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"Expression is not a method" error
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Hello,
I'm trying to calculate a variable I called CLo in the following code and it gives me these two errors: Error 1: Expression is not a method & Error 2 Method arguments must be enclosed in parentheses.
Your help is very much appreciated.Private Sub solve_Click(sender As Object, e As EventArgs) Handles solve.Click Dim W, rhu, U, B, beta, CLo As Double W = Val(weight.Text) rhu = Val(density.Text) U = Val(speed.Text) B = Val(beam.Text) beta = Val(deadrise.Text) CLo (0.0065 * beta * (CLo ^ 0.6)) = W / (0.5 * rhu * U * U * B * B) TextBox1.Text = CLo Dim lambdaw As Double Dim Dimtau As Double Dim DimFnB As Double tau = Val(Trim.Text) FnB = U / ((32.2 * B) ^ 0.5) froude.Text = FnB lambdaw = (CLo / ((tau ^ 1.1) * (0.012 + (0.0055 * lambdaw * lambdaw) / (FnB * FnB)))) TextBox2.Text = lambdaw End Sub
 Edited by Reed KimbleMVP Friday, July 13, 2018 12:36 PM placed code in code block
Friday, July 13, 2018 10:32 AM
All replies

Turn Option Strict On as the first line in the form.
For those who want to help.
Dim W, rhu, U, B, beta, CLo As Double W = Val(weight.Text) rhu = Val(density.Text) U = Val(speed.Text) B = Val(beam.Text) beta = Val(deadrise.Text) CLo (0.0065 * beta * (CLo ^ 0.6)) = W / (0.5 * rhu * U * U * B * B) TextBox1.Text = CLo Dim lambdaw As Double Dim tau As Double Dim FnB As Double tau = Val(Trim.Text) FnB = U / ((32.2 * B) ^ 0.5) froude.Text = FnB lambdaw = (CLo / ((tau ^ 1.1) * (0.012 + (0.0055 * lambdaw * lambdaw) / (FnB * FnB)))) TextBox2.Text = lambdaw
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Friday, July 13, 2018 11:30 AM 
Just curious, what are you wanting to calculate? Critical Depth?
You will need to isolate the variable to one side of the equation.
ie
CLo = some function if the remaining variables
 Proposed as answer by Reed KimbleMVP Friday, July 13, 2018 12:31 PM
Friday, July 13, 2018 11:36 AM 
Thank you for your reply.
I'm creating a tool to get some performance data of a craft.
I was trying to isolate CLo but I couldn't. it comes in two different powers in the equation (CLo and CLo^0.6)
if I couldn't find a solution for this I will try to find another expression for CLo.
Friday, July 13, 2018 12:33 PM 
Thank you for your reply.
I'm creating a tool to get some performance data of a craft.
I was trying to isolate CLo but I couldn't. it comes in two different powers in the equation (CLo and CLo^0.6)
if I couldn't find a solution for this I will try to find another expression for CLo.
Masr,
If this is for school you should tell us.
We dont know how or what you are supposed to solve or how and things etc and we try not to just give you the solution.
Not that I know the solution.
Those equations are used for lots of flow things.
When you have two equations and two unknowns what do we do? Substitute.
Also quadratic sometimes. But yes your ^0.6 is tricky. BTW should that be 2/3 ?
Power series maybe?
But that really has nothing to do with the programming part.
However some equations are easier to program than others. Most have been done already.
:)
Friday, July 13, 2018 12:46 PM