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Torque and gearbox

  1. Jan 22, 2014 #1

    I have a few torque related questions,

    But first, a basic one. Is torque Energy? Because Work=F.d (a force over a distance) measured in Joules (J). But then torque (measured in Newton per meter), sounds like it is also energy, and should also be measured in Joules, but it is not. What am I missing?

    Please take a look at this diagram of a gearbox:


    My confusion emanates from a formula I found online:

    Torque_to_differential = Torque_from_engine * GearRatio

    Where GearRatio is the ratio of the radius of the gear (gears are in blue) to the radius of the engine gear (shown in green here).

    Now I am trying to understand why that is, and so I go back to the basic torque formula:

    T = r x F ---> T= |r| * |F| * sin(theta) (If I am not mistaken)
    But since in the case of a gear box, the torque vector, the r vector and the Force vector are all orthogonal, theta=0, hence:

    |T| = |r| * |F|, therefore |F| = |T|/|r| (if I am not mistaken either....)

    But here is my confusion, first of all, let's say a certain torque T is going through the Layshaft (in red). Does it mean that at each gear on the Layshaft the same amount of torque is communicated? It feels a bit counter intuitive...

    My second questions goes back to the formula "Torque_to_differential = Torque_from_engine * GearRatio". If you look at the green gear that transmits power from the engine, my understanding of the way it communicates that torque to the first gear of the Layshaft is as follows:

    Let's call Te the torque from the engine, and Re the radius of the green gear. I compute the force Fe = Te/Re (is that even right?). That force then is applied to the first gear of the Layshaft (left most red gear, with radius R=2Re, for example), generating a torque T=Fe*R = Te/Re*2Re = 2Te. Simply put, the gear doubles the amount of torque by having twice the radius of the driving gear. Is that correct so far?

    But then things get confusing. I am assuming that the amount of torque Te that went through the first Layshaft gear, is the exact same as the one that goes through the second Layshaft gear (or radius Re/2, let's say). Is that right? Assuming it is, the force generated by that gear F=2Te/(Re/2)=4Te/Re. That force then generates a torque in the first gear of that gearbox (the large left-most blue gear, or radius R=3Re let's say) T=F*R = (4Te/Re)*(3Re)=12Te.... This means that this mechanism, assuming prefect efficiency would yield 12 times the engine torque in first gear... which sound preposterous, and not at all in agreement with the "Torque_to_differential = Torque_from_engine * GearRatio" formula, which states that the torque in first gear should be simply 3 times the engine torque, due to the 3:1 gear ratio (first gear radius R is 3 times the radius of the engine's driving gear).

    So in short, I know I am making a mistake somewhere, and I hope it will jump to you right away in my derivation.

    Finally, on the issue of torque transmission through the layshaft, if I run a troque T through that shaft, how do I know what that torque is at each gear along that shaft? If I assume it is the exact same T at each gear, then it would imply something that seems wrong to me. Namely that the forces F1 at the first gear and F2 at the second one are generated from the same amount of energy.... It would be like putting coins in a bag with two holes at the bottom, and having the exact same number of coins pouring out of both at the same time, effectively multiplying by two the number of coins you put in the bag. Clearly I don't understand what the physics are actually doing.

    Thank you very much for your time and help,
  2. jcsd
  3. Jan 22, 2014 #2


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    The units of torque are Newton-meters, not Newtons/meter. The units of torque are the same as the units of work or energy.
  4. Jan 22, 2014 #3


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    The way to look at how a transmission gear works is this:

    Let's assume for the moment that there are no losses in the gears due to friction.

    If a torque T is applied to a 2:1 reduction gear, remember that the speed of the shaft is reduced by the gearbox.
    Let's say we have an input torque of 100 N-m with a shaft speed of 1000 RPM going into this gearbox. What happens is that the speed of the output shaft will be 1/2 of the input shaft, or 500 RPM. However, the input torque of 100 N-m is multiplied by the gear ratio, so you will get 200 N-m out of the gearbox. The total amount of energy in the shafts is proportional to the product of the torque and the shaft speed, so k*100*1000 = k*200*500.
  5. Jan 22, 2014 #4


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    A torque is the twisting force on a shaft. It is NOT energy, even though it shares the same units.
    The torque applied to a shaft by a lever arm is proportional to both the length of the arm and the force applied.
    The units of torque must therefore be force * length.
    If the shaft does not rotate then no energy is expended and no work is done.

    If the shaft rotates then power is the product of shaft torque and RPM.

    A gear tooth is the end of the effective lever arm from the shaft centre to the contact point on the tooth.
    A gear pair has a ratio determined by the ratio of their lever lengths, pitch radii, or tooth counts.

    So a gear pair is a transformer that converts the ratio of shaft torque to RPM.
    If a gear pair reduces the RPM by a particular ratio then it increases the torque by the same ratio.

