come in many different shapes and styles but by in large they are nothing more than right
triangles. Figure 10-9 illustrates the simplest roof form, the shed roof. Figure 10-10 illustrates the
right triangle hiding under the roof.
Depending on the roof, the right triangle can have different proportions. Those proportions are
defined by the unit triangle, pictured in figure 10-9. The unit triangle defines the rise and run of
the roof in the simplest form. For shed and gable roofs the run will always be 12 and the rise
will vary. The unit triangle in figure 10-9 has a run of 12  and a rise of 6. In terms of geometry
we would say that the triangles base is 12 and it’s altitude is 6.
The most important thing to remember is that the geometry of the unit triangle is the
same as the geometry of the roof you are calculating.
12
6
unit triangle
Shed Roof
Right Triangle
FIGURE 10-9
FIGURE 10-10
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The gable roof (Figure 10-11) is made up of two right triangles. Sketch them next to figure 10-1
Your Sketch
Gable Roof
FIGURE 10-11
We will discuss other roof styles later in this chapter.
ROOF TERMINOLOGY AS SEEN IN FIGURE 10-12
It is important for you to know and understand the basic terminology used in
roof framing.
Span - The total width of the structure
line length
Run - One half the span (base of the roof triangle)
Rise - The height or altitude of the roof triangle
Unit run – The unit of measurement given in inches
Unit rise – The amount of rise per foot of run
Slope* -The incline of a roof expressed as a ratio
Run
of rise to run. i.e. 4/12
Pitch* – The slope of a roof expressed as rise divided
by span
Line length – the hypotenuse of the triangle
Rise
unit run
unit rise
Span
Figure 10-12
*Note: The words slope and pitch have different meanings as noted above, however you will commonly hear them used
interchangeably. We will not speak in terms of roof pitch in this book because it is something rarely used. Later we
will talk about slope angle and how it may be calculated.
13. Write in the roof properties listed below in the appropriate locations on Figure 10-13.
Properties
Span = 24-0
Run = 12-0
Rise = 6 0
unit run = 12
unit rise = 6”
Slope = 6/12*
Pitch = 6/24 0r .25*
Line Length = 13- 5
* do not include in diagram
Figure 10-13
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CALCULATING ROOF PROPERTIES
Typically, roof span, and the unit triangle can be found on your plans, but it’s line length
and rise, which must be calculated.
Armed with unit rise, unit run and total run or span we have all the information needed to
calculate total rise, common rafter length, hip rafter length, jack rafter length and more, for
any basic roof. The unit triangle is proportionate to the actual roof triangle. By
setting up ratios and factors (constants), as we did on page 67, we can easily calculate the
lengths of all our roof members. Let’s use the 6/12 slope triangle to show how easy it is to
calculate total rise and line length.
TOTAL RISE CALCULATION – In the case of the 6/12 roof, it is easy to see that the unit rise
is one-half the unit run. Figure 10-14 shows how the constant is calculated. The constant can
now be multiplied times any run and the proportional rise is obtained. For example, a total run
of 20-0 has a total rise of 10-0.
20  .5 = 10 feet.
6
6/12 = .5
.5 is the rise constant for any roof with
a slope of 6/12
12
Unit Triangle
FIGURE 10-14
REMEMBER:
The ratio is set up with the value you want to know as the numerator. The ratio above is
6/12 because we want to know the total rise. If you wanted to solve for the total run you
could set up the ratio as 12/6. 12/6 = 2 Therefore a building with a total rise of 10 feet
will have a total run of 20 feet. 10  2 = 2
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LINE LENGTH
As defined earlier, line length is the hypotenuse of the right triangle. Again, the hypotenuse
of the unit triangle is proportionate to that of the roof triangle. So, just as we did before
we can set up a ratio with the unit hypotenuse as the numerator and the unit run as the
denominator as in Figure 10-15. The resulting factor, when multiplied times the total run
yields the line length of the common rafter. All we have to do is calculate the hypotenuse
using the Pythagorean theorem and divide by 12.
6
13.41
13.42/12 = 1.12
1.12 is the line length constant
for any roof with a slope of 6/12
12
FIGURE 10-15
Homework
Calculate the rise and line length factors for roof slopes 2/12 through 14/12 and add
to your reference manual.
TRY THESE:
Where appropriate express answers in feet and inches
14. The span of a structure is 32-0  and the slope is 4/12. Fill in the blanks below.
Rise factor________
Total run ___________ Line length __________
Line length factor ____________
Total rise ____________
15. The total run of a building is 17-5 and the slope is 8/12. Fill in the blanks below.
Span ________
Rise factor________
Line length factor ____________
Line length _________
Total rise ______________
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16. Fill in the blanks based on the information given in illustration.
2x8 Rafters
12
9
29-4
Elevation
Plan View
Rise factor________
Total run ___________ Total rise ___________
Line length factor ____________
Line length __________
A CLOSER LOOK AT RAFTER CALCULATIONS
Until now we have been looking at simple wire frame models of framed roofs but in reality
rafters have thickness and height so let’s take a closer look. Figure 10-16 introduces us to
some new terminology.
See Detail 10-a
Center Line of Ridge
Line Length
Tail
Common Rafter
Layout Line
Double
Top Plate
Total Run
See Detail 10-a
Overhang
Span
Figure 10-16
In the previous examples you solved the theoretical line length and theoretical total rise. It
is important to note where the theoretical triangle is in relation to the actual rafter. Notice
in Figure 10-16 the theoretical line length or layout line is not measured at the top or
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bottom edge of the rafter. Detail 10-a on page 97 clearly shows the point to which total
rise is measured at the ridge and the top plate of the wall.
If you look carefully at Figure 10-16 you can see the overhang extends the rafter beyond the
wall plate. Note also that the rafter must be shortened at the ridge because the ridge board
has thickness.
More Roof Terminology

The Rafter Tail – The line length of the rafter does not include the tail. The
length of the tail is a separate calculation, based on the overhang dimension.
See Figure 10-16

Overhang – The overhang is always given as a horizontal measurement. If
the plan shows the overhang at 24, that means 24 horizontally from the
vertical plane of the wall. See Figure 10-16

Bird’s Mouth – This is a notch cut into a rafter to provide a bearing surface
where the rafter intersects the top plate of the wall. The bird’s mouth notch is
comprised of two cuts in the rafter: A seat cut, which is the horizontal cut
where the rafter bears on the top wall plate, and the plumb cut, which as the
name implies, is the vertical cut of the bird’s mouth. See Detail 10-a

Height at Plate (HAP) The distance measured vertically from the
intersection of the seat cut and plumb cut, to the top edge of the rafter. See
detail 10-a .
The HAP measurement is a variable determined by the carpenter on the job.
Generally speaking it should be no less than 2. Notice that the HAP
measurement can also be seen at the ridge end of the rafter.

Ridge – The horizontal framing member that rafters are aligned against to
resist their downward force. The minimum thickness allowed by the
International Structural Code for One and Two Family Dwellings is 1 nominal,
however 2 members are more typically used. Notice that the ridge has been
lowered so that its top edges align with the top edge of the rafter. Beveling
the ridge to match the angle of the rafter would eliminate the need to lower it,
however this is rarely done because of the labor-intensive nature of such a
procedure.
Note: Another point of code worth mentioning is that the rafter must have full bearing against the
ridge board. Generally, the rafter is a size smaller than the ridge board on which it bears.
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DETAIL 10-a
Total rise is measured
to this point
Ridge
HAP
?
HAP
Plumb Cut
Seat Cut
Example 10-D:
You must construct a roof with a slope of 8/12, a span of 25-0, and
an overhang of 24. The plan calls for 28 rafters a 210 ridge board and a 4 HAP.
Here is what you want to know:
 To what length will the rafter be cut, including the tail?
 What is the total rise?
 How far will the ridge be dropped?
 You must cut a 24 brace, running from the top plate to the bottom of the
ridge, to temporarily support the ridge until the rafters are in place. To what
length will you cut the 24?
Step 1: Make a sketch of the roof showing what you know. This is a good idea
until you become very comfortable with these calculations.
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Sketch Here
Step 2: Calculate the rise and line length constants
Rise Constant = 8/12 = .6667
Line Length Constant =
8 2  12 2
 1.2019
12
Step 2: Calculate the total run.
Formula: Span  2 = Total run
25  2 = 12.5 or 12 6
Total run = 12- 6
Step 3: Calculate the total rise
Rise Factor  Total Run
.6667 12.5 = 8.33
Total Rise = 8- 4 
Step 4: Calculate the line length
Line Length Constant  Total run
12.5  1.2019 = 15
Rafter line length = 15- 1/4
Step 5: Calculate the line length of the rafter tail.
