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I understand this problem, but the solution is based on the distance of the drive wheel=distance of the mixing wheel. No problem there. Thats 2 wheels, what happened to the 3rd wheel?
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Schools:HBS(08) - Ding. HBS, Stanford, Kellogg, Tuck, Stern, all dings. Yale - Withdrew App. Emory Executive -- Accepted, Matriculated, Withdrewed (yes, I spelled it wrong on purpose). ROSS -- GO BLUE 2011.
I understand this problem, but the solution is based on the distance of the drive wheel=distance of the mixing wheel. No problem there. Thats 2 wheels, what happened to the 3rd wheel?
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The third wheel "remains in the middle."
The distance the motor and mixing wheels travel is equal -- you stated you understand that.
The distance the middle wheel travels is also the same.
If the middle wheel had a radius of 6,000,000 miles, it would turn the same "distance" as a middle wheen with radius of 1 nanometer.
The number of rotations would be vastly different, but in the end it would still turn the mixing wheel the same "distance" that the motor wheel turn it (the middle wheel).
Therefore since they all turn the same distance, you can essentially ignore the middle (third) wheel.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.