    The power, (energy flow), remains the same in both shafts which satisfies conservation of energy.
  6. Jan 23, 2014 #5
    These are all helpful replies, and maybe I am missing the fact that you have already answered it, but I guess I need to refocus the question.


    Simply put, let's say we have a torque Te @ 1000RPM coming into gear A, from the engine. Can you walk me through the math and physics that eventually result in gear D having a torque 3Te (assuming the gear ratio is 3:1)? Because it seems that this is what the formula "Torque_to_differential = Torque_from_engine * GearRatio" suggests.

    Thanks again.
  7. Jan 23, 2014 #6


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    The input torque is Te @ 1000RPM
    The number of teeth on the gear wheels selected sets your speed ratio of 3.
    So the output shaft must be rotating at 1000/3 = 333.3 RPM

    Conservation of energy...
    Power = 1000 RPM * Te = 333.3 RPM * Tout
    Therefore Tout = (1000 / 333.3) * Te = 3 * Te
  8. Jan 23, 2014 #7

    What you describe is, as I understand it, what would happen if gear (A) were DIRECTLY connected to gear (D). Let me walk you through my derivation, I will break it into steps so you can easily call me out on a mistake and tell me precisely where/what it is.

    In the diagram I have attached, I have tagged 4 gears:
    GREEN gear (A): Driving engine gear (radius = Re)
    First RED gear (B): First layshaft gear (1.5:1 gear ratio)
    Second RED gear (C): Second layshaft gear (0.5:1 gear ratio)
    First BLUE gear (D): First gear in the gearbox (2:1 gear ratio)

    Let's assume that a torque Te at rpm RPMe is produced at (A). We can ask, how is that torque (and rpm, ie: power) transmitted to the next gear (B)? Because in getting the power to (D), we go through (B) and (C).

    The derivation goes as follows:

    Step 1:
    The force F generated by Te on (A) is F=Te/Re. This force is applied to (B) and generates the following torque: T = (1.5*Re) * (Te/Re) = 1.5Te. As you mentioned, in order to maintain the same energy flow (Torque*RPM), the RPM will be cut in half. At this point (now is a good time to point me to my mistakes, "quotes" are appreciated) we have at gear (B) a torque T=1.5*Te and an rpm RPM=(1/1.5)*RPMe.

    Step 2:
    Gear (C), being on the same shaft as gear (B) has the same torque (1.5*Te) and rpm ((1/1.5)*RPMe). No work or derivation here, but now is a good time to let me know if that assumption is wrong.

    Step 3:
    This is the final step, Gear (C) will transmit torque to gear (D). The force F generated by 1.5Te on (C) is F=1.5*Te/(0.5*Re) ===> F = (3Te/Re). This force is applied to (D) and generates the following torque: T = (2Re) * (3Te/Re) = 6Te. The rpm will be RPM=1/6RPMe following the "T_in * RPM_in = T_out * RPM_out" formula.

    Well the conclusion is that I am doing something wrong, because assuming perfect efficiency, due to the layshaft (and the two red gears (B) and (C)), I get in first gear (D) 6 times the torque from the engine and 1/6th the RPMs, which is not what I would expect. I would expect to get T = gear_ratio * engine_torque = 2 * engine_torque = 2Te != 6Te.

    As a final note, throughout my derivations, I have been using the following formula: F = Torque/Radius. That is because Torque = R x F = |R| * |F| * sin(theta). R and F being orthogonal, theta = Pi/2, and sin(theta)=1. Hence |Torque| = |R| * |F|, and therefore |F| = |Torque| / |R|. Now is a good time to point out my mistakes...

    Thanks again.
  9. Jan 23, 2014 #8
    Why are you using such a convoluted method?

    T = F X d

    Is an intermediate step, so you can ignore it til you understand what the answer is supposed to be.

    A gear ratio exists between a pair of gears, a gear can't have a ratio with itsself.

    Ratio of the pitch diamaters is the ratio of the teeth so just use that.

    So for ease of calculation.
    Headset input gear teeth = 10
    Headset output gear teeth = 5

    Headset ratio = 1:2
    The torque is reduced by a factor of 2. Speed increases by a factor of 2.

    For the 1st gear ratio
    Input gear = 10 teeth
    Output gear = 30 teeth

    1st gear ratio 3:1
    Torque is 3x. Speed is 1/3.

    The headset output is the 1st gear input as its on the same shaft.

    So engine torque is T.

    The 1st gear output is T*0.5*3=1.5T
    Last edited: Jan 23, 2014
  10. Jan 24, 2014 #9
    Ok that makes sense my mistake was to apply a gear ratio twice, failing to recognize the fact that all gears on the red layshaft transmit the same force to their respective gear, and that force depends on the ratio EngineGear/FirstLayshaftGear.

    Thanks all!
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