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How to find the value of R ?

Asked by dashty

dashty (view profile)

on 17 Feb 2013

If

R1 = 0.173*(E^3/2)*A1^1/3
R2 = 0.173*(E^3/2)*A2^1/3
R3 = 0.173*(E^3/2)*A3^1/3
Wi = (Ni*Ai)/M
for R1    Ni = 10  & A1 = 12
for R2    N2 = 8   & A2 = 1
foe R3    N3 = 4   & A3 = 16
  M = sum(Ni*Ri)
E = ( 1,2,...,10)

Then

1/R = sum(wi/Ri)

Find R

2 Comments

Greg Heath

Greg Heath (view profile)

on 17 Feb 2013

Please separate equations by putting them on separate lines.

Thanks.

Walter Roberson

Walter Roberson (view profile)

on 17 Feb 2013

Reformatted, but they are hard to interpret that way too.

dashty

dashty (view profile)

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5 Answers

Answer by Carlos

Carlos (view profile)

on 18 Feb 2013
Accepted answer

Is this what you mean?

>> A=27;
>> E=1:1:10;
>> R = 0.173*(E.^3/2)*A.^1/3
R =
    Columns 1 through 9
      0.7785    6.2280   21.0195   49.8240   97.3125  168.1560  267.0255  398.5920  567.5265
    Column 10
    778.5000

0 Comments

Carlos

Carlos (view profile)

Answer by Tae Yeong Kim

Tae Yeong Kim (view profile)

on 17 Feb 2013
Edited by Tae Yeong Kim

Tae Yeong Kim (view profile)

on 17 Feb 2013

It is very hard to interpret the equations. Could you form the equations neatly?

0 Comments

Tae Yeong Kim

Tae Yeong Kim (view profile)

Answer by Image Analyst

Image Analyst (view profile)

on 17 Feb 2013

Isn't it just

R = 1 ./ sum(wi./Ri)

????? Of course that assumes you already know what wi and Ri are, which none of us do.

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Image Analyst

Image Analyst (view profile)

Answer by Walter Roberson

Walter Roberson (view profile)

on 17 Feb 2013
N = [10 8 4]
A = [12 1 16];
R = ??
M = sum(R .* N);
W = (N .* A) ./ M;
Final_R = 1 ./ sum(W .* R);

Unfortunately the definition of R(i) is confused as it is based upon E which appears to be 1:10.

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Walter Roberson

Walter Roberson (view profile)

Answer by dashty

dashty (view profile)

on 18 Feb 2013

Thanks for all I need the value of R in this equation

R = 0.173*(E^3/2)*A^1/3

If the value of A = 27 and

E= (1,2,3,4,5,6,7,8,9,10)

0 Comments

dashty

dashty (view profile)

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