A motor runs at 1450 rpm

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A motor runs at 1450 rpm. Attached is a gear that has a gear module of 5 mm, and a mid tooth-diameter of 95 mm. This wheel is toothed up with a second wheel who's shaft drives a third wheel. This third wheel is then toothed up with the fourth and last wheel who's shaft runs let's say dryer, which is supposed to run at 80 rpm. This last gear has a module of 8 mm. The last gear's diameter must be less than 700 mm 'cause of space. I'm supposed to find the number of teeth and the mid tooth-diameter for each gear. Here what my book does to begin with:

The ratio is calculated as

[tex]\frac{1450}{80}=18,125[/tex]

[tex]z_1=\frac{d_1}{m_1}=\frac{95}{5}=19 [/tex] (the number of teeth on the first gear)

Now my book does something I don't get. It finds:

[tex]\sqrt{18,125}=4,26[/tex]

This number is then used to find the number of teeth on the second wheel:

[tex]U=\frac{z_2}{z_1} \Rightarrow z_2=U\cdot z_1=4,26 \cdot 19=81[/tex]

As I said earlier, I can't see why the square root of the ratio is calculated...
 
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Answers and Replies

  • #2
FredGarvin
Science Advisor
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Obviously the diameter of the first gear is 95 mm. Is there anything else you have failed to mention? How about writing out the entire problem so we know exactly what we have to deal with?
 
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I've updated the problem above to include the whole thing now...
 

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