Lecture No. 7
Key Dates
All students must complete prescheduling forms (Yellow) ASAP —
Meet Kay Hudspeth or Gladys Giron & Departmental Advisor must Sign off
Preregistration for Winter Quarter —Nov.3-Nov.17
Planning for the Big Project — Nov.2
Homework: 3.2a,c,e, 3.3a,c,g, 3.5 a&c + After Volleyball — Nov. 07
Quiz #3 — November 09
Big Project Due - November 14
October 31 Lecture Topics
EGR111 & 111A — Gladys Giron
— Report Writing Format
— Engineering Approximation
Reports
Don t turn in late
Must be good quality
-Look good
-Good content
Typical Report content
1.
2.
3.
4.
5.
6.
7.
8.
Goal & Objective
Background — result of literature & Internet research
Methodology
- Explain experiment (equipment & procedure)
- Explain data analysis method
Results
- Tables
- Graphs
- Explanation
Conclusions
- Base on results
- Refer to Goals/Objectives
Recommendation
- What would you do if you could do it again?
- What would you do extra?
- Results & Conclusions may lead to obvious recommendation
References
- Do not Plagiarize (refer to course Web Site)
Appendices (must refer to in body of report)
- Sample calculations
- Data
- Other key items not in report
9.
Other Parts of Report
- Title Page
- Table of Contents
- Abstracts
What is the following measurement using the following scale?
0
2
4
Need to use visual interpolation.
Sight
Gauge
Fuel Tank
Capacity = 16 gallons
How many gallons in the fuel tank?
Suppose you are in an airplane & are using a stopwatch to monitor the speed of cars
on Hwy 5 near Magic Mountain. Two spots on the highway are marked out and are
approximately 102.67 feet apart. You are interested in determining if cars below
are exceeding the speed limit of 70MPH. How accurate should your stop watch be?
What is the System?
0 ft.
102.67 ft = D
t = 0 sec
t = ? sec.
What are the inputs & outputs?
D = 102.67 ft
V>70MPH ?
Model
Range on t (sec)
What is(are) the model(s)?
t(secs) < D/Vlim
D = Distance (feet) & Vlim = Limiting Velocity in Feet/Sec
What are the assumptions?
You are exceeding the speed limit if your velocity is > 70 MPH
Distance measurement is accurate to nearest 0.01 feet
What is your answer?
t < D/Vlim = 102.67/(70 Miles/Hr*1Hr/3600 Sec*5280 ft/1 Mile)
t < D/Vlim = 102.67 ft /102.67 ft/sec = 1.0 sec
Therefore if t<1.0 sec then speeding.
Therefore measuring times of the order of 1.0 second.
If speed = 71 MPH then
t = 102.67/(71 Miles/Hr*1Hr/3600 Sec*5280 ft/1 Mile)
t = 102.67/104.13 = 0.98 sec
If speed = 70.5 MPH
t = 102.67/ (70.5 Miles/Hr*1 Hr/ 3600 Sec * 5280 ft/1 Mile)
t = 102.67/103.40 = 0.99 sec
Summary
Speed (MPH)
68.0
69.0
70.0
71
70.5
70.25
Time(sec)
1.029
1.014
1
0.98
0.99
0.996
Acceptable Time Error(sec)
0.01
0.005
0.002
Conclude — As you make traffic laws stricter, then you will need a stopwatch with
greater accuracy.
Significant Figures in Calculations
Some calculators display 7 digits
Some Calculators display 16 or 17 digits
One should not use all of these displayed figures — Use a reasonable number of
significant figures
Rules for Multiplication & Division
The product or quotient should contain the same number of significant digits as are
contained in the number with the fewest significant digits.
Example:
Refer back to previous problem.
How long will it take for a car to travel 102.67 feet if it is traveling 70 MPH?
Note the following exact conversions:
1 hr = 3600 seconds
1 mile = 5280 feet
V = 70 Miles/Hr * 1Hr/3600 Sec*5280 feet/Mile = 102.67 ft/sec
However since 70 MPH has only 2(two) significant figures
V = 1.0 x102 ft/sec (Scientific Notation)
(Assuming both the 7 & the 0 are significant in 70 MPH)
(If only the 7 is significant then the answer is 1.x102 ft/sec)
T(Time sec) = 102.67/(70/3600*5280)
t = 1.00003 seconds
However because of the significant figure rule
t= 1.0 seconds (Only two Significant Figures assuming both 7 & 0 are significant
figures in the speed of 70 MPH)
Significant Figures for Multiplication & Division Analysis with Dimensional
Conversion Factors
Given: t = 6.425 hrs
Find how many seconds this is equivalent to:
T = 6.425 hrs *3600 seconds/hr = 23,130 seconds (4 Sig. Figs)
The conversion factor is exact & does not effect the number of significant figures
You purchase 1.57 kilograms of roast beef in Toronto, Ontario,Canada
How many pounds of roast beef did you purchase?
Conversion factor = 2.2046 lbm/kg (not exact conversion factor)
However:
Wt. of Roast Beef in Pounds = 1.57 Kg * 2.2046 lbm/Kg
= 3.46122 Lbm
= 3.46 Lbm (Only 3 Sig. Figs)
Significant Figures for Addition & Subtraction
Rule
The answer should show significant digits only as far to the right as is seen in the
least precise number in the calculation.
Example of Sig Figs for Addition
1,725.463
189.2
16.73
67.
__________
1998.393
Answer = 1998 since 67. Is the least precise number
Example of Sig. Figs for Subtraction
897.0
-0.0922
________
896.9078
Answer= 896.9
Significant Figure for Combined Operations
Rule: Be careful with intermediate application of Significant Figure Rules
Y = 44.3 /( 771.4 — 771.35)
If apply Sig Figs. to numbers in denominator then
Y = 44.3/ (771.4 — 771.4) = 44.3/(0) (undefined)
Correct Answer
Y = 44.3/0.05 = 886 (only 3 Sig. Figs.)
Rounding
Increase the last digit by 1 if first figure dropped is 5 or greater
Example
43.6547 = 43.655 (rounded to 5th Sig Fig)
43.6544 = 43.654 (rounded to 5th Sig Fig)
End of 10/31/00 Lecture