Longitudinal and lateral load transfer

[mach11]

Hi everyone,

I was wondering what kind of lateral and longitudinal load transfers takes place in a formula 1 car.

When I was going through carrol smith, the sample problems, it had a longitudinal load transfer of 10% and lateral load transfer of 30% for a Can M car.

What kind of values are usually the limit?
Also, how does one verify the values obtained? (for a f1 car's specification)

Thanks in advance

[Jersey Tom]

One can verify the values with

(a) A Formula 1 car
(b) Half million dollars of sensors

[mep]

How about calculating it?
You can start with:

Fl=m *a
Fl * h = Fr * b
=> Fr = m * a * h / b

Fl: force longitudinal
m: vehicle mass
a: vehicle acceleration
h: centre of gravity height
b: distance from centre of gravity to rear wheel
Fr: dynamical vertical force on rear wheel due to acceleration


You can do the same for lateral weight transfer.
Sure would be nice to get exact data.

[Ciro Pabón]

As you know, it's easy to estimate the values. You only need to know the center of gravity height and the wheelbase. If you start by guessing COG is around mid engine, and maybe the COG and the center of pressure, which I assume is not far from the nose, are also aligned. I got this:



Weight transfer is 187/1384 in the arbitrary units I used in the drawing; this is around 0.14 times the braking acceleration. If you assume you deccelerate at 2 g, then you are transferring around one third (0.27) of the weight. The lower you can get the COG, the smaller the weight transfer, of course.

I imagine all members know that you can guess the static distribution of weight in a car by observing it when it is lifted. In this picture, the COG seems to be slightly ahead of the roll hoop, for a car with more weight in front.


For example, if I assume the car is inclined 15 degrees toward the front when lifted, I get that the CG is 21 arbitrary units towards the front of the car, as shown by the orange line in the first picture. From there I can estimate the static weight distribution, using the distances from the COG to the front/rear wheels. I leave that as an exercise for members with more time in their hands.

In the lateral direction the COG seems to be in the middle of the car, so you can do the same with any picture, simply copy and paste the picture in your favorite CAD and measure estimated COG height and lateral wheelbase, divide and presto (in absence of a real formula one car, sensors and the such, that is, assuming you are into the predicament JTom foresaw).

The value you'll find is a maximum figure, as teams try to lower COG as they can and ballast must have a large effect.

I also assume that dynamic conditions, with the patches moving under the wheels, the gyroscopic effects and whatnots will make all the mechanical engineers in the forum laugh to their heart content watching this feeble attempt, mumbling things about "wing loads", "flexing chassis", "downforce" and "billion dollars budgets" but hey... if you can do better, I can read.

[DaveW]

My guess would be that the c.g. is unlikely to be higher than the wheel axles, & is roughly fixed longitudinally for 2011. C.p. must, I think, be close to the c.g. normally, but is likely to shoot forward under braking (up to -5gn initially).

[Ciro Pabón]

Well, Dave, in that case the height of the axle is 125 arbitrary units and the transfer per g is 125/1384 which is 0.09 and the weight transfer at 2g would be around 0.18.

Pretty low, if you ask me. I know the cars can give you more than 2 g, let's say 3g.

In the end, my best guess then, using a COG at the height of the wheel axles (which seems pretty low to me, watching the picture) and a braking between 2 and 3g is a weight transfer between 0.27 and and 0.18 of the weight.

So, you transfer between one fifth and one third of the weight in any hard braking, which is between 200 and 120 kilos toward the front of the car. Let's call this 150 kilos and be in peace, while we buy the car to verify the values, following JTom lead.

I'd say that compared with down force, this transfer is pretty small.

An alternative would be to find a picture of a car suspended by any other point different from the hoop, and from the two pictures, the one I found and this new one, deduce the position of COG, for the "confirmation" that mach11 asks.

Fat chance, but in this forum you never know.

[marekk]

I suppose we should add a driver to this car.

[Ciro Pabón]

Well, actually my first guess was that the COG must be around the driver's buttocks (and if you believe Coulthard, this is true, at least for JP Montoya, because "when fat people start exercising that's what happens").