Torque formula
Junior User
Joined: Nov 2001
Posts: 57
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From: Renton USA
Torque formula
I'm in the process of deciding what cam to use in my 400 engine.A while ago I thought I saw a formula that tells you the amount of torque needed to move a given wieght at a given speed. I think it also factor in the percentage of grade and final drive gear ratio. Any help would be greatly apreciated. Thanks in advance. Mike B.
Junior User
Joined: Nov 2001
Posts: 57
Likes: 0
From: Renton USA
Torque formula
Thanks for the reply. I've seen that post befor, I thought that there was another formula that tells you how much torque in foot puonds is needed to move a certain wieght (like 16,000 lbs. gcvw) up a certain percentage of grade.(like a 6% grade).
Senior User
Joined: Jun 2001
Posts: 212
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Torque formula
Here is a formula that will tell you how much torque in ft-lbs will be necessary at the crankshaft to move a certain amount of weight up a certain grade. Of course, determining the aerodynamic drag, the rolling resistance, and the torque converter multiplication is the hard part. For vehicles with manual transmissions or a torque converter that is locked up you would make the torque converter muliplication equal to 1. If you make the aerodynamic drag and rolling resistance equal to 0 and run the numbers it will tell you how much more torque it will take to pull the grade than it would to run on level ground. I am sure this will all be clear as mud.
tq= (drag+roll+(wgt*(grd/100)))*tcmult*slr/(grat)
where
tq = torque (lb-ft)
drag = aerodynamic drag (lb)
roll = rolling resistance (lb)
wgt = weight (lb)
grd = grade (%)
tcmult = torque converter multiplication (unitless)
slr = static loaded radius of tire (ft.)
grat = overall gear ratio (unitless)(example transmission gear ratio*transfercase gear ratio(where applicable)*rear axle gear ratio.
tq= (drag+roll+(wgt*(grd/100)))*tcmult*slr/(grat)
where
tq = torque (lb-ft)
drag = aerodynamic drag (lb)
roll = rolling resistance (lb)
wgt = weight (lb)
grd = grade (%)
tcmult = torque converter multiplication (unitless)
slr = static loaded radius of tire (ft.)
grat = overall gear ratio (unitless)(example transmission gear ratio*transfercase gear ratio(where applicable)*rear axle gear ratio.
Senior User
Joined: Jun 2001
Posts: 212
Likes: 0
Torque formula
Sorry, made an error in that last post. You need to divide by the torque converter multiplier, not multiply.
3. "RE: Torque formula"
18-Dec-01, 12:29 PM (EST)
Here is a formula that will tell you how much torque in ft-lbs will be necessary at the crankshaft to move a certain amount of weight up a certain grade. Of course, determining the aerodynamic drag, the rolling resistance, and the torque converter multiplication is the hard part. For vehicles with manual transmissions or a torque converter that is locked up you would make the torque converter muliplication equal to 1. If you make the aerodynamic drag and rolling resistance equal to 0 and run the numbers it will tell you how much more torque it will take to pull the grade than it would to run on level ground. I am sure this will all be clear as mud.
tq= (drag+roll+(wgt*(grd/100)))*slr/(grat*tcmult)
where
tq = torque (lb-ft)
drag = aerodynamic drag (lb)
roll = rolling resistance (lb)
wgt = weight (lb)
grd = grade (%)
tcmult = torque converter multiplication (unitless)
slr = static loaded radius of tire (ft.)
grat = overall gear ratio (unitless)(example transmission gear ratio*transfercase gear ratio(where applicable)*rear axle gear ratio.
3. "RE: Torque formula"
18-Dec-01, 12:29 PM (EST)
Here is a formula that will tell you how much torque in ft-lbs will be necessary at the crankshaft to move a certain amount of weight up a certain grade. Of course, determining the aerodynamic drag, the rolling resistance, and the torque converter multiplication is the hard part. For vehicles with manual transmissions or a torque converter that is locked up you would make the torque converter muliplication equal to 1. If you make the aerodynamic drag and rolling resistance equal to 0 and run the numbers it will tell you how much more torque it will take to pull the grade than it would to run on level ground. I am sure this will all be clear as mud.
tq= (drag+roll+(wgt*(grd/100)))*slr/(grat*tcmult)
where
tq = torque (lb-ft)
drag = aerodynamic drag (lb)
roll = rolling resistance (lb)
wgt = weight (lb)
grd = grade (%)
tcmult = torque converter multiplication (unitless)
slr = static loaded radius of tire (ft.)
grat = overall gear ratio (unitless)(example transmission gear ratio*transfercase gear ratio(where applicable)*rear axle gear ratio.
Junior User
Joined: Nov 2001
Posts: 57
Likes: 0
From: Renton USA
Torque formula
Thanks again. It looks like I'm on the right track with the engine build up. Even thuogh I didn't figure in the drag our friction loss(don't know), there should be enough torque in the engine left over to over come these forces.(16,000lb, 6% grade, 1.33' radius of tire, 3.54 final drive ratio, manual trany = needs 360 ft. lb. of torque, engine will have around 460 ft. lb.)
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