I am trying to solve this equation without success so far
am I missing something here or the equation simply does not have a solution?
mathcad prime 9.0.0.0
thanks!
ACCEPTED SOLUTION
Accepted Solutions
"idx2val" is a function I wrote.
The "match" function returns a vector consisting of 1 or more 2x1 vectors which contain row and column of the position the value was found. "idx2val" uses such a 2x1 vector as input and returns the corresponding alpha and phi value.
The function definition is to the right of the right margin so you may have to switch from "page view" to "draft view" to see it.
The function definition could also be seen in the screenshot I posted here: Link to reply
EDIT: The zero after "Match" is not a function argument so it does not appear in the argument list. Simply type an opening square bracket after the closing parenthesis in "match(....)" which is the keyboard shortcut for a vector/matrix index
You will have to define BB as a function of alpha to be able to take the derivative with respect to alpha.
I can't check if Prime will be able to give you a symbolic solution for the zero(s) of the derivative (phi, L, p and Mu are undefined) because I am limited to Prime express which does not allow for symbolic operations.
Usually the symbolics ignores the assume modifier or even chokes on it.
Do you really need a symbolic solution?
If you can provide values for the undefined variables you may use the numeric "root" function or a solve block with "find" to get a numeric solution.
BTW, the symbolics does not know anything about units and so it also does not know about the pseudo-unit 'deg'.
But you can tell the symbolics what 'deg' means by writing
somewhere at the top of the sheet maybe on the right hand not printable side or in a collapsed area. 'deg' should be labelled as "Unit" for consistency.
Did you realize that in your assume-modifier you used 'psi" but 'psi' does not appear at all in the function definition of BB. Probably you meant 'phi'. But as already written I doubt that the symbolics will respect that assume-modifier at all.
'phi' appears cos(phi) but also outside a trig function at the end - is this as it should be? But I guess its value does not matter anyway if you solve for alpha. But then, I doubt that Prime will be able to find a symbolic solution anyway.
EDIT: I have to correct what I had written. It its not mandatory to define BB as a function of alpha (although it would certainly be good style to do so). It would work OK the way you did as well. So if Prime comes up with "no solution found", it could either mean that no solution exists or that Primes symbolics is not capable enough to find it. In your case, I suspect that the derivative actually has no zeros.
So your function BB should be a function dependent on TWO arguments, alpha and phi? Is this correct?
Then a partial derivative wrt alpha would not have given you the desired extremum.
You would have to solve the system you get when you set zero both partial derivatives (wrt alpha AND wrt phi).
If you don't need a symbolic solution you sure can try to numerically solve the above mentioned system.
But you would have to provide numeric values for L, M.u and p !
Guess they should not all equal to 1 as ttokoro had chosen in the absence of a specification from you.
Once you provide numeric values, a solve block with "find" should be able to find a solution.
Another way would be kind of brute force. Create a matrix using a more or less dense net of alpha and phi values and use the "min" function to fund the minimum of all values. To get the indices of the corresponding alpha and phi values you could use the "match" function.
But again - this is a numeric approach and you would need to provide numeric values for the three unknowns.
A further option is to use the "minimize" function on function BB. You may use it in a solve block so you can add the additional restrictions for alpha and phi. Of course you still need values for L, p and M.u.
An additional problem is that your function BB may also return non-real values and you have to decide what to do with them.
If you want them to be ignored by 'minimize' and 'min' you may define and use a function which sets non-real values to a very high value, like
To avoid having to evaluate BB2 twice you may use a local variable to store it.
Here in MC15
Just gave the approach using "minimize" a try in MC15.
As you can see you don't get the absolute minimum that way but rather loacal minima which depends on the guess values provides.
Some guesses fail because the function would return a non-real result using these values
And here is the approach with the minimum of the matrix of values. I replaced non-real values by 10^3 which effectively lets them be ignored.
You may notice that the minimum found also depends on the mesh size you choose (nalpha and nphi). Changing it just by +/- 1 may slightly skip over the best value. So you have to try it with several values and keep the best.
The best I could come up with (but I used dummy values 1,2,3 for L, m.u and p) was -92.233 with phi right at the end at 85° and alpha=0.66.
If the minimum is actually at the end of the range for alpha (85°), then in all probability there will not be a horizontal tangential plane there and then setting the gradient to zero cannot provide the result.
But of course the situation can look completely different with different values for the three unknowns.
I attach the MC15 file. You should be able to convert it to Prime9 format and play around with it.
Thank you for the detailed answer.
Indeed, I have values for L, p and Mu, they are:
L = 5.6 m
p = 10.73 kPa
Mu = 158.9 kN
(these values might change but they shall not be far away from that range, say +/-20%), BB will be given in kN
1. first follow up question I have is: how do I convert the .xmcdz file to .mcdx? so i can open the file you provided on mathcad prime?
