TQ question for you number crunchers
Thread Starter
Former Vendor
Joined: Nov 2005
Posts: 8,995
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From: Spring Texas
St. Jude Donor '08
TQ question for you number crunchers
out of curiosity....
Assuming you have insanely good rear traction, how much torque would be required to lift the front wheels of a C5 off the ground from a dead stop?
Assuming you have insanely good rear traction, how much torque would be required to lift the front wheels of a C5 off the ground from a dead stop?
Thread Starter
Former Vendor
Joined: Nov 2005
Posts: 8,995
Likes: 2
From: Spring Texas
St. Jude Donor '08
technically, gearing should increase torque and a TC should raise the "catch" speed of the engine. both of these accounted for, there should still be a magic TQ/hp level that give you that "maybe i should get a wheelie bar" feeling
Racer
Joined: Apr 2005
Posts: 256
Likes: 1
From: Delta BC
Somebody had best answer this as it is going to drive me nuts. My numbers just don't seem to work 
.
I would have figured the required torque would be calculated as follows:
Sum of Torques about the rear wheels = 0 (to get the fronts just off the ground would require in incremental increase in torque)
The only torques that there would be are the weight of the front end times the wheel base and assumming that the rear wheels lock, it would be the Torque that the car has at the rear wheels.
So the torque required at the rear wheels would be:
T = Mass of vehicle x .51 (weight dist) * g * wheel base
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
T = 19,343 Nm (14 266 lb*ft) which is obviously wrong. What am I missing?
Anyone know what the height of the COG for the cars is?
Thanks for keeping me up tonight RadioFlyer.
.I would have figured the required torque would be calculated as follows:
Sum of Torques about the rear wheels = 0 (to get the fronts just off the ground would require in incremental increase in torque)
The only torques that there would be are the weight of the front end times the wheel base and assumming that the rear wheels lock, it would be the Torque that the car has at the rear wheels.
So the torque required at the rear wheels would be:
T = Mass of vehicle x .51 (weight dist) * g * wheel base
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
T = 19,343 Nm (14 266 lb*ft) which is obviously wrong. What am I missing?
Anyone know what the height of the COG for the cars is?
Thanks for keeping me up tonight RadioFlyer.
Last edited by Autobot; Jul 11, 2007 at 12:26 AM.
Le Mans Master
Joined: Jul 2002
Posts: 7,370
Likes: 8
From: Slave to the evil empire
Cruise-In V Veteran
St. Jude Donor '04
Don't forget suspension setup, on a pure drag car it's setup to allow weight transfer to the rear for traction which also helps to lift the front.
Thread Starter
Former Vendor
Joined: Nov 2005
Posts: 8,995
Likes: 2
From: Spring Texas
St. Jude Donor '08
Somebody had best answer this as it is going to drive me nuts. My numbers just don't seem to work .
I would have figured the required torque would be calculated as follows:
Sum of Torques about the rear wheels = 0 (to get the fronts just off the ground would require in incremental increase in torque)
The only torques that there would be are the weight of the front end times the wheel base and assumming that the rear wheels lock, it would be the Torque that the car has at the rear wheels.
So the torque required at the rear wheels would be:
T = Mass of vehicle x .51 (weight dist) * g * wheel base
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
T = 19,343 Nm (14 266 lb*ft) which is obviously wrong. What am I missing?
Anyone know what the height of the COG for the cars is?
Thanks for keeping me up tonight RadioFlyer.
.I would have figured the required torque would be calculated as follows:
Sum of Torques about the rear wheels = 0 (to get the fronts just off the ground would require in incremental increase in torque)
The only torques that there would be are the weight of the front end times the wheel base and assumming that the rear wheels lock, it would be the Torque that the car has at the rear wheels.
So the torque required at the rear wheels would be:
T = Mass of vehicle x .51 (weight dist) * g * wheel base
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
T = 19,343 Nm (14 266 lb*ft) which is obviously wrong. What am I missing?
