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In Reply to: RE: Calling all mechanical engineers - lever problem ... :-)) posted by andyr on October 04, 2014 at 21:59:02
Now, what happens when we have different distances from each spring to the weight - say:
* 7.5"
* 8.25" and
* 10"?
Since it's no longer an equilateral triangle, the positions of the vertices are no longer fixed..... Too many unknowns to provide an answer.
Follow Ups:
And if I fix the angle of the triangle apexes, relative to the central point carrying the weight?
Regards,
Andy
"And if I fix the angle of the triangle apexes, relative to the central point carrying the weight?"
Then it would be solvable..... Some geometric calculations would be required.
And if I provide the exact angles and lengths, are you able to solve the equations, Todd?
Regards,
Andy
"And if I provide the exact angles and lengths, are you able to solve the equations, Todd?"
Either the three angles or the three sides would suffice.... Of course, with the distances you gave, you cannot just come up with arbitrary lengths.... (For example, one of the sides cannot be longer than the sum of the two longest distances you stated.) And if you give angles, you must also specify which lengths to to which angle. You'd have to do some calculating yourself to make the geometry work.
Here is a sketch of the setup:
I thought it easier to draw the 3 lines connecting the weight to the position of the 3 springs (than the triangle). The weight (8.44 lbs) is located at the central cross.
And thanks very much.
Andy
The three points you selected come very close to forming a "right triangle"..... Presume the force is still 9 lb., as stated above.
Above image has some added geometric calculations..... (Using side-angle-side, side-side-side, etc.) These calculations were to enable distribution ratio calculations. Using the opposite two points as a "fulcrum" for each ratio calculation.
The force on the upper LH point can be approximated as 9 lb x (8" / 15.5") ... [didn't calculate exact distances, the ratio if roughly the same] Comes out to 4.64 lb.
The force on the lower RH point can be approximated as 9 lb x (1.43" / 7.95") [treated like a right triangle] ... Comes out to 1.62 lb.
The force on the upper RH point can be approximated as 9 lb x (2.19" / 7.17") .... Comes out to 2.75 lb.
Now if you add the three numbers, you get 9.01 lb total distribution on the corners.... Close enough due to lost precision. (If you did precise calculations, it would add up to exactly 9.00... )
Hi Todd,
Using your modified diagram that I included in my last post - and the lever calculations I suggested (to take the arm weights into account) - I adjusted the position of the centre of the bearing slightly.
Then I did the final work on the chassis and assembled it today, and mounted the bearing & platters, and used lead shot to mimic the weights of my 2 arms. See here:
Using the diagram you most kindly provided and adjusting for the arm weights, my calculations showed that I had got the weights almost equally divided between the 3 springs - the weight on the one nearest the platter was 4.8oz heavier than the other springs (8%). So not exactly equal - but within tolerance, IMO! :-))
Today's actual measurements backed this up; here's a side-on shot of the position of the chassis:
I measured the distance from the ply base to the underside of the chassis to be:
* 26mm at the LH spring
* 27mm at the top R spring, and
* 27mm at the bottom R spring
... which indicates that the weights are almost equal! :-))
So I will send the chassis off for powder-coating on Monday ... thank you very much for your help. :-))
Regards,
Andy
OK, 9 lb weight for the bearing/platters was an initial estimation; if you take a look at my responses to Bill K, you will see that I have put the weight at 8.44 lbs.
So this makes the weights you calculated:
* upper LH point (A): 4.35 lbs (instead of 4.64 lb)
* lower RH point (C): 1.52 lbs (instead of 1.62 lb)
* upper RH point (B): 2.58 lbs (instead of 2.75 lb)
... which agrees very well with Bill's calculations. Which were:
* (A): 4.30 lbs
* (C): 1.53 lbs
* (B): 2.61 lbs
So thank you, Todd. :-))
Now, here is a slightly more complex scenario - which is the actual situation ... include the weights of the arms which are cantilevered out from the 2 RHS apexes of the triangle.
Here is your diagram - suitably modified:
I would be interested in what your opinion is of the resulting weight balance. As I see it, because the arms are cantilevered out from B & C, there is a negative action on the apex which is at the other end of that particular triangle side?
So:
* the 0.99 lb arm outboard from B reduces the weight at C, and
* the 1.65 lb arm outboard from C reduces the weight at A?
But how to calculate the resulting apex total weights, with these cantilevered weights? :-))
Is it simply according to the relative distances? So:
* at C, weight is reduced by the fraction 2.0/7.95 * 0.99 lb = 0.25 lb
* at A, weight is reduced by the fraction 1.6/16.9 * 1.65 lb = 0.16 lb.
But should this decrease in weight be added to the weight at the fulcrum apex ... so that the total weight increase along each triangle side, sums to the weight of each cantilevered arm?
Regards,
Andy
.
Andy, you are wasting people's time.
What do you need to do? Get to the point.
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