Line Length Constant  Overhang
1.2019  2 = 2.4038
Line length of rafter tail = 2-4 7/8
Step 6: Total Length of Rafter
15-0 1/4
+ 2 -4 7/8”
17 -5 1/8
Total length of rafter = 17- 5 1/8
Step 7: Calculate ridge drop. If there were no ridge board, the rafters would
meet at the centerline, but because the ridge board has thickness the rafters are
backed off horizontally, 1/2 the thickness of the ridge. See Figure 10-17
You can see in Figure 10-17 that as a result of backing the rafters away from the
centerline, the top edge of the rafter drops. You could cut a bevel on the top
edges of the ridge to continue the rafter angle to the centerline, but that would be
quite labor intensive.
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So, the question is - How much do we drop the ridge so that it’s top edge meets
the top edge of the rafter? Looking at Figure 10-17 you can see that the amount of
drop needed is the altitude of the small black triangle labeled Triangle A.
As you might have already figured out this little triangle, tiny as it is, is
proportional to the unit triangle. We can calculate the rise of triangle A by
multiplying the rise constant times its run.
The total run in this case is half the thickness of the ridge or 3/4.
Rise Factor  Total Run
.6667  .75 = .5
Ridge Drop = 1/2
Triangle A
?
Rafter
9 1/4
Ridge
1 1/2
Figure 10-17
The first element to be erected when framing a roof is the ridge board because without it, the
rafters have nothing to rest upon. The ridge is usually supported at each end by temporary
supports resting on the top plates of the wall and extending to its bottom edge. In Step 8 we will
calculate the length of the temporary support.
Step 8: Calculate the length of the temporarily ridge support.
Remember the support extends from the top plate of the wall to the bottom of the ridge
board.
Also remember that the total rise extends from the top plate to the point where the layout
line intersects with the centerline of the ridge. To solve for the support we must add the
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height at plate (HAP) measurement to the total rise, subtract the ridge drop and finally
subtract the height of the ridge board.
We already have all the information we need:
Total rise
HAP
Total Rise + HAP
Ridge Drop
Ridge Height
Length of 24 support =
8- 4
+ 4
8 -8
- 0 1/2
8- 7 1/2
- 9 1/4
7- 10 1/4
Note: Working several of these problems will bring all of this into focus. There is much more involved in actually laying
out and cutting common rafters. The purpose of this chapter is to focusing on the theory behind roof framing. An
understanding of this theory will make rafter layout much easier and fun.
Are you ready to try some on your own? Use the previous example and
your roof constants to guide you through the next three problems
17. Situation:
Fill in the blanks:
 Gable roof
 Span = 28 0
Total rise ___________
 Slope = 4/12
Line Length _________
 HAP = 5
Overhang line length ______
 Overhang = 18
Common rafter length______
 Rafters = 2X10
Ridge drop ________
 Ridge = 212
Support length _______
 The ridge is to have a temp. brace supported at the top plate.
18. Situation:
 Shed roof
 Total run = 15 9
 Slope = 7/12
 HAP = 3 1/2
 Overhang = 16
 Rafters = 2X8
 Ridge = 2X10
 No temp. brace necessary
Fill in the blanks:
Total rise ___________
Line length __________
Overhang line length ______
Common rafter length ______
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19. Situation:
Fill in the blanks:
 Gable roof
 Span = 22-8 5/8
Total rise ___________
 Slope = 6/12
Line Length ________
 HAP = 3
Overhang line length _______
 Overhang = 18
Common rafter length ______
 Rafters = 28
Ridge drop ________
 Ridge = 210
Support length __________
 The ridge is to have a temporary brace supported at the top plate.
20. Situation: Hint – find a proportion to solve total run
Fill in the blanks:
 Gable roof
 Span unknown
Total run ___________
 Total run unknown
Line Length _________
 Slope = 10/12
Overhang line length ___
 Total rise = 9 5
Common rafter length ______
 HAP = 4
Ridge drop ______
 Overhang = 28
Support length ________
 Rafters = 212
 Ridge = 214
 The ridge is to have a temp. brace supported at the top plate.
THE HIP ROOF
A hip roof is a roof shape with four sloping sides. The most basic form of hip roof is a
four-sided pyramid illustrated in Figure 10-18. Typically hip roofs look more like the
illustration in Figure 10-19.
Pyramid
Hip Roof
Figure 10-18
Figure 10-19
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G E O M E T R Y
The rafter layout of a hip roof is illustrated in plan view in Figure 10-20. Notice that the run
of the hip rafter is the diagonal of a square, having sides equal to the total run.
Let’s take the roof apart and look at the individual components. The isometric view in
Figure 10-21 shows the common rafters and ridge. Notice there are 6 common rafters on
each side of the ridge and two parallel to the ridge.
Figure 10-22 shows the hip rafters cutting diagonally across the square and rising to the
ridge board. We will come back to this later. Figure 10-23 shows the hip rafters and
common rafters together.
Finally in Figure 10-24 jack rafters have been added to complete the roof frame. Notice that
the jacks are simply common rafters that, because of their intersection with the diagonal
hip rafter, get progressively shorter in length.
Hip Rafter
Jack Rafters
Ridge
Ridge
Commons
Plan View of Hip Roof
Figure 10-20
Ridge w/ Common Rafters
Figure 10-21
Ridge w/ Hip Rafters
Figure 10-22
Roof w/Ridge Hip and Common Rafters
Figure 10-23
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C O N S T R U C T I O N
G E O M E T R Y
Framed Hip Roof
Figure 10-24
To better understand the geometry of the hip rafter, take a look at Figure 10-25. I like to
focus on the unit roof square because it has the same properties as any actual sized roof
sharing the same unit rise. There are four elements to this diagram, three triangles and a
square.
The base of the square measures 12  12, representing the unit run of the roof.
The altitude of all three triangles represents the unit rise. The two triangles with 12 bases
represent the common rafter triangles and the triangle with the diagonal as it’s base
represents the hip rafter triangle.
Common Rafter Triangles
Hip Rafter Triangle
Figure 10-25
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C O N S T R U C T I O N
G E O M E T R Y
In figure 10-25 the base of the hip triangle was calculated by applying the Pythagorean
theorem.
12 2  12 2  16.9706 
Armed with the run of the hip triangle and the unit rise of the roof, you can easily calculate
the line length of the hip. And, as we did before we can calculate a constant, which when
multiplied times the roof run (not the hip run) produces the line length of the hip rafter.
CALCULATING THE HIP CONSTANT
Step 1: Calculate the hypotenuse of the hip triangle. Let’s use the hip run that we just
calculated and a rise of 6 inches.
6 2  16.97 2  18"
The unit hip line length for a 6/12 roof is 18
For every 12 of run the hip rafter length is 18
Step 2: Calculate the hip constant by dividing the unit hip line length by the unit run.
18
 1.5
12
Hip Constant = 1.5
Step 3: Multiply the hip constant times the total run to get the total run of the hip.
Homework:
Add the hip constants for 3/12 through 14/12 roof slopes to your Reference Manual.
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C O N S T R U C T I O N
G E O M E T R Y
Let’s work an actual example :
Figure 10-26 illustrates a plan view of a hip roof with slope of 5/12. To simplify the
problem the roof has no roof overhang.
12
5
Rafters 24 OC
6-0
12-0
6-0
6-0
10 0
22-0
6-0
6-0
FIGURE 10-26
6-0
Unit Square
First lets look at what we need to calculate.




Ridge Length
Common Rafter Length
Hip Rafter Length
Jack Rafter Length
Step-by-Step Procedure
Step 1: Calculate the ridge length. If each corner of the structure is thought of as a square
with sides measuring one-half the span, the ridge measurement would measure 10- 0.
Ridge length = Building length - Span
Span = 12-0
Total Run 12  2 = 6-0
Sum of total run at each end = 6-0+ 6-0 = 12-0
Ridge length = 22-0 - 12 0 = 10-0
Ridge length = 10- 0
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Step 2: Calculate the rise constant and the line length constant and solve for total
rise and the line length of the common rafters. (Look them up in your reference
manual)
Rise constant 5/12 = .4167
Total rise 6-0  .4167 = 2-5
Line Length Constant 13/12 = 1.0833
Line Length of common rafters 6-0  1.0833 = 6.72 = 6-6
Step 3: Calculate the hip rafter constant.
