## Reduction gearing 170kv for 24v

Discussions related to motors other than hub motors.
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### Reduction gearing 170kv for 24v

Hi

I have basically got myself a working RC ebike running a 170kv Turnigy motor at 48v.

The motor runs a 16t belt which drives a 72belt, this in tern runs a 11t sprocket to a 65t sprocket on the rear wheel. This is good for around 25mph which is just perfect for me. (26"wheel)

Now I want to replicate this setup on 2 more bikes but want to try maintaining the 25mph but running the motor at 24v , I have no idea of the gearing needed as I just dont know the math. Maybe if you could try explaining it to me this would be a massive help, Im pretty sure alot of other guys out there have trouble with this aswell.

I bought this setup of a member on this forum which was basically ready to go, thats why I dont know how to calculate the gearing as it was kindly already done for me

Thanks guys
cajunjay
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### Re: Reduction gearing 170kv for 24v

If your motor has linear response (what kv is used for, a term I dislike but I digress).. then at half the voltage the motor will spin half as fast.

Now, your motor is already spinning too fast, you are gearing it to be slower. But at half voltage you are overly-gearing it to be slower, by double. So, what you need to do is cut that reduction in half, somewhere. This is easy, because you just have to undo some of what you did, or do a less extreme version. Make sense so far?

You have a double-stage geardown, if I understand correctly:

1 - 16:72 for the belt (72/16 = 4.5x slower)
2 - 11:65 for the sprocket (65/11 = 6x slower).

Since they're connected to each other output-to-input in the middle, these ratios compound and multiply. Because every 1 turn of the belt output is already 4.5x as slow as it started, and then even 1 turn of *that* sproket gets turned down 6x more. So 4.5 x 6 = 27x geardown. Still with me?

Well, instead of 27x, you want half that. ~14x.

You have 4 geardown components. Little Belt Wheel, Big Belt Wheel, Little Sprocket, Big Sprocket. Any combination of new ratios that would yield 14x is what you're after. You could change only 1 of them, you could change all of them, doesn't matter as long as the above math clears to 14x. Little ones bigger or bigger ones smaller.

Which of them you want to change and by how much, is up to you and what is simplest. Maybe change the least standard component. Maybe the most expensive. Maybe the most finicky. You might have practical minimum or maximum sizes (can't have a small sprocket with only 3 teeth, can't have a big sprocket larger than your whole wheel). It might be easier to adjust chain sizes than belt sizes. Less re-engineering. Etc etc etc. I won't give you answers, you have to think it out yourself. Teach a man to fish and all that.

Generally, since it's multi-stage, you want to split the geardown burden somewhat equally, or you might as well have done it all in one stage anyway.

To simplify, in case you're having trouble understanding this, I'll use several examples. You could make 1 change and do any of the following:

1 - Change Little Belt Wheel to 32 instead of 16. (Ratio becomes 32:72, or 2.25x instead of 4.5x)
2 - Change Big Belt Wheel to 36 instead of 72. (Ratio becomes 16:36, or 2.25x instead of 4.5x [same as above, different method to same result])
3 - Change Little Sprocket to 22 instead of 11. (Ratio becomes 22:65, or 3x instead of 6x)
4 - Change Big Sprocket to 32 instead of 65. (Ratio becomes 32/11, or 3x instead of 6x [same as above]

Again, you could redesign all 4 components and match it up, but, leaving the belt as is at 4.5x and then gearing the sprockets down to only 3x leaves each stage in a close ballpark. Makes a lot more sense than an only 2.25x geardown on the belt and a full 6x on the sprocket.

You could tune it a bit better, you could aim for 3.75x on both belt and chain side. (3.75 x 3.75 = 14x geardown), but, meh, close enough.

Fun trivia: Toothcount, all else equal, can be determined by diameter. A 6" diameter belt wheel will have double the teeth as a 3".. because circumference = pi*diameter and pi is constant. This will help you visualize any changes you make. If you cut the tooth count by half on anything, it shrinks the diameter by half. If you double, it doubles. Or anything in between.

