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Given three unequal force measurements spaced equally 120 degrees apart on a circular plate at F_R, F_C, F_L, I need to calculate what the estimated force measurements would be if taken at positions F_NW,F_NE,F_R, and F_L (90 degrees apart). All forces are perpendicular to the plate surface (straight down) and I'm treating downward as a positive force (into the screen on top view).

I have tried a few different techniques but nothing is giving me a sane answer. Any help or techniques are appreciated. Hopefully my question is clear.

Plate diagram - top view Plate diagram - ISO view

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    $\begingroup$ Welcome to Engineering! Could you please edit your question with info on what "few different techniques" you've tried, the results you got and why they don't seem reasonable to you? This helps us get a better understanding of where you're actually having trouble so we can focus our answers on that. $\endgroup$ – Wasabi Jan 25 at 23:57
  • $\begingroup$ something is missing here. The disc is supported on center? or just want 4-point equivalent of the forces and moments from 3-point? Anyway, write expressions for net force and moment about both N/S and E/W axis for both 3-point and 4-point setups, make them equal, do the algebra. $\endgroup$ – Pete W Jan 25 at 23:57
  • $\begingroup$ Hi thanks for your replies. @Wasabi - I will try to add some of my attempts. However, I think my main issue is just with the setup of the force and moment diagrams which is hard to convey. $\endgroup$ – tripleeagle10000 Jan 26 at 15:51
  • $\begingroup$ @PeteW - Sorry for omissions. The disc is supported underneath and is constrained from translation. Yes looking for force equivalents at 4 points, they are known at 3 points (F_L,F_C_F_R). I think the writing of the expression part is the part I'm struggling with :) $\endgroup$ – tripleeagle10000 Jan 26 at 15:55
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I am taking a leap of faith assuming some presuppositions including that The three force measurements represent three points on the elliptical intersection of a cylinder and sloped plane. The slope of this tilted plane is:

EDIT

After the OP's comment.

By angled surface I don't mean the disk is tilted, I mean an imaginary plane containing the endpoints of the measured forces. any 3 vectors directed orthogonally from points along the circumference of a level disk form an inclined surface. In this case because of the symmetry in the position of the measurements, 120 degrees apart, obligingly the two high measurements are equal and the minimum measurement will be the minimum on the entire hypothetical sloped disk. and the diagonal line starting from the minimum force is actually the slope of this plane.

End of edit

Therefore

$S_{plane}= \frac{1.5R}{F{max}- F_{min}} $

Because the two other measurements by symmetry have to be equal and the 120 point projects to $sin90+sin30= 1.5$

So the force at any given point is the projection of that point onto the line starting from $F_{min}$ and ending at a point diagonally across the ellipse at 180 degrees.

For example $F_{NE}= F_ {min }+ R \frac{ \sqrt 2} { 2}*S_{plane}*( F_{max}- F_{min}) $

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  • $\begingroup$ I should have clarified, the disc is not a sloped plane, it is parallel to "XY" plane. The top photo is a top view, the second is ISO. $\endgroup$ – tripleeagle10000 Jan 26 at 15:59
  • $\begingroup$ @tripleeagle10000, by the disk I meant an imaginary plane that passes through the endpoint of the vectors of the forces. I did understand the material is a level disk. If you feel you need more elaboration I edit my answer to clarify. $\endgroup$ – kamran Jan 26 at 16:44
  • $\begingroup$ Thanks for clarifications, however I'm not understanding the "R" term in your example. Also, what happens to Splane when F_C=F_R=F_L? Wouldn't the denom = 0? I'm probably missing something... $\endgroup$ – tripleeagle10000 Jan 29 at 22:08
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I think the purpose of this question is the look at the balance of forces and moments, which must be the same for the 3-force group and the 4-force group. If the 4-force group is supposed to be support forces, which isn't clear to me, multiply the results by -1.

Assume disc radius = 1. It could be any radius, but it will divide out and cancel when looking at the moments, and doesn't matter for the forces, so just use 1 for simplicity.


3 equations:

(eq1) net force

(F_C + F_L + F_R) = (F_NW + F_NE + F_SE + F_SW)

(eq2) moment about North-South axis (let east-side-down be positive)

F_R.sin(120) - F_L.sin(120) = F_NE.sin(45) + F_SE.sin(45) - F_NW.sin(45) - F_SW.sin(45)

(eq3) moment about East-West axis (let north-side-down be positive)

F_C - F_R.sin(30) - F_L.sin(30) = F_NE.sin(45) + F_NW.sin(45) - F_SE.sin(45) - F_SW.sin(45)

Note there are fewer equations than unknowns (the wobbly-table problem)! I think this means you could set one of the 4 to zero for simplicity. (but the direction of the opposite one may flip as a result). Alternatively, have the two pairs of opposites differ from the average of all 4 by the same amount (with opposite signs).


... hopefully that gets you going, bonus points if I screwed it up!

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You are trying to transform a 3 loads system (P system) to an equivalent 4 loads system (F system) that has identical pressure diagram below the disc. Assume the disc is a rigid body, it can be easily done by setting up the coordinate system, and solving by satisfying the equilibrium conditions [sum F_z, sum M_x, sum M_y]_P = [sum F_z, sum M_x, sum M_y]_F.

enter image description here

In the diagram above, if uniform pressure under the disc is assumed, by summing moment about the respective X and Y axes, it is evident that all forces must be equal, so no unbalanced pressure would result, that is P_1 = P_2 = P_3, and F_1 = F_2 = F_3 = F_4. On the other hand, the total loads in P system must equal to the total loads in the F system, so to produce an equal amount of pressure under the disc, thus sum P_i = sum F_i ---> 3P = 4F, so F = 0.75P.

Sticking to the concept stated above, you can easily transform load system from one to another with desirable end conditions.

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