wheel torque at aero limits

[kwklein]

I've been able to determine that with a rear wheel torque of 322ft/lbs my speed limit is 207mph.

All factors being equal:

cross section 5.8385 f^2
cof. of drag 0.6
rolling resistance 0.02
air pres 2124.00424 lbs/f^2
density of air 0.07411 lb-mass/f^3
temp 77

How can I compute what the needed wheel torque is at higher speeds?

Yes all of thee are actual numbers, not a simulation.

 

[kwklein]

maybe i should post this in a different section...

 

[GregLocock]

Maybe. You need the vehicle weight and rolling radius as well.

F=T*rr

F=1/2*rho*Cd*A*v^2+m*g*Crr

are the formulas you need, in consistent units.



Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[kwklein]

thank you Greg,

I do have the other two values.

weight = 762lbs
rolling radious = 12.1"

I'm still not sure where to plug these values into the formula you've given. Any help would be greatly appreciated.

 

[GregLocock]

convert them into consistent metric units


cross section 5.8385 f^2=A
cof. of drag 0.6=Cd
rolling resistance 0.02=Crr
rear wheel torque of 322ft/lbs =T
density of air 0.07411 lb-mass/f^3=rho
weight = 762lbs =m*g
rolling radious = 12.1"=rr
speed limit is 207mph=v

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[SWhit]

Speed must be in ft/sec.

 

[kwklein]

207 mph = 303.6 ft/sec

 

[rb1957]

i got some problems with this thread ...

let's start with the OP ...
torque is ft*lbs, not ft/lbs.
by torque do you meaning braking toque ?
is "rear" wheel important (for some reason not immediately obvious)

and greg's response ...
i think you were beguiled by the OP's lb/ft, shouldn't it be T = F*rr
then F = Cd*(rho/g) ... because of the units 1lbf = 1lbm*g, or 1 lbf = 1slug*1ft/sec2 (don't you just love imperial units).
then why would the aero drag force act on the ground line ?

 

[kwklein]

sorry, yes I did mean ft&lbs

the rear wheel torque statement was just to identify the source of the number so that it's known that powertrain losses have already been subtracted from the flywheel brake torque

 

[rb1957]

i'm sorry ... what are we talking about ?

i thought "rear" as in tail-dragger (the tail wheel at the back of a plane).


i don't think greg's repsonse applies (but i could easily be wrong). greg gave you the aero. drag force for something whistling along at 200+mph, somehow i don't think that's what you've got ... maybe you could explain the set-up a bit more

 

[kwklein]

you're right, I didn't make this completely clear as to the application. This is actualy for a motorcycle. I'm trying to determine how much more torque I need to overcome the aerodymaic drag at a higher ground speed.

currently I'm "stuck" at 303.6 ft/sec with 322 ft*lbs. I'm trying to determine how much ft*lbs I'll need at say 308, 311.67, and 315.33 ft/sec.

 

[rb1957]

vehicle drag is mostly a function of V^2 (assume Cd is constant). vehicle power required is a function of V^3 (D*V). engine power is T*rpm*(2pi/60). so (if i'm right) that'd be D = 1/2*Cd*(rho/g)*V^2*A
and T = D*V/(rpm*(2pi/60))

it's a bit of a different question as to whether this torque can be applied to th eground (without skidding).

 

[GregLocock]

Oops yes sorry T=F*rr

Cheers

Greg Locock

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.