by Ole SeaBee
Tags: clue
 P: 343 The Tex is not working with Win XP either. This morning I was unable to open this page at all. Now I can see the page, but the Tex does not work. I think we havee a problem.
 P: 343 I think we need to slow down about on the "no loss in power" bit. With no friction, I agree that there is no loss in power, but where in the real world do we find situations without friction? Reciprocating compressors/engines have significant friction to begin, and when we add this strange lever drive mechanism, I see no reason at all to think friction will be reduced. What is required is to do an internal force analysis, first without friction. Use the forces calculated at the contact points to estimate the friction forces, and then repeat the force analysis, the second time including friction forces. By iterating this analysis, a reasonable estimate of the friction should be obtained, and with it the power losses in operation. We are also talking about a substantial amount of engineering analysis, computation, etc. It is not a trivial effort; you have to really want the result.
 P: 14 Alrighty then.. Lets work with the slow ole man here abit ok!! First, I have a disc that is R4.5". It has 1000# of resistance to turning on it's axis (at R4.5") and I need it to turn 9 - 15 degrees. I attach a lever to the center of that disc and it is at R11.25" I think I will need 400 lbs of torque to move turn it at that point and that the disc will turn with the 400 lbs. (Forgetting friction for now.) Now I have a crankshaft that is being rotated by a piston. The piston has an equal 1000 lbs of pressure on it (at this point on, that does not change). The piston linkage is 8.6875" long the end connected to the crankshaft is at 13.3 degrees above horizontal. That linkage is connected to the crankshaft at R5.25" from the cranks CenterLine. At the time the piston linkage is at 13.3 degrees above it's horizontal center line, the crank linkage is at 22.5 degrees above it's horizonal centerline. There is a CAM on the crankshaft that is also at 22.5 degrees, it is R2.613". The cam will strike the lever on the above disc at the R11.25 mark. (The piston figures to be .375" from BDC at that point.) Do we get anywhere close to 400 lbs of torque from the cam? Thanks, ed PS. I don't know what U are referring to as Tex?
 P: 67 To answer the original post then since you don't like trig. Solving the equations i wrote in the other post. If you are interested I rearranged them for $\frac{\dot{d}}{\omega_2}$ and then used the equivalence of this equal to Torque/piston force. Pluggin in your angles gives 862.5 ft-lb of torque at the crank. Not that this is with 1000lbs at the piston. Also note that this is independent of the length of the coupler link (rod), only dependent on the angles. I also assumed that your angles were all positive and starting in the first quadrant. Cheers,
 P: 67 Thx OldEngr63 about the TeX, seems to work now. Cheers,
 P: 67 Actually, it is still not working lol. I tried to fix the other post with no luck.
 P: 343 @ FeX32: I recently wrote some TeX on another thread and it worked find. What you have written on the previous page of this thread still does not work for me. I think things are still not quite right.
 P: 67 I agree :)
 P: 67 To ed, I just realized you asked about TeX. It's a typesetting language used by this site. (and many professionals for manuscript writing). It's why when you read my other post it looked all "greek". That greek should come out as math after compiled.
P: 14
 Quote by FeX32 To answer the original post then since you don't like trig. Solving the equations i wrote in the other post. If you are interested I rearranged them for $\frac{\dot{d}}{\omega_2}$ and then used the equivalence of this equal to Torque/piston force. Pluggin in your angles gives 862.5 ft-lb of torque at the crank. Not that this is with 1000lbs at the piston. Also note that this is independent of the length of the coupler link (rod), only dependent on the angles. I also assumed that your angles were all positive and starting in the first quadrant. Cheers,
FeX32,

Not that I don't like trig. Just don't know enough of it. Many Thanks!!!

ed.
P: 67
 Quote by Ole SeaBee FeX32, Not that I don't like trig. Just don't know enough of it. Many Thanks!!! ed.
No problem!
 P: 14 ok.. might seem like I still donno nuffin an prolly don't really. But the 862# is that at the Center of the crank or at the 5.25" from it's center line? U don't know how much this means to me!! Thanks Again! ed
 P: 67 No worries. The torque is about the center of the crank. This is inclusive of the 5.25" moment arm. Basically you can think of it as 164 lbf acting perpendicular to the crank arm at that 5.25" distance.
P: 14
 Quote by FeX32 No worries. The torque is about the center of the crank. This is inclusive of the 5.25" moment arm. Basically you can think of it as 164 lbf acting perpendicular to the crank arm at that 5.25" distance.
FeX32,

That is what I thought it was, I did get that number as well as at the cam I believe it's 328 lbf. This is better than I had hoped for in this application.

It is the final answer I needed and I've worked on this machine for the better part of 6 years. I have tried to figure it many ways, never was sure of the outcome tho! Now I fully believe it will work. I had little doubt ;)!

Thank You Again!! I won't forget your help!

ed
 P: 67 You're welcome ed. Good luck with your machine. All the best to ya,

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