Explained well enough for you to solve your own problems now, or did I lose you somewhere?
Last edited by MattsAwesomeStuff on Fri Mar 30, 2012 7:57 am, edited 1 time in total.
MattsAwesomeStuff
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### Re: Reduction gearing 170kv for 24v

Nicely explained mate

I got it!
cajunjay
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### Re: Reduction gearing 170kv for 24v

MattsAwesomeStuff's explanation of the physical gearing is accurate and well spoken, as far as it goes.

However, because the motor is spinning at half the original speed it is also putting out something around, or less than, half the power...so I am not sure it will put out enough power to give you the performance you once had. Maybe others with more experience can chime in here to address this issue.
My electric vehicle projects Home of the eTownie & eCortina
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Cortina Revisited Turnigy 80x - HV 160 - 48v 15Ah
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### Re: Reduction gearing 170kv for 24v

EDIT: not sure about post below, but I'll remove this so as not to misinform
Last edited by thepronghorn on Sat Mar 31, 2012 2:39 pm, edited 1 time in total.
thepronghorn
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### Re: Reduction gearing 170kv for 24v

thepronghorn wrote:It would be closer to 1/4th the power at half voltage because halving voltage decreases amp draw by the same factor:

1/2 voltage x 1/2 amps = 1/4 watts......

Unless the motor's power was controller limited before the voltage change.

Halving the voltage does not decreases amps by the same factor. It increases the amps, which leads to faster controller failure. (IF you don't change the gearing).
Bergli
Turnigy 80-100 130kv
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### Re: Reduction gearing 170kv for 24v

the current will remain the same ( at the controllers current limit setting ) so if you want the same performance as when it was running at 48v and you half the voltage then you need to double the current. This is the big advantage of running a higher voltage that you need less current for more power.

gwhy!
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### Re: Reduction gearing 170kv for 24v

I see now what you mean, probably gonna stick with 48 volts after all but thanks for teaching me the math and theory behind it.

Learn something new everyday on this forum
cajunjay
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### Re: Reduction gearing 170kv for 24v

ProngHorn's statements are both incorrect and correct. He's just using it the wrong context.

How ProngHorn is wrong:

There is no free lunch. To move an object at a given speed takes a certain amount of energy per second which is called "power", measured in watts or horsepower. Other that slight transmission losses, it doesn't matter what you do to get that power, but that power doesn't change and you need it to move the vehicle at that speed.

Power = Volts * Amps

So, if you cut the volts in half, you have to double the amps to get the same power... and you need the same power to move at the same speed because there is no free lunch.

Volts are something you control or is constant. Amps are usually the result of what you try to force something to do.

For example, if you spin a motor up with no load and measure the amps, they will be small, because the only current is the losses in motor wires and the air resistance of the spinning motor guts against the air. But if you grab onto the shaft and try to slow it down, you'll find very little (if any) change in speed. Instead what happens is the harder you squeeze, the more amps the motor draws to maintain the same speed with the voltage it's given.

Well in the case of a bike it's not your hand slowing down the shaft, it's the tire not wanting to spin because of the bike's air and rolling resistance slowing it down.

So the result of using half the voltage is that current drawn by the motor to maintain speed and hit the power requirement doubles.

That's how ProngHorn is wrong.

---

Here's how ProngHorn is right:

Amps = Volts / Resistance ("Ohm's Law")

Let's plug in a made-up number for resistance of "12 ohms", the value isn't important, just the relationship. And pick any voltage, let's use the real ones our OP has mentioned: 48V and 24V.

Amps = 48 volts / 12 ohms
Amps = 4

Amps = 24 volts / 12 ohms
Amps = 2

Again, the exact numbers don't matter, just the relationship. The formula is true for any numbers you pick.

So, you see, you control amps by controlling voltage. When you cut voltage by half, you also drop the amps by half. The "Power Law" still applies: Power = Volts x Amps. So if volts are 1/2 as large, they make amps 1/2 as large, and thus power is 1/4 what it originally was.

So ProngHorn was right.

So why are these results incompatible?

---

Here's how ProngHorn's context was wrong:

Pronghorn assumed the resistance stayed the same in both circuits. And with the back tire lifted up, he would be correct, his results would hold true.

