0
$\begingroup$

By introducing many new gears and changing the tire size I am altering the torque delivered to a rear bicycle wheel from an electric motor. This then changes the "power where rubber meets pavement". "Forward thrust?" I don't know what the proper name is!

I have googled around, but find myself just plain confused and unable to sort through how to evaluate this by using the information I have - which consists of the ratios and wheel sizes. This web page seems to explain it all, but for the life of me I can't follow it.

Yeah, I guess I'm a dummy.

I do not have an actual measure of the motor's starting torque, yet. Have yet to figure out how to affix my torque wrench to the motor without hurting the motor or myself. Motor torque will be a variable named in the equation, so a value won't be needed at this point.

I need to determine how much MORE torque I am applying to the wheel with the new gear setup versus the original gear setup. This is so I can decide if I have increased the delivered torque enough to accomplish the goal of moving ~180lbs with relative ease compared to the original gearing. The original gearing and wheel size was insufficient for the task and by looking at the percentage or ratio of the increase I should be able to ballpark whether I have the design right to move the weight.

The current ratios were set up so that the max speed of the wheel in high gear at the motor's preferred speed of 2700RPM will be at the legal limit of 22KPH/15MPH. I thought that that should put my starting power in the lowest gear somewhere good enough to get the bike and the weight rolling along - but I cannot figure out how to verify that.

  • Original Gearing/Wheel setup:

    Motor > 5.81 reduction direct mount on a 11.5" diameter wheel (that's actual rolling diameter, it is nominally a 12") .

  • New Gearing/Wheel Setup:

    Motor > 5.81 reduction > 3-Speed transmission reductions > final-drive 1.5 reduction fixed to 20.5" diameter wheel (that's approximate rolling diameter - it is nominally a 21").

    3-Speed gearing ratios are 1.33, 1, and .75 - these are not changeable.

    New gearing combined ratios:

    1st gear - 11.05,

    2nd gear - 6.68,

    3rd gear - 8.86

    all fixed to the same 20" wheel.

If I should be calling torque 'power' when I talk about it at the wheel instead of the motor, please let me know what the distinction is and if I need to make them distinct in the equation.


Again:

I am looking for an equation to use that allows me to vary the 5.81 reduction and the 1.5 final reduction and see the changes in torque/power at the wheel.

At its core the question is:

How much more starting grunt do I get with XXX ratio and Xwheel than I did with AAA ratio and Awheel?

$\endgroup$
1
$\begingroup$

The power concept is that power in equals power out (ignoring friction losses). Power = torque * rpm. Take the original setup first. If the gear reduction is 5.81, then the speed of the $rpm_{motor}=5.81*rpm_{gear}$. Since the power has to be the same, $rpm_{motor}*torque_{motor}=rpm_{gear}*torque_{gear}=5.81*rpm_{gear}*torque_{motor}$. If you cancel out $rpm_{gear}$, then $torque_{gear}=5.81*torque_{motor}$. This is conservation of energy at work and the basis of mechanical advantage: you go 5.81 times slower but you get 5.81 times more torque. To figure out the torque gain for your more complex power train, multiply through all the gear reductions. For example, $5.81*1.33*1.5=11.6$, so the torque increase from the motor to the wheel is 11.6. Since your original torque increase was 5.81, this is double.

$\endgroup$
  • $\begingroup$ OMFG, it's that simple for the torque part of this? Wow, feeling foolish and grateful. Thank you. (That website made it seem way more complex) But how do I account for the new diameter? That must affect it somehow. Meaning if 10 FtLbs moved the 11.5" wheel barely well enough to roll the bike when loaded, then of course 19.97 FtLbs will move that same wheel and weight far better - like you said it's basically doubling it. But the new wheel is 20.5" - so how do I account for that? $\endgroup$ – 111936 Aug 19 '16 at 8:12
  • $\begingroup$ @Nixt - The torque applied to the wheel by the engine does not change with wheel diameter. What does change is the force that the wheel applies to the ground. Torque is $\tau = FL$ for a wheel in contact with the ground (perpendicular contact). That is, the force you get out is $F = \tau/L$. Convert your wheel diameters in inches to the equivalent radii in feet: (11.5/2)*(1/12) = 0.479ft, 0.854ft for the other. For the first scenario, $10ftlb/0.479ft = 21lb$ of force. For the second, $19.97/0.854 = 23.4lb$ of force. $\endgroup$ – Chuck Aug 19 '16 at 13:41
  • $\begingroup$ So, the new gear ratio helped, but doubling the torque and almost doubling the tire diameter means that you're left with basically the same amount of force at the pavement, so you won't notice any performance gains. The acceleration when you start is dictated by $F=ma$, which can be restated as $a=F/m$. So, if you lower the mass you can increase acceleration, or if you raise the force you can increase acceleration. I'll warn that any gains you get in low-end acceleration come out of top-end speed. If you keep a low gearing (+torque) and small tire, you accelerate faster to a lower speed. $\endgroup$ – Chuck Aug 19 '16 at 13:47
0
$\begingroup$

If we define the variables as $$T_m = Torque.at.motor.shaft.in.FtLbs$$ $$R_1 = original.decimal.gear.ratio$$ $$R_2 = new.decimal.gear.ratio$$ $$W_1 = radius.of.original.wheel.in.feet,$$ $$W_2 = radius.of.new.wheel.in.feet$$ $$I = increase.as.percentage$$

Then this equation should show the percent increase from the original ratio and wheel to the new ratio and wheel:

$$I = 100 \frac{(T_mR_1/W_1)-(T_mR_2/W_2)}{T_mR_1/W_1}$$


I answered this myself since the question got some activity then got ignored. Thanks to Janus and Chuck for comments that led to me building the equation. It would be greatly appreciated if anyone would confirm that this is the right way to calculate this based on info in the original question.

$\endgroup$
  • $\begingroup$ You can delete T_m altogether, it has no effect on the answer at all. $\endgroup$ – Fergus Aug 26 '16 at 10:00
  • $\begingroup$ Also, Your problem may be this: "The current ratios were set up so that the max speed of the wheel in high gear at the motor's preferred speed of 2700RPM will be at the legal limit of 22KPH/15MPH." On what basis did you make this decision? eg Have you confirmed the motor power output at that RPM is sufficient to to maintain that speed? $\endgroup$ – Fergus Aug 26 '16 at 10:07
  • $\begingroup$ 1) I thought T was needed to get the value to use to calculate the force. 2) Max speed purely on RPM and diameter of rear wheel. Have no clue if it would do it, but when the motor ran the scooter at orig ratio and 12" wheel it did about that speed on the flat. $\endgroup$ – 111936 Aug 29 '16 at 5:45
  • $\begingroup$ 1) your equation is a percent change in ratio so torque is not relevant which Is why the T's cancel (try change T and see that the answer does not change). 2) ok, sounds reasonable. $\endgroup$ – Fergus Aug 30 '16 at 5:58

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.