Hello everyone i stuck on some basic equation i think.
This is my equation, where Im looking for alpha1 and alpha6, all other date is given.
Data:
Please can anyone explain me how i should define a1 and a6 in solver block? what im doing wrong? values which im looking for should be similar to this one on the drawing below, right? but answers are totally different.
This is result of my solution but its totally wrong
In attachment im sending mathcad file, please help me and also i apologize for my bad english :)Have a nice day
You have (only) two equations. This means you CANNOT solve for three unknowns (your find has three parameters).
If you try that, nevertheless, be prepared for rubbish answers.
Now there's something else wronmg as well. The k'.1 in the find() function, is NOT the k'.1 you defined earlier. Their labels are different (see how the defined one is an italic k, while the one in the find() function is upright k.
I suggest you remove the k'.1 from the find statement and see what you then get.
Success!
Luc
Then the next thing is that your two equations are pretty similar. The only difference is that the top one has cosines, where the bottom one has sines. With all the rest the same, you're bound to find no solution at all.
Are you sure these two equations are correct?
Luc
I guess you'll have to re-check your equation.
Its not clear from the figure you posted which length you are comparing and its not clear where the negative(!?) lengths X.a and X.j can be seen in that figure.
But it seems that the first equation compares horizontal lengths and the second compares vertical lengths. Nonethless the length X.a-X.j is present in both!?
BTW, your solve block works (after removing k1' in the find command" if I arbitrarily change some signs. I don't think that any of these versions is correct and none yields result which would correspond with your sketch, but it shows what you can expect if get the equations fixed.
BTW, you can avoid angles being too large by adding additional constraints in the solve block: