mcmiu
9th February 2007 - 01:33 PM
QUOTE (mcmiu+Feb 9 2007, 07:51 AM)
Hi!
here is my problem: I have rotary ramp (knob), which is rotateing around it's own axis. Slowly, it press down a spring. I know the torque froce for rotate the ramp, and I want to find out the force wich press down the spring. How can I calculate this force?
If anyone can help me, aor can tell me another forum where can I ask about this, please tell me. If someone realy knows something about this, I could send him some more information, some pictures maybe.
thanks a lot, and sorry for my english!
Or maybe someone can tell me where can I post my question!
mr_homm
11th February 2007 - 02:10 PM
The easiest way to do this is with energy. This is called the "principle of virtual work," if you want to look for more information.
The basic thing you need to know is how steep the rotary ramp is. This is usually called its "pitch." The easiest way to measure this is to turn the knob exactly one time around (360 degrees) and measure how much the vertical height changes. Then p = (change of height)/(2*pi) is the pitch. If it is inconvenient to make one full turn, you can use any angle you want. Then p = (change of height)/(angle). The reason for using 2*pi instead of 360 degrees is because angles must be measured in radians for most rotational formulas in physics.
Now that you have the pitch, you can calculate the work done by the torque T when you twist through an angle theta radians. Work is torque*angle = T*theta. The work done on the spring will be (force on the spring) * (vertical distance it moves) = F*y. The principle of virtual work says that when the knob is in equilibrium, the sum of all the works done on it by all the forces and torques must be zero, when it moves a tiny bit. Therefore, F*y + T*theta = 0, so F = -T*theta/y = -T/p.
This is the force exerted on the knob by the spring. The force exerted on the spring by the knob is (by Newton's third law) just the opposite. Therefore, F = T/p. This is a very simple formula that you can use any time you need to convert from rotation to linear motion. All you need to know is the value of p, the pitch.
Note that there may also be some friction as the knob turns, so the actual force on the spring will be somewhat smaller that the calculated value. Accounting for the friction is more difficult, because you need to know many details, such as the amount of lubrication, materials that the sliding surfaces are made of, roughness of the surfaces, etc. If your knob is well lubricated and in good condition, the friction should be small, so the force will be only a few percent less than T/p.
Hope this helps!
--Stuart Anderson
mcmiu
12th February 2007 - 10:31 AM
Thank you verry much for your replay, mr_homm. All what you explained, was verry helpful for me, and I really understood, even I don't know much about Physics.
I would like to ask you something more.
Because it is about two plastic parts, without any lubrification, the friction between them should be 0,3-0,4(from a tabel). As I know, the dimension of contact surface is also important.(Is this true? how can I insert it in formula?)
Could you please tell me where and how can I insert this two coefficents in your formula F = T/p? Or do I need another formula?
Anyway: I thank you a lot for your replay.
mr_homm
12th February 2007 - 09:57 PM
The explanation is more complicated with friction. Imagine that you unwrap one turn of the ramp, so that it looks like a straight, inclined plane. The horizontal base of the plane = the circumference of the threads on the screw = 2*pi*R, and the height of the plane = the distance between threads = H. Then the angle of the incline will be arctan(H/(2piR)).
The thread on the screw rests on top of the thread of the hole, so the system looks like a block being forcet to slide up the incline. There will be 4 forces on the block: vertical force F coming from the spring, horizontal force T/R coming from the torque (since torque = force*radius, it is true that force = torque/radius), normal force N from the pressure between the block and the plane, and friction force f from one surface rubbing over the other. When the knob is turning, the size of the friction force is uN, where u is the friction coefficient.
Consider combining f and N together into one force. Since its component perpendicular to the surface is N and perpendicular to the surface is uN, the total force makes an angle arctan(u) with the surface. Since the surface is already tilted at an angle arctan(H/(2piR)), the total force is tilted at an angle A = arctan(u) + arctan(H/(2piR)). Since the net force is zero, the combination of T/R and F must be equal and opposite to the combination of N and f, so it points along the same angle. Therefore, (T/R)/F = tan(A). Using the formula tan(x+y) = (tan(x) + tan(y))/(1-tan(x)*tan(y)), it is easy to calculate that tan(A) = (u + H/(2piR))/(1-uH/(2piR)). Therefore, F = T/(R*tan(A)) = T*(1-uH/(2piR))/(uR + H/2pi).
The pitch I mentioned last time is P = H/(2pi), so this formula simplifies down to F = (R-uP)/(uR+P) * T/R. Notice that if there is no friction, u=0, and so this formula returns to the simple form F=T/P. So you need to measure also the radius of the threaded screw, and then you can account for friction.
By the way, since the pitch is usually pretty small compared to the radius, the numerator is approximately R, instead of R-uP, with only a few percent error. This gives a much simpler formula, F = T/(uR+P), which is almost like the original frictionless formula. I'd advise you to use this last formula, because your results are only approximate anyway, since you don't know the value of u precisely.
Hope this helps!
--Stuart Anderson
mcmiu
13th February 2007 - 08:43 AM
Thank you again mr_homm!
I really apeciate all your explanations. It helps me a lot. Right now I'm trying to solve my problem. I hope I'll se succeede somehow.
thank you very much!