Calculating gear ratios for a drivetrain

  • #36
Wilson123
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It seems your gear ratio question may irrelevant. The optimum ratio has more to do with the end user's preferences and not a formula. If you have the time consider temporarily adapting your lever design to a multi-speed bicycle, grab a chair, head someplace where there are a lot of people and test. Find out what the preferred ratios are and then decide based on your knowledge of Indonesia. Then again, sometimes you just have to make a decision based on your experience. Pursuing a formula to justify or rationalize a gear ratio choice in your situation is to me what would require justification. Please explain what you expect to achieve by calculating gear ratios when it's a fairly straight ratio? With a chain it's just a matter of counting the gear teeth isn't it? It's why I asked about sprocket or pulley. Pulleys would be just slightly more involved for ratios.

I am just trying to work out what the size of each gear should be for a use over mainly rough terrain. Is this possible? Or will the gears just be the same size with the power provided to the wheels being varied by where the user pushes the lever?
 
  • #37
jack action
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Or will the gears just be the same size with the power provided to the wheels being varied by where the user pushes the lever?
By moving his hand on the lever, chances are that the user will still provide the same power. But the torque and rpm output will change.

If the user applies the same force at the same velocity, then by moving his hand down the lever, he's only changing the effective length of the lever. Remember that Torque = Force X Radius and Angular Velocity = Velocity / Radius. If you change the radius, when one goes up, the other one goes down. That is why the power is always constant throughout a gearbox, no matter the ratio:

Power = Force X Velocity = Torque X Angular Velocity.
 
  • #38
Wilson123
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By moving his hand on the lever, chances are that the user will still provide the same power. But the torque and rpm output will change.

If the user applies the same force at the same velocity, then by moving his hand down the lever, he's only changing the effective length of the lever. Remember that Torque = Force X Radius and Angular Velocity = Velocity / Radius. If you change the radius, when one goes up, the other one goes down. That is why the power is always constant throughout a gearbox, no matter the ratio:

Power = Force X Velocity = Torque X Angular Velocity.

Is there even a point in me having two gears then? Besides from locating the lever arm away from wheel axle?

It seems that I am not getting any mechanical advantage at all from having two gears.
 
  • #39
jack action
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Is there even a point in me having two gears then? Besides from locating the lever arm away from wheel axle?

It seems that I am not getting any mechanical advantage at all from having two gears.
Of course you're having mechanical advantage with having 2 gears. Do you need it? That is another question. At one point, you will have to define your objectives in terms of Power In vs Torque/Velocity Out and do some calculations to get the Radius / Gear Ratio combination you need.

Although it is true that moving the arm along the lever will change the torque output, it will also affect the Force / Velocity input of the user's hand. I would be surprised that having an arm bent at 90° vs having a straight arm will produced the same capabilities regarding force and velocity. It's still a mechanical advantage thing, but on a biological point of view.
 
  • #40
Wilson123
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At one point, you will have to define your objectives in terms of Power In vs Torque/Velocity Out and do some calculations to get the Radius / Gear Ratio combination you need.

This is what the initial post was about, although now I'm unsure exactly where I've been helped with this due to the multiple different ideas from different users (I understand that's the point of a forum).

Are you able to clarify how I go about calculating this? Not taking into account any mechanical advantages from a biological point of view.
 
  • #41
jack action
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The torque at the lever (##T_L##) is ##T_L = FL##, where ##F## is the hand force (your 40 N) and ##L## is the lever length (your 580 mm or 0.58 m).

That torque is the same as the torque of the driver wheel ##T_r = T_L##.

The torque of the driven wheel ##T_n## is a ratio of the gear radius: ##T_n = T_r \frac{R_n}{R_r}##. Because it is a ratio, you could use the diameter or the number of teeth instead of the radius (for both of them).

The torque of the driven wheel is the same as the torque of the wheel ##T_w = T_n##.

So, putting it all together: ##T_w = \frac{R_n}{R_r}FL##. The force at the tire-road contact patch will be ##F_w = \frac{T_w}{R_w}## where ##R_w## is the radius of the wheel.

For the angular velocity (AV), it will be similar:

The AV at the lever (##\omega_L##) is ##\omega_L = \frac{v}{L}##, where ##v## is the hand velocity (should be equal to the power of the user divided by the lever force ##F##) and ##L## is the lever length (your 580 mm or 0.58 m).

