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Centre of gravity of a car

by jamesd2008
Tags: centre, gravity
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jamesd2008
#1
Mar11-09, 11:22 AM
P: 64
Hi, hope someone can help. I need to find the centre of gravity of a car (x,y and Z). I know the length of the car (5.390m) the distance between the axils (3.116m) and the height of the car (1.515m) I also no the loading on the fronat axil (151.5 N) and the rear axil (111.9 N) and the total mass of the car is (263.5 N). Any suggestions?

Thanks
James
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berkeman
#2
Mar11-09, 11:57 AM
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P: 40,658
Quote Quote by jamesd2008 View Post
Hi, hope someone can help. I need to find the centre of gravity of a car (x,y and Z). I know the length of the car (5.390m) the distance between the axils (3.116m) and the height of the car (1.515m) I also no the loading on the fronat axil (151.5 N) and the rear axil (111.9 N) and the total mass of the car is (263.5 N). Any suggestions?

Thanks
James
Welcome to the PF, James. Seems like you need more information in order to get the z part of the CG. What is the application?
jamesd2008
#3
Mar11-09, 12:01 PM
P: 64
Hi, thanks for your reply i also have the width if the axil at 1.602m. I'm trying to calculate the Breaking forces on a double wishbone suspension system. Have all the retardation acceleration information as well. But think i need to know the centre of gravity of the mass car?

Thanks

berkeman
#4
Mar11-09, 12:08 PM
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P: 40,658
Centre of gravity of a car

Can you measure the weight shift with the car tilted in either axis? That will help you figure out the z component of the CG. Don't know any other way to do it easily...
jamesd2008
#5
Mar11-09, 12:10 PM
P: 64
Sorry no all i know is the gross mass of the car and mass distribution over each axil.
nucleus
#6
Mar11-09, 12:11 PM
P: 171
One way that people do is to raise one end of the car. The procedure is explained at this site:
http://www.longacreracing.com/articles/art.asp?ARTID=22
jamesd2008
#7
Mar11-09, 12:15 PM
P: 64
nucleus, thanks for the link very interesting. James
Ranger Mike
#8
Mar12-09, 08:31 AM
Sci Advisor
PF Gold
P: 1,452
Center of Gravity (CG) is defined as that point about which, if the body were suspended from it, all parts of the body would have equilibrium ..i.e.. without tendency to rotate. it is the 3D balance point of the race car. all acceleration forces acting on a body can be considered to act through the CG of that body..we have the cg to be as low as we can get it.

Mass Centroid- is related to Cg ..sort of...ifin we slice the car into a series of sections, like a loaf of bread, each slice would have its own CG...if, in a side view, we draw a line connecting each sliced CG..we have the Mass Centroid Axis..not really a straight line but close to it.
it gives an indication of the distribution of the vehicles mass in the vertical plane.

Roll Center of the suspension is that point , in the traverse plane of the axles, about which sprung mass of that tend of the vehicle will roll under influence of centrifugal force. It is determined by drawing a line from the tire contact patch to a point in space that is the intersection of the upper and lower suspension arm linkage angles.

The CG and RC are located at different points. Usully the CG is many inches above the RC. CG can be change by moving " weight " around on the chassis.. The RC is changed by modifying how the upper and lower control arms mount to the chassis. Correct changes will improve the weight transfer to the rear tires thus improving traction.
lot of other stuff goin on here.. race car math of some is on my post in General Physics post " Race Car Physics" 31 Dec 2008

Finding the center of gravity height can be done in several ways, none of which are accomplished very easily and without some work. Presented here is the easiest method. The center of gravity height is calculated by weighing the car when level and then raising the car at least 10 inches at the rear and weighing the front again. Replace each shock absorber with a solid link to eliminate suspension travel

You need a set of electronic scales. You also need a set of blocks o eight inches tall. 10 inch is better.

First, weigh the car normally, then record the total weight (T) and the total number of pounds on the front tires-nose weight (N).

Jack up the rear of the car..CAREFULLY!! Then carefully place the blocks under the rear scales (you may find it useful to lock your rear axle). Now record your new nose weight (NI). Next, subtract your original nose weight (N) from your new, lifted nose weight (NI) and you will have the difference (Nd).

Figure it out using your calculator…

a. Multiply the nose difference (Nd) x 1660= (A)

b. Multiply the total weight (T) x 8 (the height of the blocks)=(B)

c. Divide your previous numbers, A by B, and this equals (C)

d. Add this number, C, to your axle centerline height or spindle center, and the number you obtain is your vertical center of gravity.






CGH= WB x FWc / TW x Tan of angle


Center of Gravity Height Formula


Definition of Variables

CGH - Center of Gravity Height
WB - Wheelbase (inches)
TW - Total weight
FW1 - Front weight LEVEL
FW2 - Front weight RAISED
FWc - FW2 - FW1 (change in weights)
HT - Height raised (inches)
Adj - Adjacent side (see below)
Tan q - Tangent of angle (see below)
CLF - Left Front tire circumference
CRF - Right Front tire circumference
C - (CLF + CRF) / 2 (average circumference)
r - Axle Height
redargon
#9
Mar12-09, 09:08 AM
P: 348
Quote Quote by jamesd2008 View Post
Hi, hope someone can help. I need to find the centre of gravity of a car (x,y and Z). I know the length of the car (5.390m) the distance between the axils (3.116m) and the height of the car (1.515m) I also no the loading on the fronat axil (151.5 N) and the rear axil (111.9 N) and the total mass of the car is (263.5 N). Any suggestions?

