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Mathcad help :(

SOLVED

Re: Mathcad help :(

Okay, to get started, the title of the paper.

Graphic-analytical study of the curve of intersection of two cylinders of rotation

Pic_0415_057.jpg

R - circles radius (same size)

d - distance between the cylinders axes

M (m,m') - intersection point

x = Rα (from the horizontal projection)

α = x/R

Re: Mathcad help :(

So at least a sketch worth the paper drawn on.

As far as I see its exactly the same thing we had before with one exception. The distance d is now more meaningful. In the equation of your first posting the variable d was the difference from R and the new variable d now, This would affect the equation, of course as d=0 now means we have two ellipses as intersection and therefore two sines as flat projection, and d=R would mean a curve without inflection points - exactly the opposite from the prior post.

The equation would be

equ.png

and if the goal is graphing, then Alan had already provided the way to do.

Think its time for you to fire up Mathcad and create your own worksheet. Come back if there is anything not working.

BTW, in your sketch (is it yours?) you have different x-axis with different meaning - can be confusing.

Re: Mathcad help :(

Well, from the paper i have y^2= R^2 - (d-x1)^2, but thats so close.

I'll get to work.

Re: Mathcad help :(

Alexandru Hera wrote:

Well, from the paper i have y^2= R^2 - (d-x1)^2,

The comes from the shades triangle in the vertical view. Not you would have to calculate x1. In the horizontal view you can see that x1=R*cos(alpha) and alpha=x/r (different x here) as of the unrolling of the cylinder.

So the formulas match.

Re: Mathcad help :(

So i just need to write the formulas and make a graph, or its needed something more complex like Alan did?

Re: Mathcad help :(

Alexandru Hera wrote:

So i just need to write the formulas and make a graph,

Depends on which formula you are speaking of. If its the formula we discussed in the last posting the answer is no. In a perfect world you could do so, but Mathcad unfortunately does not allow to plot implicit functions. So you will have to write in in terms of y(x):=sqrt(...) and so you will only get the upper half if you plot it. A quick and dirty way to overcome this would be to plot two curves y(x) and -y(x) and format them to look the same. Alan had chosen a more compact, elegant way in that he created vectors for the x and y values which incuded all the necessary points and the whole curve would be plotted by plotting a single expression and not two.

Second Alan had provided a vector for various values of d and plotted all the corresponding curves which might have been confusing.

You will see, no matter whether you use Alans approach or the quick and dirty one sketched above that the plotted curves will not join nicely at the left and right "ends". This is because Mathcad will distribute the points evenly on the x-axis which means the plotted points can be fairly far apart at almost vertical areas in the plot. And Mathcad won't connect the upper and lower half of thr graph - on the first appraoch because it woul not connect two different functioins plotted and in Alans approach beacuse of the NaN (respective complex values, see below) separating the halves.

Alan had provided a way to overcome this or at least lessen that effect as he provided a variable to chose the number of points.

At last the definition of the function. It look more difficult as Alan had provided a way to avoid the function to return complex results - they are substituded for the Mathcad special "constant" NaN (not a number). As far as I see, this would not be necessary, as the plot react on complex numbers pretty the same way as on NaN.

He made the function dependend on R and d, too, as to provide for more flexibility. I would strongly suggest to do so.

So you see, it can be done quick and dirty, but more comfort or more features add complexity.

So a quick and dirty approach would look like the following pic. Start from there and get more aquainted with Mathcad.

quickdirty.png

Re: Mathcad help :(

Thank you sir for this detailed explication.

One question before i start working, how can i make a 3d graph similar with the one you provided? mathcad?

Re: Mathcad help :(

Alexandru Hera wrote:

Thank you sir for this detailed explication.

One question before i start working, how can i make a 3d graph similar with the one you provided? mathcad?

I did it with Mathcad 15. Prime's 3D capabilities are rather limited compared to Mathcad 15 and so a converted sheet would not show anything. I tried it and that was the reason for not attaching the Prime worksheet. Nevertheless I am sure it would be possible to graph the two cylinders and the ciurve of intersection in Prime, too, although Prime lacks a lot of the formattings which Mathacd 15 offers and so it wont be that good looking. Also the creation of animations, like the one I attached at that post in question, is a Mathcad 15 feature which is lacking in Prime.

Best choice of course would be the use of a 3D CAD program, if you have access to.

Re: Mathcad help :(

I've tried to reproduce the results, failed at plot

Re: Mathcad help :(

Alexandru Hera wrote:

I've tried to reproduce the results, failed at plot

Whats wrong? Plots look OK here?

You don't see the lower half of y8 because above you have written y15:=stack(... instead of y8:=stack(...

You did this with other vectors, too, so I guess you are not sure what that command y8:=stack(y8,-y8) is for.

It simply stacks the points of the upper half (y8) on top of the points of the lower half of the curve and it would make more sense to assign it a variable with the same number (y8). As you have written it you would have to draw y2 to see the plot for d=14 which doesn't seem to be that comfortable.

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