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Here is a problem I have struggled on for hours. Hopefully it shouldn't be too difficult for someone more skilled at math.

Here I have my first function:

$$R_{oM} = \frac{p R_t^2 + 2 R_t R_n^2}{2 R_n^2 + p (1+ 3 c_M r_m) R_t}$$

All of the variables have positive values. This function is to be calculated individually on multiple objects, which each have their own separate values for $p$, $R_t$, & $R_n$.

Now I want to make a similar function that uses this first function as input. This formula applies when there is only one object:

$$R_{oR} = \frac{p R_{oM}^2 + 2 R_{oM} R_n^2}{D} = \frac{p R_t^2 + 2 R_t R_n^2}{2 R_n^2 + p (1+ 3 c_R r_m) R_t}$$

The formula on the right is like the first, $R_{oM}$, but with $c_M$ replaced with $c_R$. Now I must find the value of the divisor, $D$, for the above equation to be correct. $c_R$ will never be less than $c_M$.

$$0 \leq c_M \leq c_R$$

I also want $R_{oR}$ to work on a group of objects. Here is the complete formula that should work on a group of objects:

$$R_{oR} = \frac{p (\sum{R_{oM}})^2 + 2 (\sum{R_{oM}}) (\sum{R_n})^2}{D} \leq \sum{R_{oM}}$$

What must be the divisor, $D$, for the last two equations to be correct?

Electric-Gecko
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    Solve your last equation for $D$ to obtain

    $$D = \left(p R_{oM}^2 + 2 R_{oM} R_n^2 \right) \frac{2 R_n^2 + p (1+ 3 c_R r_m) R_t}{p R_t^2 + 2 R_t R_n^2}.$$

    Mathematically, not a lot is going on here. You are free to plug in the expression for $R_{oM}$ and try to simplify this / factorize it in way you want.

     – Andreas Tsevas Jun 01 '23 at 19:41
  • @Andreas Thank you. I tried this in Desmos, and it looks to be correct. I also managed to simplify the first definition I gave of $R_{oR}$, with your formula for $D$, and it became the formula I wrote on the right.

    Seeing how easy this was, I realized I needed to edit the question and add another constraint for the application of the problem. I will do some testing to see if your formula still works under this new constraint.

     – Electric-Gecko Jun 01 '23 at 20:17
  • @Andreas Unfortunately, the solution you wrote doesn't fit the last formula that I added. But I will play with this to see if I can make it work. – Electric-Gecko Jun 01 '23 at 23:33
  • Your equation for $R_{oR}$ gives a unique expression for $D$ - you just solve it for the variable as I did. There are no "other" solutions for $D$. If the expression for $D$ does not fit your new constraint, whatever it is, then that constraint is wrong / incompatible with your definitions of $R_{oR}$ and $D$. – Andreas Tsevas Jun 02 '23 at 10:41

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