Thanks
James
Thought you might want to check your numbers a little. You have a load of 151.5N on the front axle and 111.9N on the rear axle, which means the car's weight (not mass) is actually 263.4N (a rounding error, but let's take 263.5N as the actual weight). This means that your car, which is 5.4m long, 1.5m high and 1.6m wide only has a mass of about 27kg!

Ok, now once you have the actual forces, do a moment calculation about the centre of mass. ie Load on front axle*distance from front axle to cm=load on rear axle*distance from rear axle to cm. Here you have two unknowns and one equation, so you need another equation: you know that distance from front axle to cm + distance from rear axle to cm = distance between axles. Solve simulataneously and voila, cm position in terms of front or rear axle. It is true that there is not enough info to calculate the z co-ordinate of your cm, actually, you can only calculate the x co-ordinate (and the y co-ordinate, if you assume symmetry about the z axis.)
jamesd2008
#10
Mar13-09, 12:46 AM
P: 64
Thanks for your replies has been of a great help. Thanks for your time James
relish_menace
#11
May24-11, 03:06 PM
P: 1
Hi, is there any way to find out CG height i.e z coordinate without actually raising the vehicle??? please reply
jack action
#12
May24-11, 04:49 PM
P: 536
Quote Quote by relish_menace View Post
Hi, is there any way to find out CG height i.e z coordinate without actually raising the vehicle??? please reply
No. All you can do is get a rough estimate. For example, a rule of thumb for old V8-powered american car was that the CG height was at the height of the camshaft.

You can also use this calculator for a good estimation (look for h/L).
Ranger Mike
#13
May25-11, 02:40 AM
Sci Advisor
PF Gold
P: 1,452
correct..you will be close to actual if you use cam location height
El1iP3S01D
#14
Mar18-12, 11:34 AM
P: 4
Hello everyone, this is my very first post and is related to the topic at hand. This is a virtual car that i'm creating for several simulation such as rfactor and Live for Speed and GTR 2...The length of my car is 170.70 inches, now along the length of car at 32.56 is where the front track of the car starts and at 141.08 inches is where the rear track is located...I've determined the center or gravity of the car by using the calculator suggested by nucleus on post #6. That would be CG is at 16.23 inches and CG of Axle height at 3.53 inches by raising the front axle...

My question is, How do i determine the weight of front and rear tracks and find the Center of Gravity? I'm willing to pay for a video tutorial, if any one is from NYC i'm in Richmond Hill Queens...
jack action
#15
Mar19-12, 09:21 AM
P: 536
Quote Quote by El1iP3S01D View Post
Hello everyone, this is my very first post and is related to the topic at hand. This is a virtual car that i'm creating for several simulation such as rfactor and Live for Speed and GTR 2...The length of my car is 170.70 inches, now along the length of car at 32.56 is where the front track of the car starts and at 141.08 inches is where the rear track is located...I've determined the center or gravity of the car by using the calculator suggested by nucleus on post #6. That would be CG is at 16.23 inches and CG of Axle height at 3.53 inches by raising the front axle...

My question is, How do i determine the weight of front and rear tracks and find the Center of Gravity? I'm willing to pay for a video tutorial, if any one is from NYC i'm in Richmond Hill Queens...
I'm not sure of your question, but I think this will help:

http://www.team.net/TR8/tr8cca/wedge...ef/CG_def.html
El1iP3S01D
#16
Mar19-12, 09:39 AM
P: 4
@jack action, let me see if this will help...
jddj
#17
Jul16-12, 11:56 AM
P: 6
hello i am new to the forum, i came across this thread and have been trying to figure out the derivation to this formula CGH=(Wb x Fwc)/(Tw x Tanθ), but i havent had any luck. could someone please explain where this furmula came from and how to derive it.
Thanks
jack action
#18
Jul16-12, 10:19 PM
P: 536
Quote Quote by jddj View Post
hello i am new to the forum, i came across this thread and have been trying to figure out the derivation to this formula CGH=(Wb x Fwc)/(Tw x Tanθ), but i havent had any luck. could someone please explain where this furmula came from and how to derive it.
Thanks
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The load portion on the front axle when the car is leveled (Wf/W) is Lr/L [= 1 - Lf/L] (why?)

The load portion on the front axle when the rear axle is lifted (Wfu/W) is Lr'/(L cosA) [= (L cosA - Lf')/(L cosA) = 1 - Lf'/(L cosA)]

The load increase on the front axle (DWf) between the 2 states will be Wfu - Wf.

But: Lf' = Lf cosA - H sinA

So:
Wfu/W = 1 - (Lf cosA - H sinA)/(L cosA)
Wfu/W = 1 - Lf/L + H/L tanA
Wfu/W = Wf/W + H/L tanA
Wfu/W - Wf/W = H/L tanA
(Wfu - Wf)/W = H/L tanA
DWf/W = H/L tanA


Or:

H = (L x DWf)/(W x tanA)


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