Center of Gravity
Building SPEED
November 20, 2010
Center of Gravity
Early introduction is Center of Gravity is
the point where the object/figure
balances
Geometry – center of mass of a triangle
is
For a polygonal figure the center of
mass is the barycenter (under some
definitions)
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Center of Gravity
Point at which all of the weight of an
object appears to be concentrated.
If object rotates when thrown, the CoG
is the center of rotation.
When object is suspended so can move
freely, CoG is always directly below the
point of suspension.
An object can be balanced on sharp
point placed directly beneath CoG.
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Other uses
Golf
Gymnastics
Karate, Running, Swimming
Robotics (ASIMO)
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Other uses
Vehicle Rollover
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Computing
From your previous exercise you know that
k
m1r1  m2r2  mk rk
CG 

m1  m2  mk
mr
i 1
k
m
i 1
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i i
i
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Known Locations
CoG of 2 particle system lies on line
connecting them. CoG closer to more
massive object.
CoG of a ring - at center of ring
CoG of solid triangle at centroid
(average of 3 vertices)
The CoG of rectangle - at intersection of
2 diagonals.
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Balance
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CoG height
Front roll center
Roll axis
Rear roll center
Roll moment arm
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Roll Axis
Roll axis is the theoretical line which connects the
front roll center to the rear roll center.
The roll centers are the points along the axes where
the car pivots left and right and up and down when it
corners.
The roll axis is the line about which the car rolls
when you turn.
Where are roll centers located?
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Roll Moment Arm
Cornering forces affect the car at CoG.
The distance between the roll axis and the center of
gravity is called the Roll Moment Arm.
Determines how much weight is transferred in
cornering.
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Roll Moment Arm
B  RCF  A  RCR
MA  CGH 
WB
Roll Moment Arm = MA
CGH = center of gravity height
WB = wheelbase (distance between front and rear
axles)
RCF = front roll center height
RCR = rear roll center height
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Roll Moment Arm
B  RCF  A  RCR
MA  CGH 
WB
Weight of cars not evenly distributed between front
and rear axles
More weight in rear => car handles better
Optimal percentage 52% in rear, 48% in front
In Moment Arm formula A = wheelbase x rear wt %
B = wheelbase x front wt %
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Roll Moment Arm
B  RCF  A  RCR
MA  CGH 
WB
All NASCAR cars have a wheelbase of 110 inches.
Find A and B.
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Roll Moment Arm
B  RCF  A  RCR
MA  CGH 
WB
If the front roll center height is 2.5 inches, the rear
roll center height is 11 inches, and the center of
gravity height is 15 inches, find the length of the
moment arm for the car.
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Roll Moment Arm
Find the angle of depression from the line parallel to
the ground that intersects the center of the front
wheel to the bottom of the moment arm. HINT: the
wheelbase is 110 inches and the center of gravity is
not in the center since 52% of the weight is in the
back of the car and 42% of the weight is in the front
of the car.
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StockCarScience Blog
NASCAR mandates a minimum weight of 1700 lbs
(out of the 3450 lbs total minimum car weight) on
the right-hand side.
Teams like to keep as much weight as possible on
the left-hand side, so we’ll assume that they put
1750 lbs of the car’s weight plus a 150-lb driver on
the left-hand side.
The center of gravity is a little to the left of the car’s
centerline and close to the midpoint of the car
front/back.
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StockCarScience Blog
Josh Browne says that the height of the CoG in the
new car is about “at the driver’s tush”.
That’s a couple of inches higher than it used to be in
the old car.
Why does that matter?
Load transfer.
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1986 Race Car
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2001 Earnhardt Intimidator
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StockCarScience Blog
2002 Ford Taurus
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StockCarScience Blog
2007 Chevy Impala
Car of Tomorrow
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StockCarScience Blog
2007 Chevy Impala
Car of Tomorrow
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StockCarScience Blog
Josh Browne says that the height of the CoG in the
new car is about “at the driver’s tush”.
