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Nonlinear force to linear force equation 
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#1
May1012, 10:42 AM

P: 92

I need to design a cam based on the following math problem: A nonlinear force f(n) is working on a lever that is x + x' long (x+x'=constant). I need an equation for the ratio, x/x', such that the counterbalancing force, f', is constant (nonvarying).
By varying the ratio of f(n) to f', that is ratio=x/x', I should be able to find an equation describing the ratio (r) that provides a nonvarying output force f'. The distance that f(n) works through is the same as the distance that f' works through, so work in = work out. Any help will be greatly appreciated. I don't know quite how to approach this problem. I hope the attached jpeg expalins the problem in enough detail 


#2
May1212, 07:47 AM

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Not at all sure I've understood the question, but here goes.
By moments, you have f.x = f'.x'. x+x' = C, constant. So f.x = f'(Cx) (f+f')x = f'.C x = f'.C/(f+f') Is that it? (Didn't get the bit about f and f' moving through the same distance. Since f and f' are not equal, if they move the same distanvce there'll be an energy imbalance. Won't the movements be in the inverse ratio?) 


#3
May1212, 08:49 AM

P: 92

Your solution works where both f and f' are known. If you know f' and f you can find x, and then f' = f*x/(Cx). But you need x to find f' and you need f' to find x.
I believe I have found a solution: It is required that the distance traveled for f and f' be equal. Work is equal to force times distance, so if distance is equal and f is a function of n, then the work done by f is [itex]\int f(n) dn[/itex]. This yields a constant value for the area under the curve of f(n). So now the ratio(n) (which is x/x') is equal to([itex]\int f(n) dn[/itex])/f(n)*n. Then f'(n) is equal to f(n)*ratio(n), and f'(n) is constant. The net result is that the ratio varies as f(n) varies to maintain a constant counterbalance force f'. 


#4
May1212, 06:45 PM

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Nonlinear force to linear force equation
Clearly I have not understood the question  and still don't.
What is n? From your integral it would appear to be the distance through which f has moved at any time. Is that right? And when you say the two forces move through the same distance, you must mean the same overall distance over a given range, yes? It cannot be true that they have moved the same distance at each instant. 


#5
May1312, 09:00 AM

P: 92

A lever that is x+x' long has a force f on one end and a countering force f' on the other end and the fulcrum is at a position determined by x, x'. Say x = x', then the fulcrum is in the center and f must equal f'. This just a simple lever and the distance that f moves will be equal to the distance f' moves.
Now you can vary the relationship between x and x' such that the fulcrum moves to the left or right of center. The ratio x/x' will be equal to the ratio f/f' and the distance each force moves the end of the lever s/s'. Again, simple lever physics. But lets make f a function (it could be any function) such that f becomes nonlinear. Let's say the function (call it 'n') is exponential. In my example, f(n)=(1/n)^{2}. But it could be any function  that part is not important. Just that the force is nonlinear. Now, if you have a fixed positon for the fulcrum (that is, x and x' are nonvarying) then the countering force f' is simply f(n)*x/x'. My problem is to find a function so that x/x' becomes nonlinear and yields equal work, w_{f}=f*s = w_{f'}=f'*s' and f' is constant and s_{total} = s'_{total}. s is not equal to s' at all moments in time since the ratio, x/x', is varying. If area under the curve f(n)(that is, its integral) is made equal to the area under the curve (straight line) for a nonvarying f', then work done on both sides will be equal and the total distance traveled on both sides will be equal. 


#6
May1312, 04:57 PM

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f becomes a nonlinear function of what? Distance travelled? Time? It means nothing to say f is just 'nonlinear'.
Perhaps you mean nonconstant, so f = f(t). The ratio x/x' = r(t). Since f' constant, r(t) = f(t)/f'. The deltas to distance moved must be in the same ratio: ds = r(t).ds' (Check: work is conserved at all times: f(t).ds = f'.ds') The remaining constraint is ∫.ds = ∫.ds' over the range of t. I.e. ∫r(t).ds' = ∫.ds' We can reparamaterise r as a function of s': ∫r(s').ds' = ∫.ds' I.e. the average value of r(s') as s' varies is 1. 


#7
May1412, 10:05 AM

P: 92

I'm sorry  I did make it more confusing by saying that force f was a function of n where n is a series and f(n)=(1/n)^2. The force does vary as a function of time (time is proportional to n).



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