How do I "define" a semi-circle by an equation when Y is linear?
I know the cartesian formula:
x = 4 * cos ( t * 180 )
y = 4 * sin ( t * 180 )
z = 0
But I need to compensate for the trajectory (Y) which has a linear value.
I need the formula applied only to X given Y is linear 0-1.
I want to sweep a VSS to make this shape. The trajpar is the given value (t as Y):
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Dear Antonius
I have tried to create SEMICIRCLE BY SWEEP feature 
- it is working (you can investigate my model)
Note: I have used for sweep feature the Relation (for UPPER_HALF_OF_THE_SEMICIRCLE) sd4=(r^2 - (trajpar*r)^2) ^ (1/2)
Regards,
I think the Mathcad community considers us an odd bunch and I rarely get replies... go figure 
I think that is why we are "visual" people... Geometry is so much simpler to "grasp" - literally 
Dear Antonius
I have tried to create SEMICIRCLE BY SWEEP feature - it is working (you can investigate my model)
Note: I have used for sweep feature the Relation (for UPPER_HALF_OF_THE_SEMICIRCLE) sd4=(r^2 - (trajpar*r)^2) ^ (1/2)
Regards,
That's the one! I suspected I would have to divide it into quarters.
Here is the simplified "on the fly" version: sd4=sqrt((5^2 - (trajpar*5)^2))
And as a bonus, it covers ellipses 
Thanks Vladimir!
Gopal, you didn't have to remove your reply. It is definitely worthy of this discussion:
That's the formula of a circle x^2+y^2=radius^2
with value of Y varying from with trajectory parameter.
You are Welcome Antonius.
It was challenge for me too - I like Relations and Sweep feature too - it is powerfull combination.
Here is some old example
Regards,
Okay, I finally got exactly what I was after; a full semicircle.
By converting the trajpar to a range from -1 to 1; and making the initial value more than 0, it works:
sd3=5-sqrt((5^2 - (((2*trajpar)-1)*5)^2))
...((2 * trajpar) - 1)... => converts the trajpar to range from -1 to 1 and the Origin for the sweep is "10" (2 * R)
So we start off with "5" - Zero (5^2=25; 2*trajpar-1=2*0-1=-1 *5=-5 ^2=25; 25-25=0 ^2=0; sqrt(0)=0; 5-0=5)
When trajpar=0 (halfway) sd3=0 ; (5^2=25; 2*trajpar-1=2*.5-1=1-1=0; 0*5=0 ^2=0;25-0=25; sqrt(25)=5; 5-5=0)
All the "5"s are the radius value and can therefore be substituted wit relations or parameters.
I find it comforting that you can have a line go to zero within the range... but not so much when you cannot define it at the beginning of the range.
Again, thanks all for helping me picture this in my mind. Somehow I was thinking I needed the sin/cos functions to get there. Had I remembered the equation Gopal originally posted, I might have been able to work it backwards from there.
Some things just don't transfer to Creo easily. I went through 4 iterations to make this work, only to go back to the original to have it come out properly.
Since you cannot start out a VSS at zero, I have a tail on the swept line. The long edge is the semicircle we've been discussing. The sweep also has an angle change. Without the angle change, I could easily trim the surface with the initial sweep arc (origin), but once the angle was introduced, it failed the trim near the ends. Okay, merge? Merge failed! Revolve-remove? Failed! Merge flat sheet? Failed!
I went through several other scenarios to make this work including a half section mirror/merge and projecting the edge of a copy. They all created a small non-tangent deformation at the end.
Finally, I ended up extruding-removing a circle (arc didn't work either!) straight up through the sweep sketch. That finally did it.
Give it a try...
How can there be fewer mouse clicks when you have to do everything a half dozen times?
Cool stuff! I'll definately have to look at the equations, thanks!
It is the bill of a ball cap?
Good idea. I will show that.
Of course it is crazy, however It has two trajpar statements. But the sweep is a simple arc.
You are welcome, I'm glad you got it working, and with so few features, really interesting approach.
Actually, I never knew portion of a surface can be cut out with another on-laying surface in Creo.
So, just to say again, learn something new everyday.
