>Phillip Ford wrote: >>Well, I would say that it depends on the way you look at it. You have to >>compare apples to apples. Let's take an example. A see-saw or >>teeter-totter. A simple lever. Let's say it is horizontal. Put a 1 N >>force vertically down at 1 m to the left of the fulcrum. Now put a >>(Balancing) force vertically down at a point 1 m to the right of the >>fulcrum. What will that force be? 1 N. The 'leverage' is 1. Your >>'ratio', obtained by dividing one force by the other is 1. 1 mm of >>downward movement on the left side will result in 1 mm of upward movement >>on the right side. > > >All this is ok up to now... you left out a word like "balancing" above >where I inserted it in paranthesis.. but ok. OK. >> >>Now, on the right side, rather than putting the force vertically down, >>angle it at 45 degrees. Since the points of force application and >>distance measurement have not changed then the 'leverage' should still be 1. >And so it shall be. >>A 1 mm downward movement on the left side will still result in a 1 mm >>upward movement on the right side. >ok..still fine. >>But what will the force be? 1.414 N. Your 'ratio' obtained by simply >>dividing one force by the other would be 1.414. How did that >>happen? The effective ratio for speed, weight, and >>distance are supposed to be the same any way you look at it right? > > >Here you go out to lunch. That's were I spend most of my time. >If you reduce or increase the input force by changing its angle forward or >backwards by any particular degree.. then the output lifting capacity is >simliarily affected. Essentially you are just altering the amount of input >to the system. Its like placing 10 pounds on each side first, and then >placing 14 pounds on each side... to put it tritly. Either way... its >still a one to one balance. I didn't just change the 10 pound force to a 14 pound force, I also changed its direction. A 14 pound load at 45 degrees to vertical on the right side balances a 10 pound load vertically down on the left side. I don't need to change the load on the left side to 14 pounds for the system to be in balance. The vertical component of the load on the right side (14 sin 45 = 10) balances the load on the left side. The component of the load in the direction of the movement balances the load on the other side in the direction of the movement 1:1. That's the point. If you just take the magnitudes of the loads and divide them you won't get a 1:1 ratio because you're not looking at the right numbers. This is the point that I'm trying to make about taking measurements on actions. The forces measured have to be at the same points and in the same directions as the distances measured for there to be a comparison. >Listen.. this is simple Archimedis. I certainly wouldn't want to argue with Archemedes, even if he is dead. >The formula for leverage is (D1 * W1) = (D2 * W2) There is nothing in >this formula about force vectors. And even if you wish to substitute a >force vector for weight, then what you do to one side of an equation you >must do to the other. You want to change the input force, call it W1, >leave both D1 and D2 the same, and expect nothing to happen to W2. I haven't changed anything. W1 = 14 sin 45 = 10. It's still the same. So W2 doesn't need to change. By definition W1 and W2 are force components perpendicular to the lever. Try changing my 45 degree angle to 90 degrees and there's no leverage at all. The load becomes an axial load on the so called lever. Let W2 be a 10 pound load vertically down on the left side. Apply W1 at 90 degrees to the lever (in other words horizontal) and what magnitude of load do you need? It doesn't matter. Make it as large as you want and it will never balance W2 because there is no vertical component to the load. Another way to look at this is as a statics problem. For the system to be in balance the sum of the forces and sum of the torques have to be in balance. See http://www.comfsm.sm/~dleeling/physics/torque.html >>The distance ratios that we are talking about take account of the varying >>geometry as the action moves from one point to the other. The total >>downward travel of the key is x. The total upward movement of the hammer >>is y. This gives an overall ratio for distance traveled. Let's say that >>y/x = 5. This doesn't tell us that the distance ratio is 5 for every >>point in the key's travel. The overall average is 5. Perhaps for the >>first third of the key travel, if you divided hammer travel by key travel >>you might get a ratio of 5.5 and for the last third of the key travel >>4.5. But the average for the overall travel is 5. >> >Exactly...and again.. the Law of leverages dictates that this be the same >for distances, weights, and speeds. >> >>The weight ratios as currently derived give ratios for action components >>in one particular position. >This is just wrong.. see above. This is a convincing argument. What weight ratios are you talking about that account for action components at all different positions? The weight ratios that I'm familiar with (and which I thought you were talking about) are Stanwood's. These are measured with the key horizontal, the hammer shank horizontal, and the whippen horizontal. In what way does the number derived from these measurements account for changes in action geometry as the action moves through its range? >>They don't account for differing ratios throughout the stroke. >They certainly do. And in the same exact maner as they do for distances. >You find a problem below with the 52 grams at start, and 48 grams at >end... but the same kind of problems exist relatative to distance and speed. > >>And I don't see how you could come up with an 'average' or 'overall' >>weight ratio. If it takes 52 grams to get the key started moving but >>only 48 grams to get it to finish its stroke, then the average might be >>50 grams. But if you put 50 grams on the key when it's up then it won't move >See above.. the same thing happens in distances and speeds. You can come >up with the same kind of on average overall ratio ... with the same degree >of validity as we do in any other way we measure the ratio. > >Richard Brekne >RPT, N.P.T.F. I agree that you can come up with an average number, but in what way is it useful? An average distance ratio will tell me how much dip I need to achieve the blow distance that I'm using. That seems like useful information. If I establish an average weight ratio what do I do with that number? Phil F Phillip Ford Piano Service & Restoration 1777 Yosemite Ave - 215 San Francisco, CA 94124
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