I have a set of given values and then three formulas and I need to find a value for "a" when K2 = K1. How would I go about finding this value. I think loops might need to be used but I don't know how to use it. Please help.
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K1 = 43.4
b = 80; %mm
t = 20; %mm
P = 150000; %N
alpha = a/b
F = (.265*((1-alpha)^4))+((.857+(.265*alpha))/((1-alpha)^(3/2)))
K2 = F*(P/(b*t))*((pi*a)^.5)
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Antworten (1)
Walter Roberson
am 27 Mär. 2018
If a is restricted to real values, then it is the first and 13th roots of the below
-(1/5) * root(25281*Pi*Z^24 - 10112400*Pi*Z^22 - 25887744000000000*Pi*Z^15 + 54198190080000000000*Pi*Z^13 - 17537236992000000000000*Pi*Z^11 + (134217728000000000000000*K1^2+6627262464000000000000000000*Pi)*Z^6 - 25098568335360000000000000000000*Pi*_Z^4 + 27987577248153600000000000000000000*Pi*_Z^2 - 7603404763299840000000000000000000000*Pi, Z)^2 + 80
Here, root(f(Z),Z) represents the set of values, Z, such that f(Z) = 0 -- the roots of the polynomial.
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Walter Roberson
am 27 Mär. 2018
syms K1 b t P a b Pi = sym('pi'); alpha = a/b; Q = @(V) sym(V, 'r'); %convert a floating point to symbolic rational F = (Q(.265)*((1-alpha)^4))+((Q(.857)+(Q(.265)*alpha))/((1-alpha)^(Q(3/2)))); K2 = F*(P/(b*t))*((Pi*a)^Q(.5)); sol_a = solve(K1 == K2, a);
The above gives empty symbol quickly, but
sol_a = solve(K1 == expand(K2), a)
has been calculating for a while.
Maple says the result is
sol_a = -RootOf(70225*P^2*Pi*_Z^24 - 70225*P^2*Pi*_Z^22 - 140450*P^2*Pi*_Z^15 + 735110*P^2*Pi*_Z^13 - 594660*P^2*Pi*_Z^11 + (1000000*K1^2*b*t^2+70225*P^2*Pi)*_Z^6 - 664885*P^2*Pi*_Z^4 + 1853544*P^2*Pi*_Z^2 - 1258884*P^2*Pi)^2*b + b
The RootOf can be expressed as
root([70225*P^2*Pi, 0, -70225*P^2*Pi, 0, 0, 0, 0, 0, 0, -140450*P^2*Pi, 0, 735110*P^2*Pi, 0, -594660*P^2*Pi, 0, 0, 0, 0, 1000000*K1^2*b*t^2+70225*P^2*Pi, 0, -664885*P^2*Pi, 0, 1853544*P^2*Pi, 0, -1258884*P^2*Pi])
for numeric coefficients.
You will need to filter down to the positive real-valued solutions:
sol_a(real(sol_a) >= 0 & imag(sol_a) == 0)
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