Recent content by jryer

F = Moving force in the pulley driver T1 = tension on the tight side T2 = tension on the slack side e = base of the Neperian or natural logarithms (2.7182) μ = coefficient of friction α = angle subtended by contact surface at the center of the pulley T1 = F * (1 + (1/(e^ua - 1))) T2 = F...

No freewheel or over-run clutch on the pendulum. The freewheel is on the winding barrel not on the pendulum. See link below http://www.youtube.com/playlist?list=PLdw9gJ-WEjpTQnCjs96OaZ6JUhBMJIAdG

Varadha hit the nail on the head. If really need to slow the driven belt down even more, then do two things. Make the driver pulley small, like Varadha said, and make the driven pulley extra large. Maybe a 5:1 or even 10:1 ratio. This way the small driver pulley will turn 5 or 10 times, but the...

"I would like to know, does the torque required to turn a pulley increases if you increase the tension?" ... YES Here's why. If we tighten the belt, we increase the 'coefficient of friction' in the T1 & T2 equation. Coefficient of friction comes from contact between the belt and pulley. An...

Figure belt tension for tight and slack side (T1 & T2) force = power/velocity Assuming you have an motor of 2 hp (1492 N-m/s) and the belt is moving at 12 mph (5.3645 m/s). Metric units are better here so ... F = (1492 / 5.3645) = 278 Newton (62 pound force - lbf) So now T1 = 278 * (1 +...

force = power/velocity (force has other equations, we're using the one suited for this application) Assuming you have an motor of 2 hp (1492 N-m/s) and the belt is moving at 12 mph (5.3645 m/s). Metric units are better here so ... F = (1492 / 5.3645) = 278 Newton (62 pound force - lbf) So...

Need some help understanding if torque changes. Does lengthening the pendulum affect the torque on the other gears (purple, green, and plum)? In other words is torque inversely proportional to speed still applicable here? thanks

Bump. Anyone care to contribute on the 7:50pm post? thanks again

As far as negligibility, you may want to familiarize yourself with how Power is Transmitted via belts: First figure out the belt tension for both sides: F = Moving force in the pulley driver T1 = tension on the tight side T2 = tension on the slack side e = base of the Neperian or...

JBriggs, take a look at this link http://electronics.howstuffworks.com/gadgets/clocks-watches/clock4.htm What happens if we put the escapement/pendulum assembly between the blue (1st) and purple (2nd) gears? Now does lengthening the pendulum affect the torque on the downstream gears...

Was referring to the natural frequency (pitch, note) of metals frequency = speed/wavelength At the particle level, a stiff or rigid material is characterized by atoms and/or molecules with strong attractions for each other. When a force is applied in an attempt to stretch or deform the...

If we assume the belt starts at 0 on the Y axis, and is moving along the X axis of a x/y graph, then yes Y + 3 would create friction because the belt is being diverted along the Y axis. Verbally I can express this a lot better than my cheesy math skills can. I believe at this point we are...

The first one - "How would lengthening the part of the pendulum below the pivot point change the torque delivered by the escapement; would it increase or decrease?" Also, I'm assuming the change in the torque delivered is transferred (seen/realized) to the gear train, is this a correct assumption?

If the pulleys are not aligned to each other then the belt is having to make a slight turn to track onto the next pulley. Anytime a turn occurs in nature, things tend to bunch up. As they bunch up they start banging into each other and this collision creates heat. Kind of like rubbing your hands...

Yup, the energy is being dumped primarily into the ball bearings and case. Just be aware any increase of friction in the pulley system will place a greater load on the engine, thereby either reducing engine output (torque) or require the engine to generate more output (torque) to compensate...