| ▲ | rahimnathwani a day ago |
| Does one stay locked in place? Unclear.
If you set C1=A1+B1 then, when you set a value for C1, A1 and B1 are each half of that value, even if they started off unbalanced. |
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| ▲ | geon 21 hours ago | parent | next [-] |
| It would make more sense to preserve the ratio if possible. |
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| ▲ | pxx 15 hours ago | parent | next [-] | | Yeah, this concept is interesting but the fact that the simplest test case gives what's fundamentally a surprising result is very annoying. It also doesn't help that in the example, the expected outcome of 53.3333/46.6667 isn't even considered. | |
| ▲ | fouronnes3 14 hours ago | parent | prev | next [-] | | You can do this with bidicalc already! You just have to model the problem correctly. If you expect the ratio to remain constant, what you actually want is a problem with a single free variable: the scale. A1 = 1.0 // the scale, your variable
A2 = 6 * A1 // intermediate values
A3 = 8 * A1
A4 = A2 + A3 // the sum
Now update A4 (or any other cell!) and the scale (A1, the only variable) will update as you expect. | |
| ▲ | rafabulsing 15 hours ago | parent | prev [-] | | To get that, you could pass the ratio explicitly. C = 5A + 7B |
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| ▲ | idiotsecant a day ago | parent | prev | next [-] |
| What is inputs a,b,c,d,and e are polynomial coefficients? I am hoping to get a fields medal plz respond. |
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| ▲ | fouronnes3 a day ago | parent [-] | | You can actually solve fifth order polynomials with bidicalc! But it's a numerical solution, not an algebraic one, so no Fields medal. |
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| ▲ | exe34 a day ago | parent | prev [-] |
| I think it would be good if you could lock one of them. |
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