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#1
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Re: DC Motor and Gearing
OK, that changes things a bit - I was thinking you had to overcome 400 oz-in of friction and THEN 350 oz-in of torque on top of that.
The motor you selected might work, but not at 2000 RPM. To start an engine of that type (at least, those that I worked with many moons ago) needed about 500-700 RPM to start, but (as you said) only for a second or three. Looking at the photo, see how the motor gear is about 10 teeth, going to a bigger gear of about 40, which is stacked with a 20 tooth, going again to a 40? Just what you need. For the $25 it'll cost to make it happen, I'd say try it. If you can include some kind of clutch to disengage and engage the starter at will, you might consider a flywheel. A jet engine starter - thinking of it, a helicopter too - first spools up to speed with a flywheel, then uses the stored energy to turn the engine, effectively multiplying the torque considerably at the very first stage, where you need it the most. For this case, a 4 ounce 2" diameter flywheel on the fastest shaft will store considerable energy, almost enough to start the engine on its own. The motor will take a second or 2 to spin that baby up. Don Last edited by DonRotolo : 22-03-2008 at 16:16. Reason: Got my ratios wrong... |
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#2
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Re: DC Motor and Gearing
Quote:
The starter shaft has a one-way bearing on it, so it completely disengages from the nitro engine once it is started. If I have gears resting against it, it would just sit there idle. As far as the gear ratio itself goes, how does that work out? The 10 to 40 connection makes a 4:1 ratio. Then with another 10 tooth gear on the same shaft as the 40 tooth gear, connecting to a 40 tooth gear, is another 4:1 ratio. So do you simply add those up and come up with an 8:1 as the final drive? Also, what should I connect the last 40 tooth gear too (meaning, what size gear should be on the starter shaft)? Another 40 tooth gear so it makes it a 1:1 at that connection? One other thing of note; the motor specified shows a stall amp of near 92 amps. That means I need a battery that can supply 92 amps burst discharge - or it will never reach the implied stall torque we have been using as a baseline? |
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#3
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Re: DC Motor and Gearing
WRONG - Gear radios multiply. Oops. You can use the last 40 tooth gear for the output, or another one of the same size (but why?) for 1:1 if you like. The peak current from a battery may surprise you, but I^2R losses in the wiring will keep that peak current well below the theoretical 92 Amps. The 12 volt lead-acid battery FIRST uses can deliver 720 Amps for 3 seconds... So a small lead-acid will run your starter no problem. Don Last edited by DonRotolo : 22-03-2008 at 16:17. Reason: wrong ratios... |
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#4
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Re: DC Motor and Gearing
you could do a 10 to 80 as well it be bigger but possible lighter
how do you plan to make this mill, drill press?? do you know gear stuff like pitch diamter, diametral pitch, Pressure Angle you need to look at this stuff before ordering gears or making any parts |
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#5
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Re: DC Motor and Gearing
Don't gear ratios multiply?
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#6
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Re: DC Motor and Gearing
Yeah they do, lol.
3:1 and 3:1 is 9:1 |
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#7
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Re: DC Motor and Gearing
I did some quick research, and the gear stages do multiply.
I found an online calculator which can compute up to three stages: http://www.technobots.co.uk/Technica...Calculator.htm Playing with the User Defined section, I can come up with a few wild teeth combinations to equal what I need. I was browsing SDP-SI for parts, and found a simple combination of: - 14 tooth on the motor shaft - 50 tooth and 14 tooth on shaft # 2 - 32 tooth on the starter shaft. This would mean that the 14 tooth on the motor would turn the 50 tooth, which would rotate the 14 tooth also on the same shaft; which is connected to the 32 tooth on the starter shaft. This would create an 8.16:1 ratio. Until I design it all with the actual gear sizes and such, I don't know what exact formula of gear teeth I will use. But, this is an example of something that I could use. That calculator comes in handy. Next question: - How to secure the gear shafts to the bracket plate? For some reason, I am having a blonde moment here... I cannot figure out how the shafts that the gears would be secured too would attach to the bracket itself. My original planned design was to take two aluminum plates to create a "top" and a "bottom" for the gearbox. I was going to put studs on each of the 4 corners to create the box. I was then planning on having the aluminum plates drilled out in the exact same spots with flanged ballbearings placed in them. Then, I was going to place the gearshaft in these bearings with the gears attached to the shaft. But... what prevents the shaft from just falling out (other than the gear being screwed to the shaft, and the gear resting on the aluminum plate)? Or is this how it is suppose to be held together? Does that make since? |
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#8
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Re: DC Motor and Gearing
Edit: Yes, gear ratios multiply.
Let's say you have a 1 to 4 reducer - 1000 RPM gets reduced to 250 RPM, right? Now, the second reducer is also 1:4, so 250 RPM gets reduced to 62.5 1000 / 62.5 = 16, so two stages of 1:4 is the same as 1:16 In this case, for 1:8 you'd want 1:4 and 1:2 Don Last edited by DonRotolo : 22-03-2008 at 16:15. Reason: What An Idiot I am, I was just plain WRONG. |
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#9
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Re: DC Motor and Gearing
Having flown gas models in the past, I think you will find the starter motor required to be too large. Since you only need a few revs to start the engine, perhaps a smaller motor to wind a spring, with a latch to release. (like the old .049 airplane engines)
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