52 + 16.972 = 17.69
17.69/12 = 1.4743
Hip Constant = 1.4743
Step 4: Multiply the hip constant times the total run to get the hip rafter line
length
1.4743  6-0 =
8-10 1/8
Line Length of Hip Rafter = 8- 10 1/8
Step 5: Solve for the common difference deducted from jack rafter lengths
As stated earlier jack rafters are simply shortened common rafters, shortened because of their
intersection with the hip rafter. Look at Figure 10-27. As we have studied, rafters are repetitive
members usually spaces 16 to 24 inches on-center. Therefore, as a jack rafter shortens, it
shortens the same amount each time. The amount the jack rafter is shortened is called the
common difference.
The Common Difference measurement is the hypotenuse of a triangle with a base
equaling the on-center measurement and an altitude equaling the proportional total rise
(rise factor  on-center measurement = total rise). In other words, to find the common
difference in jack rafters, you multiply the on-center rafter spacing times the common rafter line
length constant.
Figure 10-27
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G E O M E T R Y
CALCULATING THE COMMON DIFFERENCE:
Determine the roof rise and the on-center measurement
Unit Rise = 5 inches
On-Center = 24
Multiply the on-center measurement times the common rafter line length
constant.
2 1.0833 = 2-1 1/16 (2 = 24)
Common Difference = 2-1 1/16
NOTE: If you multiply the constant times 16 or 24 OC the product will be in inches. It may be
easier to convert the on-center measurement to feet before you multiply times the
constant.
Step 6: Determine the lengths of your jack rafters.
This step is easy because you already have the common difference, all that is necessary
now is to subtract it from the common rafter length and you have the first jack rafter.
Next you subtract the common difference from the first jack length to get the second jack.
You will continue this until you have all the jack rafters for the total run.
Common Rafter
Common Diff.
Jack #1
6- 6
- 2- 1 1/16
4 - 4 15/16
Jack #1
Common Diff.
Jack #2
4 - 4 15/16
- 2- 1 1/16
2-3 7/8
We have now calculated all of the roof members we set out to solve. It
is now your turn to run through the calculations. Use the problem just
solved as a model for your calculations.
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21. Calculate the following dimensions of the framing members in the roof illustrated below:
Note: The number of rafters shown in the illustration is not necessarily the actual number that would be
present in a roof with the given dimensions. Your calculations will not be affected by this inconsistency.
Rafters 24 o.c.
12
8
15-0
25-0
Total run ___________
Total rise _____________
Ridge length ___________
L. L. of common rafters _____________
L.L. of hip rafters ______________
Common difference of jack rafters ______________
Jack 1 _______________
Jack 2 _______________
Jack 3 _______________
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G E O M E T R Y
22. Calculate the following dimensions of the framing members in the roof illustrated below:
Note: The number of rafters shown in the illustration is not necessarily the actual number that would be
present in a roof with the given dimensions. Your calculations will not be affected by this inconsistency.
Rafters 16 o.c.
12
10
30-5
89- 0
Total run ___________
Total rise _____________
Ridge length ___________
L. L. of common rafters _____________
L.L. of hip rafters ______________
Common difference of jack rafters ______________
Jack 1 _______________
Jack 2 _______________
Jack 3 _______________
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G E O M E T R Y
ROOF OVERHANG
Until now we have been calculating the theoretical rafter lengths measured from the
exterior wall to the center-line of the ridge but the next problem will ask you to calculate
the total length including the overhang. As you will remember the overhang is measured
horizontally from the outside of the exterior wall to the end of the common rafters.
There are two ways to approach this problem. The easiest way would be to add the
overhang to the total run and multiply times the line length factors. Another way would be
to multiply your overhang (total run in this case) times the line length constant and add
this product to the line length you calculated for the total run of the building.
Example: 6/12 slope
Overhang = 24
Total Run = 15-0
Common Difference @ 16 OC 1.118  1.333 = 1-5 7/8
LL Constant for Commons= 1.118
LL Constant for Hips = 1.5023
Total run including overhang 15 + 2 = 17
Total length of commons
17  1.118 = 19-0 1/16
Total length of hips
17  1.5023 = 25-6 1/2
Total Length of Jacks
Jack #1
19-0 1/16 - 1-5 7/8 = 17-6 3/16
Jack #2
17-6 3/16 - 1-5 7/8 = 16- 5
Jack #3
16- 5 - 1-5 7/8 = 14 - 6 7/16
and so on………………………
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Here is a roof-framing situation containing all of the problems we have
covered in this chapter. Have fun!
23. Situation:
 Hip roof
 Span = 19-5 3/16
 Slope = 9/12
 HAP = 3
 Roof over hang = 16
 Repetitive spacing = 16 o.c.
 Common rafter size = 2 X 8
 Ridge = 2 X 10
 Hip Rafters = 2 X 10
FILL IN THE BLANKS:
Total run = ____________
Total rise = ____________
Line length of common rafter ___________ + Overhang = __________
Length of temporary ridge support extending from top plate to bottom of ridge =
______________
Line length of hip _______________+ Overhang = __________
Line length of jack rafter 1
___________ + Overhang = __________
Line Length of jack rafter 2 ____________+ Overhang = __________
Line Length of jack rafter 3 _____________+ Overhang = __________
Line Length of jack rafter 4 _____________+ Overhang = __________
Line Length of jack rafter 5 _____________+ Overhang = __________
Line Length of jack rafter 6 _____________+ Overhang = __________
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24. Situation:
 Hip roof
 Span = 32-7
 Slope = 7/12
 HAP = 4
 Roof over hang = 21
 Repetitive spacing = 24 o.c.
 Common rafter size = 2 X 10
 Ridge = 2 X 14
 Hip Rafters = 2 X 12
FILL IN THE BLANKS:
Total run = ____________
Total rise = ____________
Line length of common rafter ___________ + Overhang = __________
Length of temporary ridge support extending from top plate to bottom of ridge =
______________
Line length of hip _______________+ Overhang = __________
Line length of jack rafter 1
___________ + Overhang = __________
Line Length of jack rafter 2 ____________+ Overhang = __________
Line Length of jack rafter 3 _____________+ Overhang = __________
Line Length of jack rafter 4 _____________+ Overhang = __________
Line Length of jack rafter 5 _____________+ Overhang = __________
Line Length of jack rafter 6 _____________+ Overhang = __________
I hope that you see the beauty of the roof constants with which we
have been working. They are logical and easy to calculate. All you
need is the unit rise and the total run or span and you can calculate
all the framing members of a roof. If the overhang changes or the span
changes all you have to do is multiply the new values times the
appropriate constant.
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C O N S T R U C T I O N
Chapter
T R I G O N O M E T R Y
11
TRIGONOMETRY FOR CONSTRUCTION
DEFINITION - The
study of the relationship between the angles and sides of a triangle.
Trigonometry is a most useful tool in your mathematical toolbox. It’s the kind of tool that
you don’t use often but when you need it, you are glad you made the mental investment.
This chapter will focus on right angle trigonometry, which will allow you to solve for
angles and sides of right triangles. Non-right angle trigonometry, used to solve angle and
side measurements of triangles not containing a right angle will not be covered in this
chapter. Before getting into the use of trig, I want to talk a little bit about right triangles.
We have spent a great deal of time studying triangles. One of the reasons triangles are so
prevalent in architecture is because they are the most stable geometric shape. Think about
the gable end of a house. Once the rafters are fastened to the plates and ridge it can
withstand racking, and heavy snow loads.
How stable is a framed wall with no sheathing? As in Figure 11-1, with enough force you
can rack it to the point where the studs pull loose from the nails holding them in place. It
is not until diagonal bracing is applied and nailed that the wall becomes stable, as in Figure
11-2.
RACK
FORCE
FIGURE 11-1
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C O N S T R U C T I O N
T R I G O N O M E T R Y
FORCE
FIGURE 11-2
In essence the diagonal transforms the wall-end into a right triangle, making it structurally
stable. There are many examples of triangles giving structural stability to what we build,
just look for them.
We have used triangles to square-up walls and foundations and our solutions for these
problems have been predicated on basic right triangle geometry, using the Pythagorean
theorem. But our solutions tell us nothing about the angles of the triangle. What if all we
know is the length of one side and one angle and we want to know the length of another
side? That is where trig comes in.
Let’s look at what we know and don’t know about right triangles.



We know that right triangles have three sides.
We know by definition that one of the interior angles is 90.
