I'm trying to understand pendulums. I seem to be descending into a real rabbit hole. I'm worrying about the driving weight vs the pendulum period. I suppose I'm talking about "escapement error".
(For reference, it's a 150 year-old "black forest shield clock", wooden frame, brass bushed, pendulum bob = 7g, pendulum rod = 7g, trapeze suspension, "strip pallet" escapement, 550g weight. Runs well but keeps poor time.)
I think the "escapement error" effect depends on Q. So, first question: what is the Q of my clock?
I made a video of a free-swinging pendulum and analysed the individual frames. (To get a "free-swinging pendulum" I detached the weight.) Q varied as the pendulum ran out of energy but was in the range 180 to 300. Let's say
Q = 200
Is it reasonable to measure Q with a free-swinging pendulum? Q is
2*pi*(energy at start of a swing) / (energy lost during the swing)
Some of the energy lost is due to the friction of the escapement - e.g. the pallets rubbing on the escape wheel - so if the escapement isn't running, it's not a fair test. One could calculate how much energy the weight puts into the pendulum during each swing. But to do that, you'd need to know the efficiency of the geartrain. (I did the maths and the gear train is extremely inefficient.)
In theory, does the mass of the driving weight affect the period of the pendulum? Yes. But by how much?
Consider a foliot clock. Q = 0. The foliot escapement works using the inertia of the foliot arms vs the torque from the escapement wheel - the clock will run faster as the driving weight increases and slower as the foliot inertia increases. The period of a pendulum clock is a combination of the classical pendulum we learned in high-school and the inertia of the pendulum acting like a foliot.
(Imagine the clock in space - only the inertia would matter. Or imagine removing a pendulum leaving only the crutch: the crutch is driven hard and slows the escape wheel due to its inertia. In both cases, it's working like a foliot. I think the frequency of a foliot is proportional to the square root of the driving weight.)
With a pendulum clock, Q is between 200 (for my clock) and 100,000 (for a Shortt-Synchronome). If Q is small, then the torque becomes more important. What is the equation for the period vs driving weight?
My weight weighs 550g. I added 10% to it and the pendulum period decreased by 1%. A ten-to-one ratio.
Is that a bigger than usual effect? Is it big because of the low Q? Or the light pendulum bob? Or the heavy pendulum rod? Or the strip escapement? (It's definitely not circular error.)
Thanks
Peter
(For reference, it's a 150 year-old "black forest shield clock", wooden frame, brass bushed, pendulum bob = 7g, pendulum rod = 7g, trapeze suspension, "strip pallet" escapement, 550g weight. Runs well but keeps poor time.)
I think the "escapement error" effect depends on Q. So, first question: what is the Q of my clock?
I made a video of a free-swinging pendulum and analysed the individual frames. (To get a "free-swinging pendulum" I detached the weight.) Q varied as the pendulum ran out of energy but was in the range 180 to 300. Let's say
Q = 200
Is it reasonable to measure Q with a free-swinging pendulum? Q is
2*pi*(energy at start of a swing) / (energy lost during the swing)
Some of the energy lost is due to the friction of the escapement - e.g. the pallets rubbing on the escape wheel - so if the escapement isn't running, it's not a fair test. One could calculate how much energy the weight puts into the pendulum during each swing. But to do that, you'd need to know the efficiency of the geartrain. (I did the maths and the gear train is extremely inefficient.)
In theory, does the mass of the driving weight affect the period of the pendulum? Yes. But by how much?
Consider a foliot clock. Q = 0. The foliot escapement works using the inertia of the foliot arms vs the torque from the escapement wheel - the clock will run faster as the driving weight increases and slower as the foliot inertia increases. The period of a pendulum clock is a combination of the classical pendulum we learned in high-school and the inertia of the pendulum acting like a foliot.
(Imagine the clock in space - only the inertia would matter. Or imagine removing a pendulum leaving only the crutch: the crutch is driven hard and slows the escape wheel due to its inertia. In both cases, it's working like a foliot. I think the frequency of a foliot is proportional to the square root of the driving weight.)
With a pendulum clock, Q is between 200 (for my clock) and 100,000 (for a Shortt-Synchronome). If Q is small, then the torque becomes more important. What is the equation for the period vs driving weight?
My weight weighs 550g. I added 10% to it and the pendulum period decreased by 1%. A ten-to-one ratio.
Is that a bigger than usual effect? Is it big because of the low Q? Or the light pendulum bob? Or the heavy pendulum rod? Or the strip escapement? (It's definitely not circular error.)
Thanks
Peter