- #1
shott92
- 29
- 0
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
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