For all of you math geniuses out there..

[GrumpyOne]

I'm just trying to figure out the pivots and centers so it will be completely balanced horizonatally and vertically. Please take a look at the sketch and post the answers!

Edited February 6, 2010 by benos
replaced 2600 pix wide file with 600 pix file

[Pat Miles]

Center for the horizontal length (54") would be 27"

Center for the vertical height (2") would be 2"

The balance for the Polish Plate Rack that my club has is never the same as there is always some deviation of the weight on either side of the horizontal center. The imbalance is corrected by a permanently attached sliding weight that is adjusted after the plates are put back on the rack.

CYa,

Pat

[GrumpyOne]

Center for the horizontal length (54") would be 27"

Center for the vertical height (2") would be 2"

The balance for the Polish Plate Rack that my club has is never the same as there is always some deviation of the weight on either side of the horizontal center. The imbalance is corrected by a permanently attached sliding weight that is adjusted after the plates are put back on the rack.

CYa,

Pat

I'm not a math genius by any means, but if the angle iron is only 2" high, how can the vertical center be 2" high?. Besides, I figure if you put the center anywhere in the up right portion of the angle iron, the bottom will be heavier as it will have the full width of 2" plus however far up the vertical piece of angle the hole goes. It needs to be balanced so that it spins like a "fan" blade (for lack of a better term), right? So that even when it is completely vertical or on a 45 degree angle, with no plates on it, it stays still, and doesn't move up or down.

[High Lord Gomer]

This is a completely uneducated guess, so you get what you pay for, but...

The center as far as the length would be with the holes drilled centered at the 27" mark.

If it is to pivot somewhere along the vertical surface such that the same amount of material is above the pivot point as is below the pivot point, I believe it comes down to a fairly simple geometry problem.

In the above, all of the measurements are in 16ths of an inch, so the overall width is 32/16", the overal height is 32/16", and the thickness of the material is 5/16".

The equation for the area of the material above the pivot point is: 32*(5-x) or 160-32x

The equation for the area of the material below the pivot point is: (5*27)+(32*x) or 135+32x

If those are equal: 160-32x = 135+32x

Adding 32x-135 to each side yields: 25 = 64x

So, x=.390625 16ths of an inch, or .0244140625"

That means that the pivot point in the above drawing would be 5/16 - .0244 or .3125 - .0244 or .28880859375" from the top.

Of course, I didn't even drive by a Holiday Inn Express last night so please feel free to correct any mistakes you find.

[GrumpyOne]

Thanks Gomer! That's what I came up with, but wanted a second opinion before cutting and drilling.