<!doctype html public "-//w3c//dtd html 4.0 transitional//en"> <html>   Phillip Ford wrote: <blockquote TYPE=CITE>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.</blockquote> All this is ok up to now... you left out a word like "balancing" above where I inserted it in paranthesis.. but ok. <blockquote TYPE=CITE>  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.</blockquote> And so it shall be. <blockquote TYPE=CITE>A 1 mm downward movement on the left side will still result in a 1 mm upward movement on the right side.</blockquote> ok..still fine. <blockquote TYPE=CITE>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?</blockquote> Here you go out to lunch. 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. Listen.. this is simple Archimedis. 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. But this is an equality Phil... you cant do that.  They (both sides of the equation) MUST be equal. So if you leave d1 and d2 alone, and change W1... then you change W2 in an exact equal amount. If you dont.. then you no longer have an equality, you no longer have balance, and you are no longer describing leverage. <blockquote TYPE=CITE>>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.  </blockquote> Exactly...and again.. the Law of leverages dictates that this be the same for distances, weights, and speeds. <blockquote TYPE=CITE>  The weight ratios as currently derived give ratios for action components in one particular position.</blockquote> This is just wrong.. see above. <blockquote TYPE=CITE>They don't account for differing ratios throughout the stroke.</blockquote> 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.   <blockquote TYPE=CITE>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</blockquote> 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.   <blockquote TYPE=CITE>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.</blockquote> Actually... one could look into this kind of thing...and personally I would think it would be just as interesting from all three perspectives. But all that is another story entirely... and to do so would be in the persuit of other matters entirely. Ron mentioned the desirablity to know the speed of jack travel at a certain point in the keystroke for example.     -- Richard Brekne RPT, N.P.T.F. UiB, Bergen, Norway <A HREF="mailto:rbrekne@broadpark.no">mailto:rbrekne@broadpark.no</A> <A HREF="http://home.broadpark.no/~rbrekne/ricmain.html">http://home.broadpark.no/~rbrekne/ricmain.html</A>  </html>