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Aug 20 2020 09:46am
Hi,

Ive earlier tried to "turn this formula around" to give its answere in "E" but i failed. Could any1 help me?

T = R+L+C-E
D = 180-inverted cosfi((T^2+R) /(2*R*T))

Im also multiplying it with 180 and devide it by PI or smt to get it in angles instead of radians in excel. But i think thats not relevant and can be added later.

The formula is to calculate the duration in angles pr shankshaft rotation on the ports on 2-strokes, for those who wondered xD
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Aug 20 2020 01:26pm
There's a homework help forum.

https://forums.d2jsp.org/forum.php?f=257

I also just don't understand what you're trying to solve.

This post was edited by Thor123422 on Aug 20 2020 01:27pm
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Aug 20 2020 03:56pm
Quote (Thor123422 @ Aug 20 2020 09:26pm)
There's a homework help forum.

https://forums.d2jsp.org/forum.php?f=257

I also just don't understand what you're trying to solve.


Thanks, im trying to solve the equation to give E as answere.

E =....
Member
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Aug 20 2020 04:03pm
E = R +L + C - T

If that's all you want
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Aug 21 2020 06:20pm
syms T R L C E D
T = R+L+C-E
eq1= D == 180-1/cos((T^2+R^2) /(2*R*T))
solve(eq1,E)

Solving for E equals 1 of 4 equations

E = C + L + R - pi*R + R*acos(1/(D - 180)) - (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 - 2*pi*R^2*acos(1/(D - 180)))^(1/2)
E = C + L + R - pi*R - R*acos(1/(D - 180)) - (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 + 2*pi*R^2*acos(1/(D - 180)))^(1/2)
E = C + L + R - pi*R + R*acos(1/(D - 180)) + (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 - 2*pi*R^2*acos(1/(D - 180)))^(1/2)
E = C + L + R - pi*R - R*acos(1/(D - 180)) + (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 + 2*pi*R^2*acos(1/(D - 180)))^(1/2)
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Aug 21 2020 11:49pm
Quote (TheStealthTarget @ Aug 22 2020 02:20am)
syms T R L C E D
T = R+L+C-E
eq1= D == 180-1/cos((T^2+R^2-L^2) /(2*R*T))
solve(eq1,E)

Solving for E equals 1 of 4 equations

E = C + L + R - pi*R + R*acos(1/(D - 180)) - (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 - 2*pi*R^2*acos(1/(D - 180)))^(1/2)
E = C + L + R - pi*R - R*acos(1/(D - 180)) - (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 + 2*pi*R^2*acos(1/(D - 180)))^(1/2)
E = C + L + R - pi*R + R*acos(1/(D - 180)) + (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 - 2*pi*R^2*acos(1/(D - 180)))^(1/2)
E = C + L + R - pi*R - R*acos(1/(D - 180)) + (L^2 - R^2 + R^2*pi^2 + R^2*acos(1/(D - 180))^2 + 2*pi*R^2*acos(1/(D - 180)))^(1/2)


Thanks a lot :D
But why are there 4 answeres and how can i use that in excel when there are 4 different equation? :p
Member
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Aug 22 2020 07:25am
there are 4 equations, because you are solving a 4ish order polynomial, meaning that "E crosses" the "CLRD axis" 4 times. Each cross provides a slightly different answer, some times it may not cross, and you will just get an imaginary answer. In excel, you can put each equation in a different cell, then have a final cell to check which one is real. If you want an example, inbox me
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