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#1
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Re: Drive Train design roadblocks
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I'm still rather confused on why there are two listed current draws, one being calculated as 2x the first. Just seems odd to me. (By the way, I've been reading the lower one, of course) |
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#2
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461 was running 4 motors all designed around keeping them around the 40A mark and we had no problems... we were drawing around 150A pretty regularly without problems... until we ran over a dozen back to back matches in the finals at nationals. Despite the fact of this untimeliness, they held up very well above listed capacity for said reasons. (Hint: Keep circuit breaker cool.) Overall, don't worry about the 120A breaker. Check out the spec sheet. It has no problem going 30% beyond that for 20 seconds or so. You typically won't be tripping that in a match, or even back to back matches if you make an effort to keep them cool. The 40A breakers, however, I think are much less forgiving. I would be much more concerned about keeping these guys in check. My advice is to design your motors and loads evenly around the 40A breakers. Good luck! Matt Last edited by Matt Adams : 04-01-2005 at 14:05. |
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#3
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Re: Drive Train design roadblocks
Let's go off on a slight tangent while I throw a cautionary note in here.
Everyone is all in favor of team that use the summer and fall to develop prototype robot system, and design assemblies for their team (as training exercises, demonstration units, design practice, etc.). However, your messages seem to indicate that you are well into the final designs of your competition robot for the 2005 season, and you are seeking to tweak out the last few percentage points of efficiency and power for your designs. This is a rather risky practice at this point in time. Remember, you don't even know what the game is yet! As you "finalize plans for this year's bot" you are running a chance of spending a lot of time designing a system that could be completely inappropriate for the 2005 game. If that happens, most teams historically fall into one of the classic engineering traps. Having invested a lot of time and effort in a specific design, most engineers are loathe to throw it away and start over with a clean sheet of paper - even when that is the most appropriate thing to do. Instead, they will try to modify their existing design and try to make it do something that it was never intended to do. Sometimes this works, but it is almost always at a cost in terms of lost efficiency, kludged design compromises, or brittle performance characteristics. During this last week before the kick-off, you might be better served by either letting your designs be a little "loose" and leave the determination of the fifth digit of precision for later on, or - better yet - just kicking back and relaxing while you can (because you won't be able to for the six weeks that follow!). -dave |
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#4
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Re: Drive Train design roadblocks
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I only intended to get a little more insight into the JVN calculation sheets, as well as the theory that went into them. There are certain variables (such as those mentioned above) that I have wondered about for a long time, and I'd like to learn what they are and how to make use of them. |
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#5
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Re: Drive Train design roadblocks
The Drivetrain Efficiency is kinda just what it sounds like it would be, the efficiency of your entire drivetrain, both gearbox and drive system combined (chains, belts, gears, etc that connect the motors to the wheels). Personally i would think .65 is a little on the low side, but i must admit i havn't done too much with gearbox design and building, its never bad to stick on the safe side though. As far as coming up with an overall value for your drivetrain efficiency you need to multiply together the efficiencies of each stage of your drivetrain, for example if your gearbox consisted of two sets of spur gears that were each 95% efficient and you then had a chain with an efficiency of roughly 95% also then you would multiply .95 x .95 x .95 = .857, or roughly 86% efficiency. It mostly depends on the types of gears you use in your drive system and how much your drive system consists of.
Some rough efficiencies for different types of components are as follows: Chain & Belt ~ 95% - 98% Spur Gears ~ 95% - 98% Bevel Gears ~ 90% - 95% Worm Gears ~ 40% - 70% Planetary Gears ~ 80% - 90% Hope this helps Also, please correct me if i may be wrong on something - its rather late to be posting :-) |
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#6
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Re: Drive Train design roadblocks
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This is why I have and continue to advocate that FIRST mix things up more in terms of size, weight and components allowed. Mike |
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#7
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Re: Drive Train design roadblocks
Dave and Mike did bring up 2 good points but for the sake of JVN calculation experimentation...
The 2 values for current draw relating to the two different speeds (gears) directly reflect the force it takes to rotate in each gear. low gear-more rotations (from motor)-faster rotations (from motor)-less power req. to rotate-less amperage high gear-less rations (from motor)-slower rotations (from motor)- more power req. to rotate-more amperage I am not specifically familiar with the JVN calcs. but through my own calculations and driving in matches (while tripping the 40 amp breakers in high high gear) I believe that to be your answer. -Henry |
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#8
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Re: Drive Train design roadblocks
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#9
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Re: Drive Train design roadblocks
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#10
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Re: Drive Train design roadblocks
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#11
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Re: Drive Train design roadblocks
As Henry pointed out to me just a few minutes ago (which I'm very surprised never occurred to me), it seems the second current reading is for the total draw from both gearboxes. A simple, well put answer, it would seem.
Now I am faced with figuring out what factor is causing my calculations to produce such high currents at (theoretical) max speeds much lower than those of other teams. For instance, with the gearing I'm now working with I have a current draw of 97A (194A total) at a max speed of 10.5 ft/s. Any suggestions? |
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#12
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Re: Drive Train design roadblocks
Adam-
You are right to be doing some detailed calculations like the JVN calcs as a way to figure out the potential performance of your drive system. Even if you don't end up using the design, the calculation exercise is still educational. So don't give up on that approach - just keep refining the input assumptions as you get more info or data. My first inclination is to suggest that your assumed efficiencies (you quoted 60% overall, I think) are too low. This definitely will drive your top speed to a lower value and your currents (at a given speed) to a higher value. The efficiencies posted earlier in this thread are pretty accurate, and if you do a good design job (rolling elements, get rid of sliding contact wherever possible, minimize misalignment under load) you can hit the high end of those values. One other source of efficiency loss is the tire-to-carpet interface. This is maybe a 93-95% type of number. Soft sticky tires are lower (90-93% maybe). I think a good system with a simple spur gear drivetrain (no planetaries, no belts, just gears and chains and wheelchair wheels) should be able to be in the mid-80's efficiency-wise. Advanced topic: For people who are doing more than a closed-form calculation (i.e., they are using calculus to derive a simple equation that they can numerically integrate - the type of stuff you would do in 1st year of engineering school) I would suggest adding a loss term that is a constant with respect to speed. For a *very* efficient gearbox I use a 0.2 Nm spin loss applied to the output of the gearbox, and a 12 N drag loss applied to the robot mass. Hope this helps, Ken |
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