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QUESTION: 1.If shaft 1 turns at 1750rpm, then at what speed would wheel 1 turn?
2.what size whould wheel 3 be to turn shaft 2 at 3450rpm?
3. What size wheel would wheel 2 need to be to turn wheel 5 at 3600rpm?
4. What size would wheel 4 need to be to turn shaft 3 at 3600rpm?
5. What size whould weel 4 need to be to turn wheel 6 and turn shaft at 3600rpm?
6. What size would wheels 4 and 6 need to be to turn shafts at 3600rpm?

ANSWER: Hi. This question seems to be referring to a figure. Can you attach it in a follow-up?. At any rate, the key relationship to use in solving this type of problem is to use the fact that the speed (in terms of rpm) is inversely proportional to the gear radius. In other words, if a drive gear, A, is attached to another gear, B, then if B is half the size of A, then B will rotate twice as fast. You can use any ratio to determine radius given speed (or vice versa)

BTW, the term "size" is the same as radius (they are proportional). Using this relationship, you should be able to work out the answers.

Randy

P.S. Please show your work on the follow-up. It will be easier for both of us to explain what you are having trouble with.

---------- FOLLOW-UP ----------

QUESTION: I apologize for the confusion. my girlfriends son sent you the information but left out the diagrams that go with the actually design.

Please open the attachments and you will find 2 diagrams that might make things a little eaiser to work out the sizes with.

I am a disabled Veteran and I can't do math worth a durn. Thank you for your time and effort concerning this problem...

Respectively,,,,
G. Lamb

1. Wheel 1 (W1) is attached to shaft 1 (S1). Since S1 turns at 1750 rpm this means W1 turns at 1750 rpm too.

2. W3 is attached to S2 and is driven by W1. We want W1 & S1 to turn at 3450rpm so we need to use the formula expressing the inverse relationship between wheel sizes (diameters, D) and rotation speed (rpm, R): using the notation above

D1/D3 = R2/R1.

Solving for D3 and putting numbers in gives

D3 = D1(R1/R3) = (3")(1750/3600) = (3)(0.51) = 1.52".

As you can see, if we want a wheel to turn FASTER relative to its driving wheel, then it has to be SMALLER.

3. W2 is attached to S2 so it is turning at R2 = 3450 rpm. We want W5 to turn at R3 = 3600. This is slightly faster than W2 so W5 should be slightly smaller. Note that D2 hasn't been specified yet so let D2 = 3". Then

D5/D2 = R2/R5 or D5 = D2(3450/3600) = (3)(0.96) = 2.75" = 2 7/8".

4. W4 doesn't drive S3, W5 does (accordng to your drawing), therefore the size W4 doesn't matter.

5. Since you want S4 to turn at the same rate as S3, W6 has to turn at the same rate as W4, which is on S3 and so turns at R3 = 3600rpm. So R4 =R3 = 3600. According to the formula above, D6 = D4. Since the sizes aren't specified yet, you can make them any size you want (such as 3").

6. As just stated, they would be the same size.

Hope this helps. Its really just a matter of a little book keeping.

Randy

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randy patton

Expertise

college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

Experience

26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

Publications
J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

Education/Credentials
M.S. MIT Physical Oceanography, B.S. UC Berkeley Applied Math

Past/Present Clients
Also an Expert in Oceanography