Reverse a formula to control an actuator between 0% and 100% with finer control rather then using a lookup table

Zyg

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Mar 28, 2019
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Hi All,

I have a work problem that I require a formula to be reversed it is a graph and I need to solve this so I can put the formula in to control an actuator between 0% and 100% with finer control rather then using a lookup table and picking 20 or 30 random points. I have used this formula to plot look up tables in excel but the control is not smooth as you can imagine every time I have a "gap" in the points i have chose my control is essentially stopped until the next point the controller is not smart enough to graph it.

y = (0.00000256x^4) - (0.0008x^3) + (0.0932Xx^2) - (4.95x) + 110

if someone could reverse that to find x or show me how to solve one part of the brackets that would be a great help.

Thanks in advance,
 
… need to solve this so I can … control an actuator between 0% and 100% …

… y = 0.00000256x^4 - 0.0008x^3 + 0.0932x^2 - 4.95x + 110 …

… reverse that to find x …
Hello. I'm not sure whether symbol y represents the percent (in your formula) because the smallest value of y generated by that formula is slightly more than 2.84 (i.e., y never reaches zero for any value of x).

However, if you desire a formula for x in terms of y, here's one:

x = 25/24*sqrt(12*(1159-192*y)/((3*sqrt(3)*sqrt(262144*y^3-4724032*y^2+27633984*y-46406965)-792*y+17432)^(1/3))+12*(3*sqrt(3)*sqrt(262144*y^3-4724032*y^2+27633984*y-46406965)-792*y+17432)^(1/3)+33)-25/24*sqrt(-12*(1159-192*y)/((3*sqrt(3)*sqrt(262144*y^3-4724032*y^2+27633984*y-46406965)-792*y+17432)^(1/3))-12*(3*sqrt(3)*sqrt(262144*y^3-4724032*y^2+27633984*y-46406965)-792*y+17432)^(1/3)+13230*1/(sqrt(12*(1159-192*y)/((3*sqrt(3)*sqrt(262144*y^3-4724032*y^2+27633984*y-46406965)-792*y+17432)^(1/3))+12*(3*sqrt(3)*sqrt(262144*y^3-4724032*y^2+27633984*y-46406965)-792*y+17432)^(1/3)+33))+66)+625/8

For example, assuming y is the percent and you're interested in 33%, then substitute y=33 above and evaluate to find x=25. In other words:

0.00000256(25^4) - 0.0008(25^3) + 0.0932(25^2) - 4.95(25) + 110 = 33

For the lowest percent generated by your formula (roughly 2.8434%), let y=2.8434 and find x≈104.29 (rounded)

For 100%, let y=100 and find x≈2.1 (rounded)

Again, I'm not really sure what you're doing with x and y, so let us know if I've misunderstood something. Cheers

?
 
Are you asking how to solve for x? The thing is practically a linear function so I would think that you could just keep the terms up to second order:
\(\displaystyle y = 0.0832x^2 - 4.95x + 110\)

and write it in standard form to solve:
\(\displaystyle 0.0832x^2 - 4.95x + (110 - y) = 0\)

and now use the quadratic formula to solve for x. in terms of y. It should be reasonably accurate.

-Dan
 
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