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How much Torque need to lift front end?

27K views 22 replies 12 participants last post by  woodz428 
#1 ·
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.
 
#5 ·
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 ]</p>
 
#10 ·
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? :D
 
#12 ·
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 no-nos!
 
#13 ·
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
 
#14 ·
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 wheelie-ability. of course, i'm still so dazzled by willys brilliance, i cant see anything yet.
 
#16 ·
I used the symbol "W" for weight. "W" is the weight of the car, "Wr" is the weight of rear of the car, etc. "Wt" is the force in pounds of the traction of the tu=ires on the pavement. "L" is the symbol for lever arm length, "Lr" is the lever arm length from center of gravity to rear axle. All of the symbold are defined in the text of my posts, takes a little digging to find them sometimes!
 
#17 ·
You cannot predict how a car will react on the starting line.You can have a car that is a monster wheelstander,and if you limit the front end travel by as little as one inch the car will never leave the ground.Aside from weight distribution,there are chassis settings such as instant center.Pinion angle has no effect on how a car launches,it is often confused with instant center,but nothing you can do with the driveshaft will cause the car to lift the front end.The instant center is the point that the rear axle pivots on the frame of the car.On a leaf spring car it is usually considered to be the front spring eye mount,the same goes for a ladder bar car.On a 4 link car,both factory and racing 4 links,it is considered to be the imaginary point where the upper and lower links would intersect at the front most point.
Once you have found the instant center you need to know what it does for your chassis tuning.A shorter instant center will hit the rear tires harder and lift the rear of the car more while a longer one will lift the front of the car more and not apply force to the rear tires as much as a shorter I.C..Lowering the instant center in relation to the center of gravity of the car will cause the rear suspention to both hit the tires and lift the front end less,while raising the instant center will make the suspention more agressive in both respects.
To further complicate all of the above,shock absorber rates cant be used to further tune the way the suspention reacts.If you need to lift the front end on a 4 link car a good setup might be a high instant center that is long,but if you couldnt get the instant center to be long enough,you could stiffen the rebound rate of the rear shocks to stop the rear suspention from raising on launch and the loosen the rebound rate of the front shocks to aid in weight transfer.
Another matter is spring rate,which in the case of a wheelstanding car is nomally regarded as being a very low rate front spring that has a very long free standing height and a very high rate rear spring which will prevent the car from squatting and allow the rear of the car to apply more leverage toward the front of the car.
I guess my point is that the only thing I made clear with this post is that you cannot clearly determine how a car will react on the starting line.
Other factors that you need to consider are what type of wheelstand you are after.Typically,a light car with a high winding low torque engine and tall rear end gears will lend itself well to a chassis tuning that promotes a tall equal wheelstand,where both front tires are evenly off the ground.This is because most of the torque is hapening in the rear axle,not in the engine compartment,so the car wont twist very much as it rises.A heavy car with a high torque engine and milder gears will create a lot of force on the driveshaft and will cause the car to twist on the starting line.
So now that you are all confused,you can give me more info on your car and if it is a common combination I can tell yo what you need to do to put sunshine under the hoops.
 
#20 ·
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. :D 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!! :D :D :D
 
#22 ·
I might add to this discussion that race tires regularly exceed greater than 1.0 coeffiecient of friction.

How? The theory can be distilled down to molecular bonding of the rubber to the asphalt surface.
 
#23 ·
"Give me a fulcrum and a long enough lever and I can move the earth". While torque is the topic a good gear ratio can overcome the lack there of. I had a Morris Minor in the early seventies that was my beater car, 998cc or there abouts with a 5.38 rear gear and it would pull the front wheels about 8-10". Seemed impossible, it was so under powered but raise the r's and drop the hammer and she would pull them right up, won many a $20 bet off guys that thought I was yanking their chain, I'd have them lined up by the curb to watch. About 25 of those was all the tranny could take so I drove it the next several months minus first.
 
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