We know that the side opposite the 90 angle is the hypotenuse, the longest side
on a right triangle.

We can calculate the sum of the interior angles of any triangle using the formula
(n-2)180, where n = the number of side of a closed polygon. (3 - 2)180 = 180 =
sum of interior angles

Because one angle is 90 and the sum of interior angles is 180, the sum of the
other two angles must total 90. And by this definition the other two angles must
each be less than 90 making them acute angles.
30
90 + 60 + 30 = 180
90
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60
C O N S T R U C T I O N

T R I G O N O M E T R Y
If we know one of the other angles besides the 90 angle, we can calculate the third
angle.
X
180 = 90 + 25 + X
180 = 115 + X
180 - 115 = 65
90
25
Because the sum of the two acute angles must total 90, simply subtract 25 from
90 to ascertain the other angle.

We know the relationship shared by the triangles sides.
a2  b2  c
c
a
b

We know nothing about the relationship between the lengths of the sides and the
interior angles of the right triangle and that is where trigonometry comes in.
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Before going any further, let’s assign letters to the sides and angles of our right
triangle to make it easier for us to communicate.
C
b
a
A
c
B
Figure 11-3
As in Figure 11-3 the angles of the triangle are labeled with capital letters and the sides are
labeled with lower case letters. The important thing to notice is that the side directly across
from the angle has the same letter only in lower case. (See Figure 11-4) The angles can be
assigned any letter and in any order as long as the side opposite is assigned the same letter.
This form of identification is mostly used when solving non-right angle trig problems. As
we begin to solve right angle trig problems, triangle sides will be identified by the capital
letters at their ends. i.e. AB, BC, CA
Note:
You may also see angles described by the letter at the end of each line segment,
such as BAC, where the center letter, represents the vertex of the angle.
A
b
C
c
a
Figure 11-4
119
B
C O N S T R U C T I O N
T R I G O N O M E T R Y
Complete the labeling of the triangles below by adding the lower case letters representing
the sides. Remember it is solely up to the person labeling the triangle to choose the letters
and their position on the triangle.
9.
A
B
C
10.
x
y
z
11.
B
a
c
12.
X
120
y
Z
C O N S T R U C T I O N
T R I G O N O M E T R Y
Trigonometry is of no help unless a certain amount of information about a triangle is
available. Besides knowing that one angle is 90 we must have more information to
determine angles or side lengths. If we want to know one of the acute angles of a triangle
we must know the lengths of at least two sides. If we want to know the length of a side we
must know one of the acute angles and the length of one side.
Here are some typical problems that can be solved using trigonometry.
Example 11-A:
What is the length of side AB?
A
?
10
32
C
Example 11-B:
B
What is angle C?
C
?
12
A
15
B
In Example 11-A, we are given one angle and the length of one side and asked to calculate
the length of one of the other sides. In Example 11-B, we are given the lengths of two
sides and asked to solve for one of the angles.
Before we begin solving problems with trigonometry we must learn another method of
labeling triangles.
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LABELING THE TRIANGLE FOR TRIG
In both examples 11-A and 11-B, we either had information about the sides and needed to
know the angle, or we knew an angle and a side, and needed to know the length of another
side. What is common to both examples is the angle. The sides all have a relative position
to that angle. In Figure 11-5 we are given angle A, which equals 28. To carry out
trigonometry we must understand how the sides of the triangle relate to that angle. Side
CB is opposite angle A. Side AC is next to or adjacent to angle A and side AB is of course
the hypotenuse.
A
28
28
adjacent
C
B
hypotenuse
opposite
Figure 11-5
Figure 11-6 illustrates how the relative position of sides to angle changes, within the same
triangle, when the other acute angle is given. When angle B is used side AC becomes the
opposite side and CB becomes the adjacent.
A
Opposite
C
62
Hypotenuse
62
Adjacent
B
Figure 11-6
Identifying the relative positions of sides to a given angle takes some practice so until you can
simply look at a triangle and see the relationship, draw your own triangle and label it for
reference. Label only what you know and what you need to know.
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T R I G O N O M E T R Y
Follow along with the step-by-step procedure given below.
Step 1:
angle
Identify what you know about the triangle. In the triangle below
B and side AC are given and we are asked to solve for side AB
A
10
?
C
32
B
`Step 2: Label the side you know and the side you want to know as
they relate to the angle.
Hypotenuse
Opposite
32
Note:
Do not label the adjacent side in this problem; it will create a point of confusion.
Label only what you know and what you want to know.
Let’s go through the steps again, using another example.
Step 1: Identify what you know about the triangle. Angle C has
been assigned the letter X until its value is known.
Remember - Always start the labeling by identifying the angle then determine the relative positions of the sides.
C
X
17
A
15
B
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T R I G O N O M E T R Y
Step 2: Based on the position of the angle, label the two given sides.
X
C
Adjacent
A
Opposite
B
Explanation: Side AB is opposite the angle you want to know and side
BC is next to or adjacent to it. Again, notice only the sides that are given
in the problem are labeled.
Note:
Do not label the hypotenuse in this problem, as it will create a point of confusion.
Label only what you know and what you want to know.
13. You want to know the length of side CA. Sketch a similar triangle next to the one below
and label the angle, the side you know and the side you want to know.
B
10
20
C
A
15. You want to know angle B. Sketch a similar triangle next to the one below and label
the angle you want to know and the two sides.
C
12
A
15
B
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C O N S T R U C T I O N
T R I G O N O M E T R Y
16. You want to know the length of side CA. Sketch a similar triangle next to the one below
and label the angle, the side you know and the side you want to know.
C
68
A
25
B
17. You want to know angle A. Sketch a similar triangle next to the one below and label
the angle you want to know and the two sides.
C
8
B
10
A
Make up some problems of your own, just like these, and practice so
you can quickly identify the opposite, adjacent and hypotenuse sides
of any triangle at a glance.
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T R I G O N O M E T R Y
TRIG FUNCTIONS - SINE, COSINE & TANGENT
SINE, COSINE and TANGENT are trigonometric functions each representing a ratio of
one triangle side over another. In each case, dividing one side into the other derives the
function. Once the sine, cosine or tangent is determined and through the magic of the
hand held calculator the angle can be calculated.
Here are the six ratios:
OPPOSITE
SINE of  =
HYPOTENUSE
COSINE of 
ADJACENT
=
TANGENT of  =
HYPOTENUSE
OPPOSITE
ADJACENT
HYPOTENUSE
COSECANT of  =
SECANT of 
OPPOSITE
HYPOTENUSE
=
ADJACENT
COTANGENT of 
ADJACENT
=
OPPOSITE
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T R I G O N O M E T R Y
Here they are again stated verbally:
Dividing the opposite side by the hypotenuse gives you SINE
Dividing the adjacent side by the hypotenuse gives you COSINE
Dividing the opposite side by the adjacent side gives you TANGENT
Dividing the hypotenuse by the opposite side gives you COSECANT
Dividing the hypotenuse by the adjacent side gives you SECANT
Dividing the adjacent side by the opposite side gives you COTANGENT
We will be working with all six ratios however the first three are all
you need to memorize.
HERE ARE TWO EASY WAYS TO REMEMBER THESE RATIOS:
SOH CAH TOA…..When pronounced quickly it sounds like sock-a-toa. Just say it to
yourself and write the ratios down in the margin of your paper.
This one works best for me:
OSCAR
HAS
(OPPOSITE)
(HYPOTENUSE)
A
HAIRY
(ADJACENT)
(HYPOTENUSE)
OLD
ARM
(OPPOSITE)
(ADJACENT)
That is pretty easy to remember and the only other thing you must remember is to pair
SINE with OSCAR HAS, COSINE with A HAIRY, and TANGENT with OLD ARM.
Say “ SIN, COSINE, TANGENT” several times and you will have the order memorized.
Also, noticed they are in the same order on your calculator.
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USING THE CALCULATOR FOR TRIGONOMETRY
Calculating angles from trig functions goes way beyond the scope of this book and so we
will rely on the SIN, COS and TAN keys on our calculators to do the dirty work. Here is
how the trig function keys work on the calculator.
The keys labeled SIN, COS & TAN can be used two ways
1. If an angle is keyed into the calculator followed by pressing SIN, the SINE of
the entered angle will be produced.
2. If the SINE value is entered into the calculator followed by pressing the shift
key and the SIN key, the angle will be produced.
Try this:
Enter 45 into your calculator followed by pressing the TAN key.
Did the calculator produce the number 1?
