How Can I Calculate the Minimum Value of n2 in Tractor Power Transmission?

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robax25
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Hi, I have a problem to design traktor power transmission.Can anybody tell me How to calculate minimum speed for output shaft.I need to calculate number of rotation(maximum and minimum) of the driven shaft
Here is the equation: n(driven) max=(n(drive) + n2 (max). (z3/z1)) /(z3/z2 +1)
so I get 2500 rpm
n(driven) min=(n(drive) + n2 (min). (z3/z1)) / (z3/z2 +1)
So I need to get n2 mimimum value. Here is given only n2max but n2(min) is not given.How can I calculate the n2 min value?

I tried to calculate and get n2(min)= 1875 rpm but it should be 2000 rpm.
Here is the formula,
n(motor)/n(pump)=V(pump) / V(motor)
At the end, the minimum driven shaft rotation is 1437.5rpm but it should be 1500 rpm
 
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It is not homework. It is a question from mobile hydraulics and I have all solution.However, I don't understand how to calculate minimum rotation motion for n2(min).This is my question.I understand a lot how it works.It is not instructed by professor, I bought personally mobile hydraulics book and from there, I get the question.
 
Can you please post a title, author, year and ISBN reference to the mobile hydraulics book.

Is the ratio of V1 to V2 variable?

Notice that the rotation rates, n, are specified as counterclockwise when viewed from the left. That makes it possible for some n values to be negative, which may make a difference when +1.
 
you are right. I solve the problem
 
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