That’s a couple of inches higher than it used to be in
the old car.
Why does that matter?
Load transfer.
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StockCarScience Blog
Braking creates torque transferring some weight from
rear to front
This means there is more weight on front tires than
on the rear tires when the car is braking
Acceleration causes weight transfer from front to
back
Cornering causes weight to shift from the inside
wheels to the outside wheels.
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StockCarScience Blog
Car’s grip is proportional to how hard the wheels are
being pushed into the track.
Braking you’re transferring weight from the back
wheels to the front => losing grip in rear and gaining
grip front
Accelerating => losing grip in front and gaining grip
rear
Amount of weight that shifts is proportional to how
high off the ground the center of gravity is
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StockCarScience Blog
Weight transferred 
weight of car  acceleration  CG height
track
Acceleration is in g’s.
acceleration  CG height
Fractional weight transferred 
 100
track
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StockCarScience Blog
Fractional weight transferred 
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acceleration  CG height
 100
track
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StockCarScience Blog
1. Appeal to symmetry, that is assume a 3600 lb
race car with the weight equally distributed on
each side
2. Track (or tread width) = distance between the
two wheels. In a NASCAR car, the track must be
between 61-1/4” to 61-1/2”.
3. With a CG height of 15”, at a lateral acceleration
of 1g, the weight transfer leaves you with about
920 lbs on the left-side tires and 2680 lbs on the
right-side tires
Left Turns ONLY
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StockCarScience Blog
1. Raise the CG to 17.5”.
2. Keeping everything the same acceleration and
track, there are 770 lbs on the left-side tires and
2830 lbs on the right-side tires.
3. Lost 150 lbs of grip on the left side just by raising
the CG.
4. You can only go as fast as the tire with the least
amount of grip, so more weight transfer means
less grip.
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StockCarScience Blog
Some drivers have suggested lowering CoG
Would make a big difference in how the cars handle
Height of the CoG determined by mass distribution
To lower CoG, must increase the total mass of the
car (for example by adding mass to the frame rails,
but then the engine has to move a larger mass)
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StockCarScience Blog
Or must move mass from the top of the car to the
bottom without compromising safety
Could make cars wider to decrease the weight
transfer
What would that do to the car’s side force?
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Determining CoG Height
Weigh the car in a known configuration using four
scales – one under each tire.
Assume the car weighs 2500 pounds.
Since weight is distributed 48% in front, the front
scales should read 1200 pounds when the car is
level.
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Determining CoG Height
Raise the rear end of car a fixed height, E
The weight of the car will be redistributed and we
front tires now carry more weight, say1225 pounds.
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Determining CoG Height
Find the angle at which rear end raised.
E is the opposite side and the wheelbase, WB, is the
hypotenuse
E
sin  
WB
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Determining CoG Height
For example, if you raised the car with the wheelbase
of 110″ up to a height of 24″
then sin(α) = 24/110 = 0.21818 and α = 12.6˚.
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Determining CoG Height
WB  D  cot 
CGH 
W
D = angled front weight minus the level front weight
W = total car weight.
Compute height above the ground, add this distance
to the height of the center line of the wheels above
the ground
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Determining CoG Height
WB  D  cot 
CGH 
W
WB = 110
D = 1225 — 1200
W = 2500
α = 12.6˚
WB  D  cot  110  25  cot 
CGH 

 1.1  4.4729  4.92
W
2500
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Determining CoG Height
CGH = 4.92 inches
Ground to center line of the wheels = 12.75 inches
 CoG is 17.67 inches above the ground.
Compare that to the center of gravity height of an
SUV, which will be in the 30 inch range.
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Determining CoG Height
Find the center of gravity height of the following car.
The wheelbase is 109 inches, level front weight is
1230 pounds, the angled front weight is 1310
pounds, the amount you elevated the rear is 27
inches, and the total weight is 2500 pounds. Assume
that you have 12 ¼ inches from the ground to the
centerline of the wheels.
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