That AV is the same as the AV of the driver wheel ##\omega_r = \omega_L##.

The AV of the driven wheel ##\omega_n## is a ratio of the gear radius: ##\omega_n = \omega_r \frac{R_r}{R_n}##. Because it is a ratio, you could use the diameter or the number of teeth instead of the radius (for both of them).

The AV of the driven wheel is the same as the AV of the wheel ##\omega_w = \omega_n##.

So, putting it all together: ##\omega_w = \frac{R_r}{R_n}\frac{v}{L}##. The velocity at the tire-road contact patch will be ##v_w = \omega_w R_w## where ##R_w## is the radius of the wheel.

You will notice that ##Fv = F_w v_w##, which means power in = power out.

You should used SI units with those equations (N, m, N.m, m/s, rad/s).
 
  • #42
Wilson123
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Thanks a lot for your help, I have a few questions regarding this method.

1. Am I correct in thinking I determine my own gear ratios to begin with and carry out the equations, deciding if my end result is correct and changing the ratio accordingly if required?

2. For calculating hand velocity, how do I know the power of the user to divide over the lever force (40N)?

3. How can I justify what torque and velocity I am trying to achieve with these equations in order to know when I have the ratio correct?

Besides from them 3 questions, the rest of it makes sense to me.
 
  • #43
jack action
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I think you need to re-read this thread and start playing with numbers in the equations, such that the info sinks in. Some of the questions you asked are already answered in posts #21 and #24.
 
  • #44
Wilson123
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I think you need to re-read this thread and start playing with numbers in the equations, such that the info sinks in. Some of the questions you asked are already answered in posts #21 and #24.

I've had a re-read through of the thread and answered most of my questions, thank you.

I initially did the calculations using a gear ratio of 0.2m diameter for the driven and 0.1m diameter for the driver. I worked it out as getting an output of Fv=301.7N with the method you have shown. Looking at previous posts I only need this output to be 53N, therefore I do not need as much of a difference between my gears and I should recalculate with this in mind... Am I thinking about this correctly?
 
  • #45
jack action
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Without more detailed info about the power input and the road conditions, that seems to be the most reasonable guess (53 N @ 1.4 m/s), at least as a first attempt for a prototype to be tested. But that is just my opinion and I have no expertise in wheel chair design.

Surely 300 N @ 0.25 m/s seems to be way to slow. It will be like trying to pedal in first gear with a 21-speed bicycle: The legs go very fast but the bicycle doesn't.
 
  • #46
Wilson123
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Without more detailed info about the power input and the road conditions

What kind of information would be required to take rough terrain in account?

Surely 300 N @ 0.25 m/s seems to be way to slow

Can you clarify why this would be at 0.25 m/s? I thought I was calculating the output power, and my initial calculation had found the output to be too high at 300N?
 
  • #47
jack action
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I'm sure 0.008 for rolling resistance is too small. How much higher should it be, it is difficult to determine for rough terrain. I doubt it would be below 0.02-0.03. Even then, this represents an average and since the power input will be very inconstant, these variations may have a greater effect. Especially if there is lots of pebbles and other irregularities on the road.

300 N times 0.25 m/s gives 75 W, i.e. the power that a human can easily produce (again, read the information already given and let it sink in). And 0.25 m/s is 0.9 km/h which is very slow. This means that, to go at a more normal pace, the user will have to push and pull the lever very fast (even if it will be extremely easy). With the little information I'm given here (Hand force = 40 N and hand power = 75 W) and the extrapolation I've done (Hand speed = 75 W / 40 N = 1.875 m/s), it is the best I can do for predictions.
 
  • #48
Wilson123
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I took the rolling resistance from the below table taken from this link http://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html

upload_2017-3-17_19-5-10.png


Perhaps a higher one is more suited for rough terrain?

300 N times 0.25 m/s gives 75 W, i.e. the power that a human can easily produce (again, read the information already given and let it sink in). And 0.25 m/s is 0.9 km/h which is very slow. This means that, to go at a more normal pace, the user will have to push and pull the lever very fast (even if it will be extremely easy). With the little information I'm given here (Hand force = 40 N and hand power = 75 W) and the extrapolation I've done (Hand speed = 75 W / 40 N = 1.875 m/s), it is the best I can do for predictions.

Therefore, 75/3000=0.25

I want my answer to be 1.4 (walking speed), so I need to decrease the power out to around 50N? (75/50=1.5)
 

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