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#1
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Re: pic: 2011 "Nova" Swerve Drive (Front)
I suggest looking into worm gears, they can get you a lot more reduction in that space, and I bet it will be easier to find a set that small. with a worm gear you may be able to remove the cimplebox entirely.
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#2
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Re: pic: 2011 "Nova" Swerve Drive (Front)
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#3
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Re: pic: 2011 "Nova" Swerve Drive (Front)
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#4
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Re: pic: 2011 "Nova" Swerve Drive (Front)
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According to the JVN mechanical design calculator; 7 feet per second after frictional losses. 196 LBS of pushing force with rough-top tread. |
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#5
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Re: pic: 2011 "Nova" Swerve Drive (Front)
Is it possible to reduce the CIMple box further, I was hoping to get right around 10 FPS
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#6
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Re: pic: 2011 "Nova" Swerve Drive (Front)
You mean reduce it less. And I would just decrease the reduction on the bevel gears.
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#7
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Re: pic: 2011 "Nova" Swerve Drive (Front)
With all due respect, if you can't figure out gear ratios, you really should not be designing a swerve drive.
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#8
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Re: pic: 2011 "Nova" Swerve Drive (Front)
With all due respect, I didn't post myself on this website, I posted my design. If you have a comment or criticism about that, I'd be happy to hear it. But if you only want to discourage me from experimenting and learning, I'd prefer if you kept that to yourself.
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#9
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Re: pic: 2011 "Nova" Swerve Drive (Front)
From looking at it, you have a fairly robust looking module with most of the issues attended too, aside from that bevel gear looking tiny.
What i'd be concerned with in this module is the amount of power necessary to steer those wheels (depends on the number of modules and what motors are steering). Also depending on your steering arrangement some method of angle adjustment between the steering sprocket and the module may be necessary. Do you have the machining capabilities to make that one piece bracket and securely mount it to the vertical support? It also looks rather huge, but i like small compact packages if you look at our recent swerves. Chris's statement has some merit, and I've learned to like when people are straight up blunt with me, as it makes things much easier to gauge and get a realistic evaluation of your current state. I would suggest listening to his advice and getting a firm grasp on gear ratios and the efficiency of various methods of power transmission, this will only help your design. Exercises like this are a chance to develop further, and as has been pointed out you could use some work in the gearing arena. Don't immediately think someone is attacking you when in reality they're just trying to bump you back to the path you've strayed from. He gains nothing from tearing you down. |
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#10
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Re: pic: 2011 "Nova" Swerve Drive (Front)
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As far as the module is concerned: -The module is designed to be tied (via 25 chain) to the other three modules. (All I know is a banebots motor was used to rotate, but I wasn't in charge of that system, so I don't know how much power was needed. We also used 1" wide wheels, and rotating wasn't a problem) -following some earlier advice, I'm working on interfacing a more robust and larger set of bevel gears. -We do have access to the machining necessary to create the lower chassis. -The size may be deceiving, but the wheels are only 3" in Diameter and 1/2" wide. I don't know if that still constitutes a "large" design, but if it does, I'd sure like to see Winnovation's swerve design ![]() Thanks! Last edited by Jedward45 : 12-08-2011 at 19:33. |
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#11
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Re: pic: 2011 "Nova" Swerve Drive (Front)
Not everything you want is going to fall into your lap, sometimes you have to go get it. The world belongs to those who show up. In this case showing up might mean searching CD's White Papers for JVN's Design Calculator.
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#12
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Re: pic: 2011 "Nova" Swerve Drive (Front)
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#13
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Re: pic: 2011 "Nova" Swerve Drive (Front)
My concern with the dual wheels is that every turn guarantees one or the other will have to scrub, and hard - based on my assumption that the wheel axle is solid and common to both wheels.
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#14
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Re: pic: 2011 "Nova" Swerve Drive (Front)
And incidentally, it was posted above that the JVN gear calculator would give you 7 ft/s. That's probably a decent estimate, but it's worth discussing some of the theory behind that result.
First, you take your wheels, which are designed to be Ø3.00 in. (From examining that drawing by eye, based on the size of the CIM, which is Ø2.50 in, I'm guessing the wheels are actually slightly larger than that. Did you account for the thickness of the tread backing, and maybe some of the tread?) We'll stick with Ø3.00 in for simplicity. Then find the circumference per revolution: C = π × d = Ø9.42 in/rev = Ø0.785 ft/rev. That's the distance the wheel travels in one revolution. Though I'd really rather work in SI, it would just confuse the Americans—so were going to calculate the speed of the robot in ft/s. That means we need rev/s, or its familiar cousin, rev/min. This is determined by the speed of the motor, at its operating condition. Note that this is distinct from free speed. Most of the time, when we talk about robot speeds, we're either talking about the theoretical free speed (when calculating), or the actual speed (when measuring). Since the free speed of the CIM motor is ωfree = 5 310 rev/min = 88.5 rev/s, and we've got a 10:1 gear ratio (Z = 0.1) a little multiplication gives: ωfree × C × Z = vIn JVN's calculator, there's usually an arbitrary factor of 81% corresponding to operating condition (graphed on a torque curve)—in other words, the motor is operating at 81% of free speed. It's arbitrary in the sense that there's no physical reason why it has to be 81%, but based on JVN's testing and experience, it was a reasonable value. (I've occasionally picked a slightly higher value for this—around 85% or even 90%. Again, it was just a guess, but I've designed several drivetrains with lots of motors, which tend to be less heavily loaded under ordinary driving, so the motors presumably run closer to their free speed.) In the calculations presented by Isaak* above, I think he's either using a 100% (i.e. at the free speed) calculation, inputting the correct speed and rounding up to 7 ft/s—in other words, using JVN's calculator to do the math I showed above—or else he's using the rough free speed JVN preloads it with (5 500 rev/min) and a weird speed loss constant of 97.25%. In any event, remember that the speed loss constant is just a way of saying "when the wheels are off the ground and spinning under power, their speed is 81% of what it would be in a frictionless gearbox." That's different from, but related to efficiency. Anyway, in my estimation, your wheels are probably more like Ø3.25 in (due to the tread thickness), so that might skew things a bit. Using speed loss of 81% and efficiency of 90%, using 1.2 as the coefficient of friction, substituting in the correct values for CIM motors instead of the rough/old ones in some versions of the JVN calculator, and taking robot weight to be a rather plump 167.5 lbf (hostbot + minibot + battery + bumpers), I'm actually estimating that you'd see something closer to 6.10 ft/s and 201.0 lbf pushing force. By the way, remember that all of this is predicated on your battery actually being able to deliver that performance—which it won't. As the load on the battery increases, its voltage drops, basically shrinking your speed curve (and hence your power). The more advanced versions of the JVN calculator will let you model motor performance over time at some voltage, from which you can determine the (constant-voltage) load on the battery. If the load on the battery is significant for a long enough period (e.g. when accelerating), it may be time to look at a few cases where the applied voltage is lower, in order to estimate mid- and late-match performance. For now, though, don't worry too much about the battery voltage and power consumption, because your design isn't crazy enough for it to matter too much. (I've had a few designs where these absolutely did matter.) Estimate a desirable speed, then try to find gears to match. * I just realized he's not "Hawaiian" or even "Hawaiʻian", but rather "Hawiian Cadder"...I'm not sure what to make of that. |
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