- #1
maximiliano
- 43
- 0
I think I have a rough idea how to get there...but I'm not sure.
Let's assume I'm riding a bicycle, and I'm peddling at a constant pressure (torque). This is a peddling pressure which would be the exact amount which would just lift a 50 pound weight off the ground and hold it there.
Okay...so as I'm peddling along, with precisely this much pressure on the chain...can I assume that is the EXACT pressure (in pounds) that each link of the chain is experiencing? IF SO, if I then calculate the surface area of each link that is exposed to peddling pressure to be 1/20" on one side of the pin and 1/20" on the other side, thus a total load bearing area of .1"...would I just then multiply the 50 pounds of total pressure by 10 (1 / .1)...for a total pounds per square inch being loaded on each link to be 500 psi??
Or...am I all wet on this calculation? What I'm attempting to do is to describe to some people how much actual pressure, in psi, each chain link is loaded by when someone is climbing a mountain at a constant speed. I used the 50 pounds of pressure as a guestimate of the torque required to do this. It's in that ballpark, but that number is not as important as it is to be able to calculate the PSI per link once the overall torque is known.
Thanks!
Let's assume I'm riding a bicycle, and I'm peddling at a constant pressure (torque). This is a peddling pressure which would be the exact amount which would just lift a 50 pound weight off the ground and hold it there.
Okay...so as I'm peddling along, with precisely this much pressure on the chain...can I assume that is the EXACT pressure (in pounds) that each link of the chain is experiencing? IF SO, if I then calculate the surface area of each link that is exposed to peddling pressure to be 1/20" on one side of the pin and 1/20" on the other side, thus a total load bearing area of .1"...would I just then multiply the 50 pounds of total pressure by 10 (1 / .1)...for a total pounds per square inch being loaded on each link to be 500 psi??
Or...am I all wet on this calculation? What I'm attempting to do is to describe to some people how much actual pressure, in psi, each chain link is loaded by when someone is climbing a mountain at a constant speed. I used the 50 pounds of pressure as a guestimate of the torque required to do this. It's in that ballpark, but that number is not as important as it is to be able to calculate the PSI per link once the overall torque is known.
Thanks!