Note:
If you get a different answer, check the calculator display for the letters GRAD or RAD.
If you see either, consult you manual on how to get out of grads or radian modes.
With the number 1 still on the calculator display, press the shift key (purple shift of the HP48G)
and then TAN and you should get 45 again.
Try converting these values on your calculator:
17. What is the SINE of 34.65? ______________
18. What is the TANGENT of 46.765 _______________
19. Convert the COSINE .4568 to an angle. ______________
20. Convert the SINE .5 to an angle. _____________
21. What is the COSINE of 67.99? _______________
22. Convert the TANGENT 3 to an angle. ______________
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You already know something about trigonometry if you have noticed signs along steep,
sloped mountain highways warning trucks to slow down because of a 6% Down Grade. If
you take away the percentage value, leaving .06, that value represents the tangent of the
slope you are descending. See Figure 11-7 to see how this works.
Slope angle
100
6
Figure 11-7
Percent grade on a highway is simply the ratio of vertical rise in feet, divided by a
horizontal run of 100 feet.
6
 .06
100
.06  100 = 6%
The reason .06 is the tangent of the slope angle is because the 6 side is the side opposite
the slope angle and the 100 side is the side adjacent to the slope angle. Refer back to the
three ratios given at the top of page 123 and notice that tangent is calculated by dividing
the opposite side by the adjacent side. Convert the tangent to an angle and now you know
the angle of the slope.
Angle = 3.4336
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Now it is time to put together everything we have learned and solve some trig problems.
Practice along on the three examples below. Don’t just read the examples but do them so
that you can get a feel for the step-by-step approach.
Example 11-C:
Follow along and do each step as you read.
What is angle A ?
C
6
B
?
A
12
Step 1:
On the side of your work sheet or on anther piece of paper write out the
ratios as follows:
O
SIN * H
O = OPPOSITE
H = HYPOTENUSE
A = ADJACENT
A
COS * H
Abbreviate to simplify
O
TAN * A
NOTE:
The reason SIN, COSINE and TANGENT are placed below the line in the ratios
is to help you determine whether to multiply or divide to obtain the unknown
quantity. Notice that a times sign has been placed between the two quantities
below the line. Multiplying their values gives you the value above the line. If you
know one value below the line and the value above - divide the bottom value into
the top value to obtain the unknown bottom value. If the angle is given in the
problem, convert it to SINE, COSINE or TANGENT before multiplying or
dividing.
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Step 2: Sketch a similar triangle and label the sides and angles. Label only what you
know and want to know.
O
A
X
Step 3: Look at the letters representing the triangle sides (O & A), relative to the
angle and Identify the ratio that includes both. It can’t be SIN because A is
not present in that ratio set. It cannot be COSINE because O is not present.
Therefore, it must be tangent because both O and A are represented.
Step 4: Set up the ratio formula next to the triangle you sketched, and substitute the
values you know, which in this case are 6 for the opposite side and 12 for
the adjacent side.
6
TAN * 12
Step 5: Look at the known values and decide whether you will multiply
or divide. In this case you divide 12 into 6.
6
 .5
12
.5 is the tangent of angle X
Step 6: With .5 entered into your calculator, press shift TAN. Answer = 26.5651
Angle X = 26.5671
Note:
The rise constants that you calculated in chapter 10 are also the tangent values of the roof
slope. If you want to know the roof slope, enter the rise constant and press tangent. In the
example above, the roof slope for a 6/12 roof is 26.5671
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Example11-D: Follow along and do each step as you read.
What is the length of Side AB?
C
35
15
A
B
Step 1: You should already have your ratios set up for the last problem.
O
SIN * H
A
COS * H
O
TAN * A
Step 2: Sketch the triangle and label what you know and want to know.
35
H
O
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Step 3: Identify the ratio containing both O and H, which would be SINE.
Step 4: Set up the ratio formula next to the triangle you sketched, and substitute
the values you know, which in this case are 15 for the hypotenuse and
35 for the angle.
O
.5736 * 15
Step 5: Look at the known values and decide whether you will multiply or divide. In
this case you multiply the sine of the angle times 15.
Step 6: Multiply the SINE of 35 (.5736) times 15.
.5736  15 = 8.6036
Side AB = 8.6036
Let’s do one more problem together and then I will cut you loose on your
own.
Example 11-E: Follow along and do each step as you read.
What is the length of side BC?
C
28
A
22
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C O N S T R U C T I O N
T R I G O N O M E T R Y
Step 1: You should already have your ratios set up for the last problem.
O
SIN * H
A
COS * H
O
TAN * A
Step 2: Sketch the triangle and label what you know and want to know.
H
28
A
Step 3: Identify the ratio containing both A and H, which would be COSINE.
Step 4: Set up the ratio formula next to the triangle you sketched, and substitute
the values you know, which in this case are 22 for the adjacent and
28 for the angle.
22
.8829 * H
Step 5: Look at the known values and decide whether you will multiply or divide. In
this case you divide the sine of the angle into 22.
Step 6: Carry out the division
22
 24.9165
.8829
Side BC = 24.9165
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Now it’s your turn. Solve for the unknown quantities labeled X .
All triangles are right triangles
23.
42
34-0
X
Answer ____________
24.
15
X
100
Answer ____________
25.
46-0
Answer ____________
X
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26.
X
150- 0
70
Answer
____________
27. How tall is the flagpole below?
X
62
Answer ____________
10-0
28.
16-0
7-0
X
Answer ____________
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29.
14-0
X
49
Answer ____________
30.
X
18-0
51
Answer
____________
31.
9
Answer ____________
21
X
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32.
69
36
X
Answer ____________
33. You are framing a house with a 8/12 roof slope. What is the slope angle?
Show work
34. A Staircase has a rise of 7 1/2 and a run of 11. What is the slope angle of the
staircase.
Show work.
35. Statement: You are stationed at point A. You walk 45 , in a straight line, to point
B. You turn 90 to your right and walk 35 to point C.
Question 1: If, you walk a straight line from point C back to point A, how far
would you walk?
Show work
Question2: What is angle BAC?
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MORE ON TRIG
Now that you have mastered the mechanics of performing trig problems, here is look at
the logic behind trigonometry.
The relational values of the trigonometric functions can be seen graphically by building a
model of a circle with a radius of 1. Just as pi is the circumference of a circle with a radius
of 1 unit, the trigonometric functions, SINE, COSINE and TANGNT are proportional to
a radius of one unit. Radius rules in trigonometry!
Here is how the relationship between radius and trig functions works:
Step 1: A 60 angle can be seen at the center of the circle in figure 11-8.
Step 2: In figure 11-9 a vertical line segment has been drawn down from the end of
B, and perpendicular to radius A.
Step 3: In figure 11-10 a tangent line has been extended vertically so that its end
intersects with radius B, which has been extended beyond the circumference
of the circle.
radius B
60
radius A
FIGURE 11-8
FIGURE 11-9
FIGURE 11-10
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Step 4: Three line segments in Figure 11-11 have been highlighted and named,
TANGENT, SINE and COSINE. If you compare the length of each line
segment to the length of the circles radius, what would you predict the
relationships to be?
COSINE looks like it is about half the length of the radius
SINE looks like it is about three quarters the length of the radius
TANGENT looks maybe one and three quarters longer than the radius
Find the actual values on your calculator.
SINE = .8660
COSINE = .5
TANGENT = 1.7321
TANGENT
SINE
COSINE
Figure 11-11
Let’s look at another example:
What do you predict the values of SINE, COSINE, and TANGENT to be for a 45 angle?
SINE = _________
COSINE = ___________
TANGENT = ___________
45
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Check out your predictions on the calculator
SINE and COSINE are the same aren’t they?
One last example:
What do you predict the values of SINE, COSINE, and TANGENT to be for this
30 angle? Ignore the B for now.
SINE = __________
COSINE = ____________
TANGENT = ____________
B
30
Check out your predictions on the calculator.
The values of SINE and TANGENT aren’t that different are they?
Look at the example above and imagine that line B is the hand on a clock.
As you move the clock hand towards 3 o’clock the tangent and sine values decrease
and the cosine value increases. The value of COSINE at 0 is 1 where as the values of
SINE and TANGENT are 0.
As you move the clock hand toward 12 o’clock the COSINE decreases and the SINE and
TANGENT increase. At 90 the COSINE is 0 and the SINE is 1. TANGENT cannot be
calculated because the tangent line is parallel to the clock hand at 90 .
