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- Thread starter JGM
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Trying to determine the loads on an output shaft of a transmission when parked on a grade with different axle ratios. Not for school work. Just general knowledge.Welcome to the PF.

Is this question for schoolwork? What is the context of the question?

- #4

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The wheel torque produced is ##T_w = F_w r##, where ##r## is the tire radius.

The gear ratio reduces the torque seen by the driveshaft, so ##T_d = \frac{T_w}{GR}##, where ##GR## is the axle gear ratio.

##\theta = \arctan\frac{16}{100} = 9.1°##

##F_w = (16000\ lb)\sin9.1° = 2530\ lb##

##T_w = (2530.5\ lb) * (1.583\ ft) = 4006\ lb.ft##

##T_d = \frac{4006\ lb.ft}{7} = 572\ lb.ft##

- #5

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The wheel torque produced is ##T_w = F_w r##, where ##r## is the tire radius.

The gear ratio reduces the torque seen by the driveshaft, so ##T_d = \frac{T_w}{GR}##, where ##GR## is the axle gear ratio.

##\theta = \arctan\frac{16}{100} = 9.1°##

##F_w = (16000\ lb)\sin9.1° = 2530\ lb##

##T_w = (2530.5\ lb) * (1.583\ ft) = 4006\ lb.ft##

##T_d = \frac{4006\ lb.ft}{7} = 572\ lb.ft##

Shouldn't this include the acceleration of gravity?

F_w = 16000 Lbs* 32.174 ft/s2

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Baluncore

Science Advisor

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I used SI units.

Mass = 8000 kg.

Force due to gravity = 8000 * 9.8 = 78400 newton.

Wheel radius is 39” = conveniently 1 metre.

16% grade = 9.09 deg; Sin(9.09°) = 0.158

7:1 axle ratio.

Drive shaft torque = 1m * 78400N * 0.158 / 7 = 1769.5 Nm

1769.5 Nm = 1305.1 ft.lbs

This is quite different to jack action's 572. ft.lb

Mass = 8000 kg.

Force due to gravity = 8000 * 9.8 = 78400 newton.

Wheel radius is 39” = conveniently 1 metre.

16% grade = 9.09 deg; Sin(9.09°) = 0.158

7:1 axle ratio.

Drive shaft torque = 1m * 78400N * 0.158 / 7 = 1769.5 Nm

1769.5 Nm = 1305.1 ft.lbs

This is quite different to jack action's 572. ft.lb

Last edited:

- #8

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That is because we don't use the same numbers:I used SI units.

Mass = 8000 kg.

Force due to gravity = 8000 * 9.8 = 78400 newton.

Wheel radius is 39” = conveniently 1 metre.

16% grade = 9.09 deg; Sin(9.09°) = 0.158

7:1 axle ratio.

Drive shaft torque = 1m * 78400N * 0.158 / 7 = 1769.5 Nm

1769.5 Nm = 1305.1 ft.lbs

This is quite different to jack action's 572. ft.lb

- I use 1 ton = 2000 lb and you use 1 tonne = 1000 kg;
- I use a wheel radius of 19" and you use a wheel radius of 39".

- #9

Baluncore

Science Advisor

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Mass 8 short ton = 7257.5 kg.

Force due to gravity = 7257.5 * 9.8 = 71123.5 newton.

Wheel radius is 19” = inconveniently 0.4826 metre.

16% grade = 9.09 deg; Sin(9.09°) = 0.158

7:1 axle ratio.

Drive shaft torque = 0.4826 * 71123.5 * 0.158 / 7 = 774.75 Nm

774.75 Nm = 571.5 ft.lbs

We agree.

In Australia the standard Ton was a Long Ton = 1016.05 kg = 2240 lbs.

In 1966 that was replaced by the metric Tonne = 1000 kg = 2205 lbs.

Are vehicle weights in the USA always specified in Short Tons = 907.18 kg = 2000 lbs ?

- #10

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https://en.wikipedia.org/wiki/Short_tonAre vehicle weights in the USA always specified in Short Tons = 907.18 kg = 2000 lbs ?

I'm in Canada and a ton has always been 2000 lb around here. But we're with the SI system as well, so this is not a usual unit for us nowadays (unless we have to do business with our neighbor).

As for truck classification, it is done in pounds, so no problem there:

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