5 ms·
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
by rahimnathwani 11mo 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.
- exe34 11mo agoI think it would be good if you could lock one of them.
- hamandcheese 11mo agoYou can.
- shomp 11mo agoWith # (octothorpe) #Val
- idiotsecant 11mo agoWhat is inputs a,b,c,d,and e are polynomial coefficients? I am hoping to get a fields medal plz respond.
- fouronnes3 11mo agoYou can actually solve fifth order polynomials with bidicalc! But it's a numerical solution, not an algebraic one, so no Fields medal.
- geon 11mo agoIt would make more sense to preserve the ratio if possible.
- pxx 11mo agoYeah, 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.
- rafabulsing 11mo agoTo get that, you could pass the ratio explicitly. C = 5A + 7B
- fouronnes3 11mo agoYou 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.