Hopefully this will give you better insight into the meaning and derivation of the trig
functions. Use this model when you answer a trig question on a test. Does your answer
make sense? Draw this model or picture it in your mind and predict what the trig function
should be. If your calculated answer is way off base from your model, you have probably
made a mistake somewhere.
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Chapter
12
PERCENTAGES, DISCOUNTS & MARK UP
As a contractor you will encounter percentages, discounts and mark-ups on a regular basis.
Most suppliers give customers discounts for opening accounts and paying their bills on
time. Discounts are also given for purchasing materials in volume. For example, a
cabinetmaker buying plywood by the unit will pay less than the cabinetmaker buying by
the sheet.
Contractors must mark-up the cost of each job they perform. This mark-up covers
expenses not directly associated with each individual job, such as phone bills, utilities,
insurance, rent, etc. Mark-up also generates profit for the company.
Having a good understanding of how to calculate discounts and mark-ups may make the
difference between making money or loosing money.
PERCENTAGES
Discount rates and mark-up rates are usually expressed as percentages and when
estimating materials, waist factors are often used and expressed as percentages.
What does percent mean? As most words in the English language, percent is made up of
two root words.
Per meaning “for each” and cent meaning hundred (century, centennial, centigrade). The
symbol used to express a percentage is %. Let’s look at an example to bring real meaning
to the word percent.
Example 12-A: A carpet layer must lay carpet in a room measuring 250 SF. To insure
that he has enough carpet he adds in an extra 25 SF. Two hundred
and fifty square feet represents 100% of the amount of floor to be
covered and could be called the WHOLE. The added twenty-five
square feet represents a portion of the whole. We can setup a
proportion, which states that 250 S.F. is to 100% as 25 SF. is to X%.
250 25

100
x
250X = 2500
X = 10
25 SF is 10%
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Another, much easier way to solve this percentage problem is to
simply divide the portion of the whole (25) by the whole (250).
And multiply time 100.
25
 .10  100 = 10%
250
Any fraction, whole number, mixed number, pure decimal or mixed
decimal may be expressed as a percentage.
To express a fraction as a percentage, divide the numerator by the denominator and
multiply by one hundred and display the percent symbol.
1/4 = .25 .25  100 = 25%
To express whole numbers, pure and mixed decimals as a percentage, simply times one
hundred and add the percent symbol.
.6793  100 = 67.93%
Convert the following whole numbers, mixed numbers and fractions into percentages:
1. .4333_____
6.
12.65 _____
2. 5/16_____
7.
8 3/13 _____
3. 7/32 _____
8.
167.90 ____
4. 0.045_____
9.
.0023 _____
5. 1 4/5 _____
10.
15/8
_____
Try this:
12. A carpenter must side 600 S.F. of wall area. Experience tells him that he will need to
add an extra 120 S.F. of material to his order as a waste factor. What percentage did
he add for waist?
Show work below
How do you think the carpenter knew to add 120 SF in the previous problem? He knew
from experience, when he uses that type of siding, 20% extra is needed because of waste.
With that in mind let’s restate the problem.
Example 12-B:
A carpenter must side 600 S.F. of wall area. He will add in a 20% waste
factor. How many square feet of siding will he order?
Setup a proportion: 600SF. is to 100% as X SF is to 20%
600 X

100 20
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100X = 1200
X = 120 SF for waste
600 sq. ft. + 120 sq. ft. = 720 S.F. = Total amount of siding ordered.
An easier way to solve this problem is to change 20% in to hundredths, by moving the
decimal place to the left two places (.20), multiply times 600 and add the product to 600.
600  .20 = 120 600 + 120 = 720 sq. ft.
TIP:
Here is an even faster way to make this calculation. Add one to the pure
decimal (1 + .20 =1.20) and multiply times the whole (600)
1.20 600 = 720
600 is automatically added into the solution when you multiply 600 times one.
Try these:
12. How many square feet of floor felt should be ordered for a job when the area of the
floor is 2,248 square feet and a 15% waist factor is added.
Show work below
Answer _______________
Change the following percentages their decimal equivalents.
13.
23%
_____
18.
99%
_____
14.
4%
_____
19.
7.8 % _____
15.
.87%
_____
20.
.059% _____
16.
23 5/8%_____
21.
9.32% _____
17.
125% _____
22.
174 3/8%_____
23. It is estimated that 12,390 concrete blocks are need to complete a certain job. How
many more blocks are required if a 14% waist factor is used?
Show work below
Answer ____________
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24. You have calculated that 842 board feet of 2 4s are needed to complete an addition
on which you are bidding. If 15% is added for waist how many board feet will you
actually order?
Show work below
Answer _____________
25. If 6 bevel siding is overlapped 1 1/2 what percentage of the siding remains exposed?
Show work below. Drawing a quick sketch may be helpful. Setting up a proportion may also help.
Answer ____________
DISCOUNTS
As stated earlier suppliers often give their valued customers discounts for purchasing materials
in large quantities and for paying their monthly bill in a timely manner. A discount is a sum of
money subtracted from the original price. Discounts are usually referred to as percentages,
i.e. 5% discount or 25% discount. The result of subtracting the discount from the original or list
price is termed, the net cost.
Terminology:
L = List price
R= Discount rate
D= Discount
N= Net cost
Example 12-C: The list price on 12 2x4’s is 690m. If you buy by the unit you receive a
5% discount. What is the discount per thousand?
L  R= D
$690.00  .05 = $34.50
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What is the net cost per thousand board feet?
L - D
=
N
$690.00 - $34.50 = $655.50/M.B.F.
TRY THIS:
26. You have been purchasing oak plywood for $44.00 per sheet. Your sales
representative suggests that you purchase by the unit as a way of saving money. A unit
consists of 20 sheets and costs 7% less per sheet. How much will you pay per sheet when
buying by the unit and how much will the unit cost?
Answer_________________
DOUBLE DISCOUNTS
Many times suppliers offer several discounts for example, one for volume purchasing and
one for paying your bill on time.
Example12-D: You buy a unit of 2x4’s 690M with a unit discount of 5%. You receive
an additional 2% discount for paying cash. How much would you pay for
an 800 bd. ft. unit at the end of the month?
800.00  .690 = $552
$552.00 .05 = $27.60
$552.00 - $27.60 = $524.40
The cost of 800 bd. Ft.
Volume discount
Volume discounted cost.
To calculate the discount for paying cash, use the discounted cost as below.
$524.00 x .02 = $10.48
Subtract the cash discount from the already discounted price
$524.40 – $10.48 = $513.92
Net Cost
TRY THIS:
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27. The list cost for a new table saw is $2500.00. The supplier offers a 12% discount off
of the list price. If you pay cash he offers an additional 2% off. Find the net cost.
Show work below
Answer________________
WORKING BACKWARDS
Example12-E: You purchased a hammer for $26.99. The hammer had been
discounted 15% off of the list price. What was the list price?
Look at what you know:
LR=D
L-D=N
If the list cost times the discount rate is subtracted from the list cost, the
net cost (what you want to know) is obtained as below.
L - (L  R) =N
In order to solve this problem algebraically, one (1) must be placed in front
of list cost (L) because it is unknown, and has no place holder. What you
are actually doing is multiplying the unknown times one. See below.
1L -(1L .15) = $26.99
Next carry out the multiplication within the brackets.
(1L  .15 )= .15L
This simplifies the equation
1L - .15L = $26.99
Subtracting .15L from 1L further simplifies the equation.
1L - .15L = .85L
Now the equation looks like this:
.85L = 26.99
Divide both sides by .85 and you have your answer.
L = 26.99/.85
L = $31.75
SIMPLY STATED:
STEP 1: Subtract the discount rate from 100
STEP 2: Divide the answer in step 1 by 100
147
100 – 15 = 85
85
 .85
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STEP 3: Divide the answer from step 2 into the discounted price.
26.99
 $31.75
.85
Example 12-F: You purchase a Skill saw for the list price of $125.00. The next day you
see it on sale for $118.99. What was the rate of discount?
What do you know from reading the problem?
LR=D
$125.00  R = D
L-D=N
$125.00 - D = $118.99
D can be substituted by the formula (LR)
$125.00 - (125  R) = $118.99
Simplify
$125 - (125R) = 118.99
Solve by subtracting 125 from both sides of the equation. Don’t worry
about the negative signs. When you multiply or divide two negatives your
answer will be positive.
-125R = -6.01
R=
 6.01
 .048
 125
R = .048
.
