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#1
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pic: Gearbox
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#2
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Re: pic: Gearbox
No, your gearbox is not geared to X fps.. It's got a reduction. With a certain sized wheel it would drive a robot X fps.
What's the reductions? What size wheel is it designed for? |
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#3
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Re: pic: Gearbox
Quote:
First Stage - 11 to 54 Low Stage - 36 to 40 High Stage - 20 to 56 |
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#4
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Re: pic: Gearbox
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It is designed for a 3" to a 3.25" wheel. It can work with the 3" colson, and can go faster if the CIM gear is swapped to an 12, 13, or 14 tooth gear. The overall reductions are 5.45:1 and 13.75:1, which are a little fast for 2 CIMs driving a 3.25" wheel. I plan to add an rs-775 to the gearbox. |
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#5
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Re: pic: Gearbox
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Of course, I could also just be completely wrong here ... I'd love someone to confirm/deny what I'm saying with a better explanation. |
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#6
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Re: pic: Gearbox
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If you are interested, here's an example of a single CIM and one rs-775 gearbox, with free speeds not matched, then matched. To get these, you need to find what the rpm of each motor will be, use the motor data to calculate the torque at the shaft of each motor, then you need to find the torque at the output, which is the CIM motor shaft in all these examples. Code:
Free Speed Stall Torque Stall Current Free Current (Amp) Power (W) CIM 5310 2.43 133.00 2.70 337.81 RS-775 (12V) 13000 0.78 86.67 1.80 266.60 CIM torque (Nm) vs. rpm R/5310 + τ/2.43 = 1 τ/2.43 = 1 - R/5310 τ = 2.43 - R*0.000457627 RS-775 torque (Nm) vs. rpm R/13000 + τ/0.78 = 1 τ/0.78 = 1 - R/13000 τ = 0.78 - R*0.00006 with same gear ratio, both at 12V, at 6000 rpm (this is possible) τ775 = .78 - 6000(.00006) = 0.42 τCIM = 2.43 - 6000(0.000457627) = −0.315762 The CIM is providing resistance, and reducing availible torque Total torque = 0.104 Nm geared so that free speeds match at 5310 (2.45:1), at 12v, at 3000 rpm (nothing higher than 5310 is possible) The rs-775 will be geared down. rpm of the 775 will be (2.45 X 3000) = 7350 τ775 = .78 - 7350(.00006) = .339 this is at the shaft of the motor at the CIM, it will be 2.45 times more = .83055 τCIM = 2.43 - 3000(0.000457627) = 1.057119 Total torque = 1.888 Nm It makes sense because when free speeds are matched, no motor ever has to push the other past its free speed, so the torque contributed by each motor will always be positive, and the CIM will always contribute more. The ratio between the contributed torque of the CIM and the contributed torque of the rs-775 will always be equal to the ratio of their power, if the free speed is matched. The ratio is about 79:100 for the rs-775 to the CIM. Last edited by Jared : 06-07-2014 at 10:47 AM. Reason: left out a part. |
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