It is a seven ft diameter wheel, with 4 1/2 ft arm, I did not push the wheel I let go, of a arm, which raised another to its point of drop. I had been using steel cable to keep the wheel from turning, by a loop over a bolt, but it had come loose
How much force does it take to move a weight on a 2 degree incline? If you double or triple that force, for the down stroke, with mechanical advantage.
The amount of force does not interest me as much as distance! what is the lineal distance (in circumference) required to move a weight in 2 degree increments back to its point of working potential leverage? Note: you only have 3.5' radius to accomplish this?
Fcdriver wishes to play games! He wants to see who can come closest to winning the brass ring.
"So what and how much was the offset? What degree was the incline?"
The offset can be anything past the balance point. The incline could be zero
and/or irrelevant.
The driving force in a 180 degree fall is only lifting the same amount if not more weight 2 degrees. Meaning you have 90 drivers to get one weight back to 12:00 for a ratio of 1:90
This of course does not account for the fact that the driving weight lifting its counter part 2 degrees does not pass six and stays there.
rlortie wrote:Fcdriver wishes to play games! He wants to see who can come closest to winning the brass ring.
"So what and how much was the offset? What degree was the incline?"
The offset can be anything past the balance point. The incline could be zero
and/or irrelevant.
The driving force in a 180 degree fall is only lifting the same amount if not more weight 2 degrees. Meaning you have 90 drivers to get one weight back to 12:00 for a ratio of 1:90
This of course does not account for the fact that the driving weight lifting its counter part 2 degrees does not pass six and stays there.
Do me a favor take a bathroom scale, attach a 48" arm to a pivot( on the end) put a 10 lb weight on the (other)end, put a 1x2 along the length, (pushing down on to your scale)what does the scale read each time you more the 1x2 along the length ? What is the max force of a 10 lb weight?
Last edited by Fcdriver on Sun Jan 03, 2016 7:10 am, edited 2 times in total.
Forget your lust for the rich man's gold
All that you need is in your soul
And you can do this, oh baby, if you try
All that I want for you my son is to be satisfied
I will play your game only because cold weather keeps me from doing what I should be doing.
"Attach a 48" arm to a pivot". What does the arm consist of and at what point do I place the pivot? "Put a 10 lb weight on the end". Which end if the pivot is not centered, or does it matter as the weight will cause it and the arm to rest on the floor. "Put a 1x2 along the length". Which side of pivot do I place the 1x2, on the side with the weight resting on the floor or the end up in the air? "what does the scale read each time you move the 1x2 along the length? depends on which side of the pivot and weight it is place upon. "What is the max force of a 10 lb weight?" 10 pounds if it is sitting on the floor!
If I move the 1x2 outward from pivot and on the opposite side of the weight the scales will increase by leveraging the 10 pounds resting on the floor, but it must be a very heavy 1x2 to show any significant gain on a 24" pivot providing the pivot is centered.
Try one more thing for me would you, take a pencil hold by the end with it parallel to the floor try to hold a ball against the wall,,now hold the pencil vertical, and try to hold a ball against the wall. This is the difference between, a oscillating cam follower and a translating cam follower.