>Phillip Ford wrote: > > > Just because something is a lever doesn't mean that the leverage ratio > is going to be the same at every point along the lever. You pick > different points on the lever and you get different numbers. Richard Brekne wrote: >I think it goes without saying that if you measure different points you >will get different arm lengths and essentially a different lever. I have >made that exact point several times now. But changing the force vector >alone at either the input or output point will not alter the levers >leverage. It just will change the net force either applied to the lever, >or to the thing being leveraged. A force vector is a seperate, independent >component............. > >The point is that the factors that influence the effective ratio of the >system are the exact same for both speed, weight, and distance anyway you >look at it............... 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 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. 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. A 1 mm downward movement on the left side will still result in a 1 mm upward movement on the right side. 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? >What is needed is a common language for describing the overall action >ratio... "overall" meaning the net effect of all discrete points. We seem >to have no trouble conceptualizing this if its distance we are talking >about... but in reality its exactly the same thing for distance, as it is >for speed, as it is for weight. 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. The weight ratios as currently derived give ratios for action components in one particular position. They don't account for differing ratios throughout the stroke. 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. I think you have to be concerned about what the weight ratio is at certain points in the key's travel. You don't have to be concerned about what the distance ratio is at any particular point in the key's travel. >So it just remains to find a convention of measureing the distances of the >arms of the levers that corresponds to this SW ratio. To back up a bit, why do you think that the SW Ratio is the overriding thing that everything should be corresponding to? >Let me turn this all around a sec....Using the formula Ron Overs gives. I >could calculate the amount of weight needed at the key to balance 10 grams >placed on the tip of the hammer. I thought he said that you couldn't. According to his calculations the amount of weight would be 58 grams. According to Stanwood it would be 55 grams. > Now of course I could choose to do that for any given point through the > keystroke, and of course that would vary... but the overall action ratio > is not just any one of these points... it is the net effect of all of them. >Now on to a point of order.. more or less removed from the discussion >above............. > > >Richard Brekne I'll respond to this on your next post which had the diagram attached. Phil F Phillip Ford Piano Service & Restoration 1777 Yosemite Ave - 215 San Francisco, CA 94124
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