2. I was able to use the minimize function as you wrote it and obtain the correct results (I have the results to check them)
3. I am wondering how to combine the function B2 and pass it to the minimize function so that it always ignores non-real numbers (although it already seems to do depending on the initial guesses, as you showed
4. the approach using the matrix of values seems to be the most 'complete' (as minimize may result in a local minimum as you stated, and this approach seems to incorporate the B2). In trying to write this solution I ended up with the following error:
Using the values with the correct units showed me that I had typed the formula wrong in MC15 - the square root extended to far 😉
ad 1) When you installed Prime 9 you were given the option to check the box to install the legacy files converter. If you didn't you can run the installer again and add the converter. When the converter is installed you should be able to use it from the Input/Output tab. Its greyed out in the picture because I am limited to the free Express version which does not allow to use the converter:
A .xmcdz file is just a compressed (zipped, hence the "z") .xmcd file. I guess that the converter should not have problems using it, but I can't check so. If it chokes on the file, I sure can repost it in non-zipped format. But it looks like you already retyped most of the vital part anyway.
ad 2 & 3
To use B2 you simply replace BB by B2 😉
But when I do so in MC15, things go havoc and the values for alpha and phi suddenly are outside the desired range.
No wonder, we did not tell Mathcad that it should respect these ranges. To do so we have to use Minimize in a solve block. A solve block looks differently in Mathcad and in Prime. In Mathcad its simply started by the keyword "Given" and ends when you use a function like "find", "minerr" or "Minimize". All in between is part of the solve block and you can add additional constraints. In Prime the same functionality is achieved by inserting a solve block from the Math ribbon via "Solve Block".
I was surprised to see that a minimum was found at the lower end of the phi-range even though the guess was fairly large with 75°.
ad 4)
This method seems to give reasonable results but chances are that you miss a minimum with a "wrong" grid size.
As an example a grid size of 1500 for both values yields a minimum of -150.43 kN, but increasing the grid size to 2000 for each in the hope that a denser grid would give a more precise, even lower result, just gives us -144.49 kN.
Interesting, that phi is now at the higher end of its range near 85°.
But once such a value is found it might be of value to zoom in. Here we just look for alpha between 0.1 and 0.15 and phi between 83° and 85°.
Here again we get different, sometimes higher, results when we change the grid size, but usually all results are below -150 kN.
According the error you get: You have to use 10^8 N (or just 1 kN as I did in MC15) in the "if" function and not just the dimensionless 10^8. The "min" function can't work with matrices with mixed units. Whats smaller, 2 N or 5 V ? 😉
Modified MC15 in xmcd format attached
"idx2val" is a function I wrote.
The "match" function returns a vector consisting of 1 or more 2x1 vectors which contain row and column of the position the value was found. "idx2val" uses such a 2x1 vector as input and returns the corresponding alpha and phi value.
The function definition is to the right of the right margin so you may have to switch from "page view" to "draft view" to see it.
The function definition could also be seen in the screenshot I posted here: Link to reply
EDIT: The zero after "Match" is not a function argument so it does not appear in the argument list. Simply type an opening square bracket after the closing parenthesis in "match(....)" which is the keyboard shortcut for a vector/matrix index
I still cannot get the idx2val to work, sorry
I setup the idx2val as a function (see highlighted), same for the definition of idx2val itself
and yet I get this:
You could put the matrix creation and the calculation of alpha and phi in one program to get rid of the function "idx2val" altogether and keep all in one place. simply chose the dataset (2x3 matrix) to use and off we go
You can omit the division of Bmin by N in the program and the addition of N to BB.min in Prime becaus eon contrary to Mathcad Prime allows mixed units in matrices.
Finally here is a generic function for finding the minimum of a function in two variables by brute force.
Arguments are:
1) the function in question (should be unit-less in arguments and result)
2) a 2x1 vector with the low and high bounds of the first function argument ("x")
3) a 2x1 vector with the low and high bounds of the second function argument ("y")
4) either a 2x1 vector with the grid line numbers in x- and y-direction or a single scalar which is then used for both directions
5) a number which determines how often the algorithm should run. With each run only the small space around the minumum found in the previous run is considered, so this number indicates how far we zoom in. If the number is 1 the result is similar to what we had achieved so far by creating a matrix and using the min function.
To save memory the program does not create huge matrices and uses min on them but rather cycles through the grid, calculates the function value and keeps the smallest so far.
Examples show that its still important to use a larger number for the grid size because if the first grid misses the local min, further zooming in will not bring it back again. Its also quite sensible with respect to the grid number. 499 might give a good result while 500 may return a value not that good.
So experimenting and trial and error still applies.
Here is the function
And here some usage examples
MC15 sheet attached, the function is on the last page