Change to a percentage by multiplying by 100
48  100 = 4.8%
SIMPLY STATED:
STEP 1. Subtract the net price from the list price: $125.00 – $118.99 = $6.01
STEP 2. Divide the answer by the list price. $6.01/$125.00 = .048
STEP 3. Multiply by 100 and add the percentage symbol .048 100 = 4.8%
NOTE:
The Simply Stated methods work great if you can remember them. If you are in a situation where you do
these kinds of calculations frequently this is the way to go. However, if you don’t, the algebraic method is
something to fall back on. By remembering one formula, L  R = D you can solve any discount problem.
MARK UP
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When a construction company bids on a job an accurate estimate is carried out to
determine the approximate costs. Some costs are directly associated with the job such as
materials, labor, rentals, permits, and waist removal; and are termed direct job costs.
Others are indirectly associated with the job such as rent, utilities, office staff, and sales
commissions and are termed indirect job costs. These indirect job costs are known as the
cost of doing business or overhead.
Markup is added to the direct job costs to pay the indirect job costs and to generate net profit.
Net profit is what is left over after all the direct job cost and indirect job cost are paid. Most
construction companies generate between 2% and 7% net profit. That means if a company does
$500,000.00 in total sales for the year, between $10,000.00 and $35,000.00 in net profit should
be generated.
Here’s How Mark-up is Calculated:
L = List cost (what the job sells for or costs the customer)
C = Your cost (direct job costs)
R = Markup rate (a percentage multiplied times the direct job cost to cover
indirect job cost and net profit).
M = Markup or C  R (the actual dollar amount added to the direct job
costs to cover indirect job costs and net profit)
L=C+M
or
L = C + (C  R)
Example 12-G: You are able to construct an entertainment center for $1,200.00 (labor
and materials).To cover your overhead you use a 30% markup. After adding the markup
what will you sell the entertainment center for?
C  R= M
$1200.00  .30 = $360.00
C + M = L
$1200.00 + 360.00 = $1560.00
Tip: You can carry out this calculation all in one step if you place a one in front of the rate (R). By doing this,
the cost and the markup are automatically added together.
$1200.00  1.30 = $1560.00
Example 12-H: You are selling a bookcase for $399.99. The rate of markup is 40%.
What is your cost? Look at what you know.
L= $399.99
R= .40
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C= ?
L = C + (C  R)
L = 1C + (CR)
or
Substitute what you know into the equation.
$399.99 = 1C + (.40C)
$399.99 = 1.4 C
$399.99/ 1.4 = C
C=$285.70
SIMPLY STATED:
STEP 1: Add 1 to the mark-up rate 1+ .40 = 1.40
STEP 2: Divide the mark-up rate plus one into the list cost
399.99
 $285.70
1.40
Example 12-I: You are selling a custom-made toolbox for $359.99. You
remember the cost being $294.20 but you can’t remember what the
markup was. What was the markup?
L = $359.99
C = $294.20
M= ?
R= ?
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L = C+(C*R)
359.99 = 294 .20 + (294.20  R)
359.99- 294.20 = 294.20R
$65.79 = 294.20R
R
$65.79
 .2236
294.20
Convert .2236 into a percentage by multiplying by one hundred.
R = .2236  100
R = 22.36%
SIMPLY STATED:
Step 1: Subtract the cost from the list price $ 359.99 - $294.20 = $65.79
Step 2: Divide the answer in step one by the cost. $65.79  294.20 = .2236
Step 3: Change into a percentage .2236  100 = 22.36%
NOTE:
The simply stated solutions are easy to calculate if you can remember them. Memorizing formulas works
fine if you use them all the time.
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T H E
H P 4 8 G
C A L C U L A T O R
GETTING STARTED WITH THE HP48G CALCULATOR
This calculator is like no other calculator you have ever used. It has more features and can
do higher math than you will ever need. It is fully programmable so you can write your
own programs. Imagine having all of the mathematical formulas you use for construction
at your fingertips. The calculator prompts you to input what you know and then it
crunches the numbers to give you the answer. You will see the virtues of this feature later
on.
The HP48 can convert feet and inches to decimal feet; it can add and subtract feet and
inches without conversions. When used for surveying the HP48 can add and subtract
degrees minutes and seconds without conversions and much more.
TUTORIAL
Getting Started – Read along and do exactly what is asked. Do not try to second-
guess and jump ahead. Words in caps refer to keypad labels. Open the HP48G Users
Manual to Chapter 1. There are some illustrations that may be helpful in working through
this tutorial.
One of my favorite features on the HP48G is the stack. The stack position is found on the
left side of the view screen and stack data is found on the right. The stack allows you to
save data as you put it into the calculator. The data is always there for you to use in your
next calculation.
There are easier calculators to use but none can automate construction math like the
HP48G. The secret to using it is just that, USING IT! I know it is intimidating at first but
the more you use it the easier it gets.
To turn on the HP48 simply press the ON button at the bottom left of the keypad.
To turn off press the
key directly above the ON key. Notice the word OFF written
above the ON key. It is written in green to guide your attention to the right shift key
directly above.
Turn the calculator on again.
To adjust the display contrast, press and hold down the ON key and press the – key to
lighten or the + key to darken the display.
This is a good time to look over the keypad. We have already discovered the right shift
key, so now look directly above it and you will see the left shift key. Very soon you will
be using these shift keys to move to different layers within the calculator, to access green
and purple key functions and much more. Notice that when you press the
key or the
key an annunciator appears at the top of the screen. Try it. Press
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and
. Notice that when you press a shift key twice the annunciator goes away.
Annunciators are simply visual reminders that you have selected a certain function.
Notice the upper most keys on the calculator. These are called menu keys. Next, notice the
dark rectangular bars directly above the menu keys. These are called menu labels. We will
be accessing default menus and creating our own menu labels very soon. You should also
see the numbers 1,2,3,4 on the left side of the display. This is called the Stack. When you
key in a number and press enter, the number goes onto the stack. You could place 1000 or
more numbers on the stack and have access to them all.
We will come back to the keypad but first let’s enter some numbers onto the stack. Press
123.24. Notice that the numbers are in the bottom left corner of the display. You can edit
the number at this point. For example let’s change the number to 125.25 by pressing the
DROP key four times and keying in 5.25
Now press ENTER and notice that the entry moves to the right bottom corner of the
display. This works much like a computer in that the calculator does not recognize your
entry until you press ENTER and place it on the stack.
Press ENTER again. Notice that the entry in the #1 position on the stack is duplicated.
You could duplicate it as many times as you want. Nice feature! Press ENTER ten
times more.
Next press the UP ARROW (found to the left of the NXT key). Notice the pointer that
appears on the left side of the display at stack level 1. Press the up arrow again. Keep
pressing it until you get to stack level 12. Next, press ON which will bring you back to the
entry mode. We will talk more about the stack levels later.
Let’s get rid of one of our entries by pressing DROP (found written in purple at right
center of key pad). Nothing visible happens on the display but if you arrow up the stack;
you will now find only 11 entries. The DROP key not only lets you edit your entry but
also allows you to remove whatever occupies the #1 stack level. The #1 stack entry was
removed and the entries above moved down. Press ON to return to entry mode.
Now press the key to the left of DROP, labeled DEL (delete). Notice by pressing the
DEL key you loose everything on the stack. You get one chance to recall what you just
deleted by pressing
UNDO (the EVAL key) Cool?
Clear your stack by pressing DEL
One of the greatest features of the HP48G is the SWAP key. SWAP is written in purple
but it can be used without pressing the
key. Here is how it works. Often times you
calculate a number and then need to divide it into another number. So for example, put 25
on your stack (don’t forget to press ENTER). Next put 525 on your stack. If you press
the  key you will divide 25 by 525. However we want to divide 525 by 25. You guessed it,
press the SWAP key and then the  key. Did you get 21 as your answer? Good!
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BACK TO THE KEYPAD
Notice the alphabetical letters printed next to the keys. We can write words on the HP by
pressing the  (Greek symbol for alpha) key, found directly below ENTER.
Notice the alphabetical letters printed next to the keys. We can write words on the HP by
pressing the  (Greek symbol for alpha) key, found directly below ENTER.
Let’s type in a letter but first we must clear the stack. You know how! (DEL)
Next press . Notice the alpha symbol () appearing at the top of the display. This is
simply a reminding that you are in alpha mode. Now press the F key. When you keyed in
the letter the alpha symbol disappeared telling you, the calculator is no longer in alpha
mode. In other words you would have to press  again to display another letter. To display
more than one letter without pressing  every time, press  twice. Do it! After typing
, type in the letters UN. Notice that the letters are all in upper case. If you want to
write in lower case simple press
before each letter you want in lower case. To do this
press DROP twice, then key U N. Fun should now be on stack level one. As a final
note, don’t forget to press  after you have finished entering the word or words on to the
stack. Pressing  a third time takes you out of alpha mode. Finally, press ON to remove
Fun from the display.
