huntoon,
Thank You. I have viewed that site and many like it before. Was hoping there was something new in the works.
My work does not utilize wind/wave/solar power. I am referring to a machine using magnetics that will run it's self. Parts will wear out and all that so it won't 'Run...
Hello Again,
I hope I don't come across as an a@# or worse! I have talked to a couple physicists and when you get to this subject they just shut down. "It is impossible so no more conversation!"
I know it has been tried by much smarter people than me. I might look at things differently...
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...
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
Alrighty then.. Let's 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...
FeX32,
ok.. As I said earlier. Sin and cos are not something I understand in any way. I am not certain of the amount of torque I will have at the point/position I listed in my first post here. I know there are ways to calculate it, just do not have that kind of education.
Thanks,
ed
Well, Thanks for your help.. U have put me in my place! I didn't ask the right questions, so I get greek for an answer.
Torque at the above listed position is what I was looking for!
Thank You OleEngr63,
This is being driven abit differently than an IC engine. The cam strikes at both 22.5degrees before TDC then again @ 22.5 before Bottom Dead Center. (In this application it's actually 22.5before Horizonal on both left and right strokes.) My calcs say the lever is going...
Thanks, That might be what I am thinking of, or read about. I am not even sure the differences there!
The crankshaft will be turning a flywheel like in a car transmission, a cam is also mounted on the crankshaft. The cam turns a lever a few degrees. The Crankshaft arm is R - 5.25" out from...
Q_Goest, You have just made my day! I am working with 2 different forms of leverages in my project and this was the one thing that I didn't totally understand. If I'll have little or no loss in the crank shaft turn there, I should be able to turn the main lever with less than 16% of the power...
Really, I thought I had read (and understood) somewhere in my searches, that there is a loss of force/power that is not going into turning the crankshaft, at 90 degrees from TDC is the only place where the 1000 lbs is going into turning the crank and the rest of the 180 degree turn has a loss...
Hello,
This is for sure not a 'Homework' problem. I have no clue! I understand that a crankshaft looses power from a piston because the force is pressing the Crank outward verses around it's shaft. I have no clue as to how to figure this out. I have looked all over the web and could find...