Anyone know what the height of the COG for the cars is?
Thanks for keeping me up tonight RadioFlyer.
I think you're calculating for the total mass of the car. You forgot you you're trying to lift the front part of the car (everyginth in front of rear wheels) and the mass in the rear then acts as a counterbalance.
...oh and you're welcome
Race Director
Joined: Mar 2007
Posts: 18,308
Likes: 21
From: Marlton. Increasing performance one speeding ticket at a time! NJ
Somebody had best answer this as it is going to drive me nuts. My numbers just don't seem to work .
I would have figured the required torque would be calculated as follows:
Sum of Torques about the rear wheels = 0 (to get the fronts just off the ground would require in incremental increase in torque)
The only torques that there would be are the weight of the front end times the wheel base and assumming that the rear wheels lock, it would be the Torque that the car has at the rear wheels.
So the torque required at the rear wheels would be:
T = Mass of vehicle x .51 (weight dist) * g * wheel base
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
T = 19,343 Nm (14 266 lb*ft) which is obviously wrong. What am I missing?
Anyone know what the height of the COG for the cars is?
Thanks for keeping me up tonight RadioFlyer.
.I would have figured the required torque would be calculated as follows:
Sum of Torques about the rear wheels = 0 (to get the fronts just off the ground would require in incremental increase in torque)
The only torques that there would be are the weight of the front end times the wheel base and assumming that the rear wheels lock, it would be the Torque that the car has at the rear wheels.
So the torque required at the rear wheels would be:
T = Mass of vehicle x .51 (weight dist) * g * wheel base
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
T = 19,343 Nm (14 266 lb*ft) which is obviously wrong. What am I missing?
Anyone know what the height of the COG for the cars is?
Thanks for keeping me up tonight RadioFlyer.
But if you run with the numbers:
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
Equivalent mass of a uniformly distributed load = .25 x (1459-150-150-100)=264.75kg
T = 6894 Nm (5085 lb*ft)
Now using an Auto transmission (1st gear 3.06, with a 3.42 gear upgrade) gives:
5085ft*lbs/3.06/3.42 = 486ft*lbs
Too many assumptions, but I bet that's pretty close.
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Thread Starter
Former Vendor
Joined: Nov 2005
Posts: 8,995
Likes: 2
From: Spring Texas
St. Jude Donor '08
Yikes, so many variables, so little time. Some of the weight is behind the rear wheels, this will act as a counter-weight. So estimating 10%, that leaves about 150kg axle-back. Subtract that from the axle-forward weight and that leaves approx 1160kg. Some of the unsprung weight (rear suspension) isn't being lifted, so that goes out of the equation, plus the CG isn't located directly between the wheels. Take another 100kg off the estimate. Last, and most important, the weight distribution isn't concentrated over the front wheels. For just a quick analysis, considering uniform weight distribution, the uniformly distributed weight "acts" like a concentrated load equivalent to 1/4 of the total load, concentrated on the front tires. There's so many other things, like, since the car is rotating, its cg is changing, so weight distribution is shifting, etc.
But if you run with the numbers:
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
Equivalent mass of a uniformly distributed load = .25 x (1459-150-150-100)=264.75kg
T = 6894 Nm (5085 lb*ft)
Now using an Auto transmission (1st gear 3.06, with a 3.42 gear upgrade) gives:
5085ft*lbs/3.06/3.42 = 486ft*lbs
Too many assumptions, but I bet that's pretty close.
But if you run with the numbers:
Mass of car = 1459 kg
g = 9.81 m/s^2
Wheel base = 104.5 in = 2.6543 m
Equivalent mass of a uniformly distributed load = .25 x (1459-150-150-100)=264.75kg
T = 6894 Nm (5085 lb*ft)
Now using an Auto transmission (1st gear 3.06, with a 3.42 gear upgrade) gives:
5085ft*lbs/3.06/3.42 = 486ft*lbs
Too many assumptions, but I bet that's pretty close.
now wasn't that fun