But unlike most electrical circuits, we don't have constant resistance, we have varying electrical resistance based on the fixed power demand of traveling a certain vehicle at a certain speed because there is no free lunch.

When you cut the voltage by half but insist on demanding the same power, the electrical resistance gets cut to 1/2 which spikes the amps by double.

This might seem counter-intuitive in common language. You're thinking "But this makes it harder on the motor, wouldn't the resistance go up?" Yes, in common language. But electrical resistance is a specific term that means something different. Any voltage will cause infinite current (amps) unless you have some resistance. So for example, the filament of a 100 watt light bulb has half the resistance of the filament on a 50 watt light bulb. A high load is one that *doesn't* limit the current to be lower.

This might be too many new concepts all at once, but to relate it back to motors... you can think of jammed motor like short circuit. Like if you jammed a paperclip into a socket. And in fact, if you give a seized motor a normal amount of voltage, it will melt. A motor is in the constant process of pushing the short circuit away from itself, like a solenoid or railgun. Only we cheat and make it a circle so when the motor tries to push part of itself away, it just circles right back around and draws more current again. The more hard we make it for the motor to push the rotating part away from the non-rotating part, the more time it stays in the low-resistance close-contact area and the more time the current is "allowed" to spike.

This is how a motor's resistance would be non-constant depending on the voltage given, and how that resistance drops with less voltage and causes the average amps to rise when you have a set power requirement.

That's dumbed down a bit (and parts of my explanation are close to being outright false, to limit the number of new concepts to cover), but conceptually it all holds true.

Any of that make any sense?

---

The real-world consequences may or may not be okay. You get away with half as high of a battery voltage, but double the battery current. The controller and wiring will also need to support double the current, but only half the voltage.

In terms of wiring, voltage rating = how thick the insulation is. Almost all wire insulation is rated for 120+V (if not 300+) before electricity will jump through it so that's moot, no saving there.

In terms of controller voltage rating, that depends on the the way the innards of the component are designed, but generally at modest voltages like these they're all in the same ballpark anyway. So, probably no savings.

Higher current on the other hand, wire will require double the cross sectional area (50% increase in diameter) and weight (negligible). And for controller, generally devices have a max current (internal wire and magic stuff thickness) so there is a penalty. You might need components twice as big, or if bundled, twice as many of them side by side. It depends how your controller is built.

If your controller had excess capacity by double, there's no penalty in going this route.

Back at batteries... unless you're limited by pack size (and can't pick smaller cells), technically there'd be no savings here either, either way.

For example, if you had imaginary tiny 2V cells, perhaps you had 4 in parallel (for capacity) by 24 of those in series. So, 96 total batteries at 48V. To go down to 24V, you could just cut the pack in half and have 4x12, for 48 batteries at 24V. But then you got rid of half your battery pack and half your range. Maybe you want this, it's half the price and an option.

To maintain your old range, you need to put 8 in parallel, 8x12, still the same 96 batteries. For equivalent range, your battery neither shrinks in size nor weight nor cost nor anything. Just changes in wiring.

The other concern is to make sure your batteries can supply double the current. This usually isn't an issue unless your cells are truly tiny. Most batteries anyone would even consider for riding can spit out plenty more current than the motor could take anyway. It's a side-effect of picking high enough capacity batteries to have useful range anyway.

One final caveat... suppose you wanted to do the reverse. You wanted to go from 24V to 48V, using the same batteries with different wiring. That *MAY* not be possible, if your batteries are too big. For example if you have 2V batteries and only 1x12 to get 24V... you can't just take a pair of scissors and cut single batteries in half to get half the capacity, then chain up 1x24. For LIPOs, this may be an actual limitation since individual cells are often ordered quite large, and it's why you see people often doubling their range when they double their voltage.. because there's no rearrangement of cells possible.

Data Dump Complete. Confused yet? Probably too much at once, but, if you learned at least some of it, some is better than none.
MattsAwesomeStuff
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### Re: Reduction gearing 170kv for 24v

MattsAwesomeStuff wrote:Any of that make any sense?
Nope

Miles
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### Re: Reduction gearing 170kv for 24v

miles wrote:Nope

MattsAwesomeStuff
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