Aerodynamics -- pitching to find additional aerodynamic resistance

[shott92]

hi guys
i know the title... (sigh not again) and this may be the case but i was hoping one you you clever guys on here could help me to the most apporpriate answer is not. thanks

basically we know from skydiving and motorsports if the center of aerodynamic pressure is not equal to the center of gravity we get a torque, moment.

im trying to find out though how much additional resistance can be obtain from this,

if braking force is limited to 1388 Newtons, by forward roll of the vehicle, PLUS the aerodynamic resistances, we say at 25m/s this is equal to 79 Newtons, we now have a total of (1388+79=1467) so far simple :D

but let's say this aerodynamic pressure (CoP) acts at a point of 1.05 meters from the floor, while the Center of Gravity (CoG) is only 0.88 meters... firstly i did struggle to work out this torque thinking
(N*m)-(N*m) = N*m (seemed logical)
so i did (79*1.05)-(79*0.88) = 13.43 Nm ??

so now we have this torque i guess acting upon the CoG with a height (CoG_y) of 0.88m and (CoG_x) at 0.82m from the front contact patch along a wheelbase of 1.5m

i know that torque and force at difference but i just feel that there must be a way to get from this torque to find out how much the braking force can be increased before the vehicle will roll... my initial attempt was to try dividing this by the difference in distance then it dawned on me that i was just getting N (obviously)

please if it can't be done could somebody help me figure out how it affects front and rear loads or pitching/roll anything else that can be gain from the above information because i have spent so long getting this far to waste it all now would be a heart braker ;) lol

thanks in advance guys

 

[Tom_K]

Weight transfer always occurs acceleration and in particular during braking. The forces involved in weight transfer during braking are orders of magnitude greater than the force of aerodynamic drag, making aero drag during braking an insignificant consideration.
To minimize weight transfer the car should have the cog as low as possible and the front shocks need to be adjusted to prevent the bottom of the car from scraping on the road.
There is just too much to cover in one post but you can find a lot of information here.
May I suggest this post be moved to the automotive engineering forum where it will no doubt receive better attention?

 

[shott92]

hi, it was the assumption that the weight transfer is what limits the braking to 1388N i am trying to get the most accurate results possible however there is not much i can find out there equation wise regarding this unfortunatly.
i am happy for this to be moved i wasnt sure where to place this post TBH as in my opinion although featured on a vehicle, i have most of the figures needed, making it a maths question based on Newtonian physics and also regarding engineering (automotive and in general) figured an aerospace engineer would probably have the most knowledge on this topic hence the post here

 

[jack action]

The forces involved in weight transfer during braking are orders of magnitude greater than the force of aerodynamic drag, making aero drag during braking an insignificant consideration.

That is actually not true. The reason why drag (& lift) forces are generally not included into calculations for weight transfer is because the CoP is assumed to be at the same place as the CoG (especially height wise). This really simplifies the equation for braking weight transfer. Expand equation (3b) from this web page to see the math.

 

[shott92]

i understand that it is a common assumption that CoP is considered to be the same height as the CoG and this is my struggle, i found your link rather good has almost everything i have done so far in one place, seems to make the same considerations as else where about the CoP i gather this is considered CL_A here and the co-efficient of lift, as explained in equation 2 the author continues his calcualtion with the assumption of it being = 0