


How much Torque need to lift front end?
Does someone have a formula to determine how much torque, trans gears, rear end gears would be needed to get the front wheels off of the ground?
I'm not a mathmagician.

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are there any guidelines at all? a nice way of guesstimating whether a particular setup will get you off the ground?



Why would you want the front to come off the ground? You are just losing precious time.
If you want a showboat, just move the engine back, rear axle forward and ballast the trunk/rear face bar. 


You can come close if you know some details about your car. First you must know where the center of gravity of your car is in relation to the rear wheels. Since torque is force acting on a lever arm, you must be able to generate enough torque so your rear end can lift the weight "W" of your car acting on a lever arm "L" equal to the distance in feet from the rear wheel center to the center of gravity. Once you know that and the gear ratios in your tranny "Rt" and rear end "Rr", you can do an easy calculation. Also, if you are running a torque converter, you gain an approximate 2.5 torque increase "Rc" at the instant you begin to move which is applied in the equation.
Assume, W = 3000# L = 48" = 4ft Rt = 2.75 Rr = 4.56 Rc = 2.5 Torque at the flywheel required is, (L x W) / (Rr x Rt x Rc) = (3000# x 4ft) / (2.75 x 4.56 x 2.5) = 382ft.lb. [ February 18, 2003: Message edited by: willys36@aol.com ]</p> 





that counts for sure
thanks willys36, you are a mathmagician for sure. 


One more item I didn't mention B4, to find the center of gravity of the car all you need to do is weigh the front end "Wf" seperately and the whole car "W". Now, measure the wheelbase "Lc".
Here is how to use this info to determine the length "L" to use in the previous equation, Assume, Lc = 100" = 8.33' Wf = 1800# W = 3000# L = (Wf x Lc) / W = (1800# x 8.33') / 3000# = 5' Remind me and I'll post a picture in my Photo Album tonight when I get home that shows how to weigh your car with a 200# bathroom scale. There are some issues with tires that will determine whether you lift the front end or spin the tires. If you have enough traction, it won't matter what size your tires are, you will pop a wheelie. Since they will be fixed to the ground during a wheelie, they could be 20" or 30", it won't matter because they are essentially a part of the ground and your car will be trying to rotate around the axle that is anchored to the ground. However, if the tires can't generate enough traction to anchor them to the ground, the torque you apply will break the tires loose, they won't be 'part of the ground' anymore and they will spin preventing the front end from raising. Traction "Wt" is a force (like weight) resisting the torque applied to the rear axle and is a function of the coefficient of friction "Fc", weight pressing down on the tire food print "Wr", area of the tire foot print "At", and diameter (radius) of the tire "Rt" which is the lever arm for the traction force. Traction can be calculated if you know the weight of the rear of your car, the exact area of the tire in contact with the pavement, and the coefficient of friction of the tire/pavement combo. the equation is, Wt = Fc x Wr x At This is the force available to resist breaking the tires loose, thus causing a wheelie. If the it is at least Wt = (W x L)/Rt Then you can wheelie. If it is smaller, you will spin the tires. Coefficient of friction is a fraction from 0.0  1.0 and is experimentally determined for two surfaces in contact. Polished steel skate blades on ice have an Fc of near 0.0 (impossible to reach perfect 0.0) while 80 grit sandpaper on 80 grit sandpaper has an Fc of 1.0. I don't remember what racing tires on pavement Fc is but it must be around 0.9. As you can see from the equations, wheelies are enhanced by more weight on the rear tires, center of gravity toward rear wheels, tall tires, wide tires with a lot of tread lying on the ground. Spun tires are result of opposite trends. Both spun tires and wheelies benefit from low gear ratios and high engine torque. Drag racers perform a balancing act of getting as much traction on the rear tires (move weight over rear tires, biggest tires possible, etc.) while keeping the center of gravity far enough forward to prevent wheelies. They want ALL of the torque to be transmitted into the traction force which will propel the car forward as fast as possible. Spun tires and wheelies are both nonos! 


There is another factor that will really help if you have the torque and power and that is your pinion angle. If it is set correctly this is what determines where the car gets it's lift. The pinion angle can be determined with a angle guage, if it is set at about 2 degrees negative(approx.) at launch the drive shaft will be at 0 at launch this will make the car leap out of the hole instead of spinning (if thats a prob.) and will use the straight line of the driveshaft to properly lift the front end. If you need more lift you can go back and set your pinion angle at more of a neg. degree. This is why you really need some aftermarket control arms and some adj. upper control arms, to be able to set the pinion angle. As well as what everyone else has said in these replies.
Hope this gives you a better feel for all the factors involed. Dragrace87 


while i would agree that pinion angle should be as close to zero as possible i cant see that a few degrees of dif. is going to do anything at all to help or hurt wheelieability. of course, i'm still so dazzled by willys brilliance, i cant see anything yet.


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