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 math-magician.
I'm not a math-magician.
Originally posted by willys36@aol.com:
<strong>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 fly-wheel 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 ]</strong><hr></blockquote>
Ow, ow, ow...what is it HK says? Grape ache!
Does this count for front tire totin'? <a href="http://users.superford.org/ylobronc/melissa/duster800x600.jpg" target="_blank">CLICKY</a> See the light under the tire?
Originally posted by Super Streeter:
<strong>The formulas posted above would be only somewhat applicable in a car with no suspention.Once the car is set upon springs and shocks and all the force of the engine is applied to levers the entire game changes.</strong><hr></blockquote>
Well . . . actually the equations are exaclty right 'cause they are based on physical laws. If the process were static, and the perameter inputs were correct, they would be dead on accurate. Unfortunately, the process is not static, it is very dynamic thus complications come in when your tires squat, springs wrap, loads shift, coefficients of friction change with tire and track temperature, etc., etc. To do a precise analysis, you would need to do a dynamic calculation on a computer that accounted for all the transients over time. All that being said, the equations should do a great job of sizing the components of the system for the purposes here. This is the apporach an engineer would take in intial setup. In fact, every engineering calculation has some error involved since we can never be 100% correct in our measurements. It is not the equation's fault, it is a human inability to gather good information. To account for this, engineers look at every input into an equation and using some very rigorous methods, estimate the error in each input. For example in the equations I showed, tire tread contact area is an input. An engineer would do his best job estimating the length and width of the tread patch. Then he would estimate how far wrong he could be on that measurement. A result might be 10" x 10" = 100sq", +5sq"/-3sq". He would do this for every measurement, combine the errors of each and calculate an 90% error band of something like 356.3ftlb - 383.5ftlb. The 90% error band means the actual torque will be within the calculated range 90% of the time. The higher the percentage error band, the wider spread that is required on the estimated torque to insure the true value falls within the smaller range. On machines like the space shuttle, the error band percentage is extremely high and the allowable tollerance is very tight so engineers must put in a huge effor to insure extreme accuracy in the measurements to achieve the required precision. Conversely, unstable's torque estimate can be very unprecise and still be accurate enough achieve acceptable results. I wager that when his car is standing vertical on it's rear bumper he will not be quibbling whether it did it with 358.234ftlb or 374.665ftlb of torque!!