Lets make some changes to the way your HP is configured.
SYMBOLIC MODE OR NOT SYMBOLIC MODE THAT IS THE QUESTION.
Press
 ( SPC key to the left of the + key) and look at what was placed in the stack. You will
either see 3.14159265359 or . If the numeric value of pi is displayed your in good shape. If the
 (Pi ) symbol is displayed your calculator is in symbolic mode, which you don’t want.
To get out of symbolic mode do the following:
Press
MODES, find MISC in the menu labels at bottom of display and press
the menu key directly below. Next you should see SYM with a square box next to
it. Press the menu key below SYM and the box will go away. You are now out of
symbolic mode. Press VAR to restore the menu labels. Enter  again and you
should see its numeric value on the stack.
NOTE: For some unexplained reason from time to time your HP48 will get itself into symbolic mode. You enter pi
expecting to see 3.1415…. and in fact you see . Knowing this procedure will get you back into action in
a hurry.
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SETTING THE NUMBER OF DECIMAL PLACES AND MORE
As you saw when you placed the numeric value of  on the stack, the HP48 displays 11
decimal places. If we were studying Quantum Physics we might need numbers displayed to
the one hundred billionth, but for surveying and carpentry four decimal places will serve
us well. Here is how to set the decimal places.
Press
MODES, which will produce a screen showing the calculator modes.
This is a good time to introduce the use of the side to side and the up and down
arrows on your keypad. Notice the dark highlighted area is over Std. Press the
down arrow and notice that the highlighted area or cursor moved to Degrees. If
you press the down arrow again, Rectangular will be highlighted. Press the
down arrow one more time and your cursor is on FM (Fractional Mode). Press
the left arrow to highlight beep. There should be a check next to it. You have
probably heard the beep when you have hit the wrong key. If it annoys you press
the menu key below CHK and the check mark will disappear and so will the
beeping sound. Arrow right to clock. If you want a digital clock on your display
put a check next to clock. We will cover setting the clock later. Do not put a
check next to FM, which stands for fractional mode.
Arrow back up to Std. Change the number format from Standard to Fixed by
pressing the menu key below CHOOS (Choose) and the down arrow to Fixed.
Press the menu key below OK. Notice that Std has been replaced by Fix 0. Arrow
right to place the cursor over the zero and key in the number 4 and press the
menu key below OK.
Angle Measure should be set to Degrees and Coordinate system should be set to
Rectangular.
Press OK.
As a test enter  and see if the decimal places have changed to 4 places.
ADDING SUBTRACTING MULTIPLYING AND DIVIDING W/THE HP48G
As you know from the previous exercises, in order to manipulate a number it must be
keyed in followed by ENTER to place it on the stack. Consequently multiplying 5  25 is
done differently on the HP than on the standard calculator. On a standard calculator you
would enter 5, press times, press 25 then equals.
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On the HP you press 5 then ENTER, then 25, then the times symbol. This is probably
backwards from the way you are accustomed, but if you think about how the HP operates
it makes perfect sense.
Here is the logic – You key in 5 and then press ENTER. The calculator now knows that
you want to do something with 5. Next you key in 25 and press ENTER. The calculator
now knows you want to do something with 5 and 25.
You press the times symbol and now the calculator knows what you want to do with 5 and
25.
The same method of entry applies to addition, subtraction and division.
Try these simple problems on your HP48G
1. 45 + 6 = (51)
Enter this way: 45 ENTER 6 +
2. 293.45  395.59 = (116,085.8855) Enter this way: 293.45 ENTER 395.59 
3. 23.4 + 30.2 = (53.6000)
4. 32.6 – 21.596 = (11.0040)
5. Add 48, 20, 193.93, 928.99, 193.039 = (1383.9590)
Note: With addition you can enter all the numbers on the stack and then press the plus symbol
repeatedly until the numbers are all added. 48 ENTER 20 ENTER ………. and so on. Try it.
6. (((21.51 + 69.87) – 54.44 ) 32.99)  14 = (87.0465) Remember to key in the
numbers and then press the operator (+ -  ) Enter this way: 21.51 ENTER 69.87 +
54.44 – 32.99  14 
7. Try squaring a number. To square 5 (52) press 5 on the keypad, then
X
8. Notice that by pressing the square key the answer is placed on the stack. Now
press
X to get the square root of 25, which is of course 5.
8.
9.
29 2  38 2 
(47.8017) press 29 then
X , 38 then
X ,+,
X
25.4 2  14.4 2
 Did you get .8369?
25
10. Divide the answer from #9 into 1067.98. Remember the SWAP key. Did you
get 1276.0428?
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Note: In problem 9 the calculator has rounded the answer to 4 decimal places. However, if you use the calculated
value for a subsequent calculation the calculator will use 11 decimal places to compute the answer.
SETTING THE CLOCK, DATE AND ALARMS
You have probably already checked clock in the HP Modes menu. If not, press
MODES and arrow down to clock and place a check by it.
SETTING THE CLOCK AND DATE
Press TIME. Arrow down to Set time, date… and press OK. Enter the correct
time by inputting the correct hour and pressing OK, then arrow right to minutes, etc.
Next set the date in the same fashion. Press OK to finish. Check out the date and time
on your display.
SETTING ALARMS
You can set an alarm, for example to remind you that BCT 104 meets at 10:30 a.m. Lets
do it just for fun. If you are out of the TIME mode press
TIME once again and arrow
down to Set alarm, press OK. The cursor should be to the right of message. Type in the
following: Go to Math, You know how to type alpha characters. Note: When typing words
you will need to place a space between them. This can be done by pressing the SPC key, found to the left of
the + key.
After typing in the words press  to get out of alpha mode and press OK. Set the time for
3 minutes past your current time. The date should be the current date.
Press OK and notice the ((.)) symbol at the top of your display. This tells you an alarm has
been set.
Hopefully you heard the alarm and saw the message at the top of the display.
EDITING & REMOVING ALARMS
To remove the alarm get back into the TIME mode and with Browse alarms highlighted
press OK. If you wanted to keep the alarm but change the time press the menu key below
EDIT and make whatever changes you need. To set a new alarm in addition to the one
which you previously created press the menu key below NEW and put in the new alarm.
If you want to remove the alarm press the menu key below PURGE and press OK.
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TRANSFERING DATA TO AND FROM THE HP48G
One of the neatest features of the HP48G is its ability to transfer data and programs to
other HP48G calculators. Look at your calculator, just above where 48G is printed, and
you should see a small triangular pointer. Directly below you will find an infrared sensor
for transferring and receiving data. Here is how it works.
The first thing to do before you begin to transfer or receive data is to make sure you are in the
right directory.
SENDING
If you are transferring data from your calculator to another press
I/O (the 1 key).
You should see a menu showing Send to HP48, Get from HP48 and so forth. With
Send to HP48 highlighted, press OK.
Now you get a new screen showing NAME:
The
calculator wants to know what you want to transfer, so press CHOOSE. A new directory
appears displaying all of the programs in the directory you have selected. Scroll down to
the program you want to transfer and with the program highlighted press OK. If you want
to send more than one program put a check mark next to each program by pressing the 
CHK menu label and then press OK. Now you are ready to send, so press OK.
After you press OK, you will see the word Connecting at the top left portion of your
display. Both calculators should be placed together with the little triangular pointers lined
up. You will see the packets of data being transferred on your display.
RECEIVING
First make sure you are in the right directory. Next press
I/O and select Get from
HP48 and press OK. After you press OK, you will see the word Connecting at the top
left portion of your display. Both calculators should be placed together with the little
triangular pointers lined up. You will see the packets of data being transferred on your
display.
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Index
159
A
Index 1, 1
Index 1, 1
Index 1, 1
T
Index 1, 1
Index 2, 2
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 2, 2
Index 1, 1
Index 1, 1
Index 3, 3
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
B
K
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
L
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 2, 2
Index 2, 2
Index 2, 2
W
Index 1, 1
Index 2, 2
Index 1, 1
C
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 2, 2
Index 2, 2
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
D
M
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 2, 2
E
N
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 2, 2
Index 2, 2
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
G
R
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
S
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
H
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 1, 1
Index 2, 2
160
161
162
163
164