Finding Center of Gravity 1

[EMc]

Physics questions

Can anbody give me a formula by which to find a car's center of gravity?

Also a formula to calculate lateral grip? I know this one depends in part on tires, but I would be happy with an approximate figure.

A web site where I can go for all such math problems would be best of all (center of gravity, lateral grip, braking distance, ramp angles, etc.)

Thanks much

 

[GregLocock]

With respect to the centre of gravity you need to specify what information you have.

I do it for race cars by working out the effect of every single component in the car and adding it up. I don't think that will work for you.

Heightwise (which I suspect is what you are after) it's about 500 mm off the ground for a small car, a little more for a big one.

Lengthwise you just need to consider the axle weights, statics then gives it to you.

Lateral grip is 0.7 g for bad cars on bad tyres, 0.9 g for good cars on good tyres and what 1.1 g ??? for well set up street cars on sticky tyres.

Cheers

Greg Locock

 

[EMc]

Thanx Greg!

I was after height, I should have specified that.

A friend was telling me about a process for figuring the weight of the car at regular distances along it's length, and converting that data into a line graph, then finding the center of each basic shape inside the graph. He finally said he would simply loan me his book on statics.

I can see some of the work you guys go through, and I have to give you credit for having more patience than me!

Thanks again.

 

[Zpeedstr]

FINDING THE HEIGHT OF THE CENTER OF GRAVITY BY MEASURING WEIGHT CHANGE WHEN TILTING THE VEHICLE:

-----------------------------------M
------------------------------------\-----------------Q'
-------------------------------------\------------------\
--------------------------------------M-----------------\
--------------------------\--------c--|---------------h--\
----------------------ang X----------|-------------------\
O------------------------------------C---------------------Q
[--------------------- a ------------]
[----------------------------------- b --------------------]

b = wheelbase or track O-Q from axel to axel
a = length to center of gravity O-C when level
X = angle of tilt ang QOQ'
h = height of tilt vertically at Q'
c = height of center of gravity above the axel line
M = total weight
W = weight at O

1. W = M (b-a)/b = M (1-a/b)
when tilted through angle X, Q is raised by h and more weight is transfered to O
increasing W to W'
2. b' = b CosX
3. a' = a CosX - c SinX
substitute 2 & 3 in 1
4. W' = M (1 - (aCosX - cSinX)/bCosX) = M( 1 - a/b + (cTanX)/b)
rearranging 1 gives
5. a = b(1-W/M)
substitute in 4 gives
6. W' = M( 1 - b(1-W/M)/b + (cTanX)/b = M( W/M + (cTanX)/b)
solving for c
7. c = (W'/M - W/M) b/TanX =
8. bCotX (W' - W)/M =
9. b^2(W'-W)/hM

To get the center of gravity above ground add the axel height above ground to c.

 

[GregLocock]

Note that you must lock the suspension out when doing the above, and repeatability is abysmal.

Cheers

Greg Locock

 

[unclesyd]

You might want to checkout the following website.

http://www.auto-ware.com/software/ccw/ccw.htm