How to trim lines on a 3d sketch

I have a sphere. It is 7.125" in diameter. The origin is right in the middle, and I have work planes offset from the origin planes, say, 8 inches. There are 4 such planes, and there are 2d sketches of 5.25" dia. circles on each. I made a 3D sketch and projected them to the surface of the sphere, using the 'project along vector' option in the project to surface command dialog box. Here's the question: how do I trim portions of that 3d sketch on the surface of the ball? See pic. Any input is appreciated.

First question is why do you need to trim?

When you project a circle onto a sphere - don't you end up with a 2d circle? (no 3D sketch needed)

Ah, yes, good point. Is there an existing algorithm for the diameter of a circle on the surface of a sphere with radius X, and Y units away from center point of sphere?

I don't know much about 3D CAD, I'm very early in my learning. So...for what it's worth....

 

 

I think this would be calculated not on 3D design but on 2D design. I'll be brief (or try to be): years ago I wrote a VB6 program to calculate saddles on pipe. That is, I wanted a program to calculate and draw a line I could print, trim physically the paper, and then wrap around a piece of pipe to mark so I could then cut the pipe to fit on another. In the process I had to find a formula to calculate the distance from a plane to the intersection of the curved surface of the pipe along a vertical line rising from the plane.

 

 

Picture it this way: if you lay a pipe on the ground and want to know the distance from the ground to some point along the curved surface of the pipe you need some algorithm to do so. I found one (but this was about 8 years ago and I don't recall where I found it). You can, too. What you could do is this: calculate the distance from the plane to the curved surface of the "pipe" (sphere's circumference) and then subtract that from the sphere's radius and find the radius of the circle which would, at that particular radius for the sphere and distance from the center point, rest/lie on the sphere.

 

 

I'm attaching an AutoCAD drawing.

 

 

The green circle represents the sphere as seen as a cross section at it's center point. The violet/purple line represents the horizontal distance from the sphere's center point and the vertical radius of a circle which would rest on the sphere's surface. The white line represents the tangential line which is what you measure from at some distance (the horizontal violet/purple line) from the sphere's center to the sphere's surface. The blue line represents the distance at that point which is calculable. Take the length of the calculated blue line and subtract from the sphere's radius to get the radius of the circle which would, at that distance from the sphere's center, rest on the sphere's surface.

 

 

I will go look for the formula...but hopefully this will point you in the right direction for now.

 

Remember, I'm not an expert and I could be wrong or I could have misunderstood your question.

 

--HC

Sample Radius Determination.dwg

Frustrating. I found an elegant solution to this years ago...if I remember correctly.

 

 

Regardless...here's some more help.

 

 

Any circle which will rest entirely upon the face of the sphere will, by definition, be perpendicular to an axis drawn from the center of the sphere.

 

 

The cross section of a sphere at its center is a circle.

 

 

A line which is perpendicular to a line drawn from the center of a circle will, by definition, be two thirds of a right triangle which is completed by a diagonal line.

 

 

Attached is another drawing.

 

 

Line C, the hypotenuse or distance is the radius of the sphere. Line A is the distance from the center of the sphere where you want to calculate the radius of a circle which will rest entirely upon the surface of the sphere at that point.

 

 

I swear there was an easy way to do this in one simple formula...but...this will work.

 

 

Calculate angle "a": ACOS (A / C)

 

 

Calculate angle "b": 90° - angle "a"

 

 

Calculate leg B (the radius of the circle you're seeking): A / TAN ("b")

 

 

I could be wrong about a simpler solution...it's been years. But the aforementioned formulas will get you there.

 

 

Rots o' ruck.

 

 

--HC

Right Triangle.dwg

Arrgh. Simpler still:

 

 

B (your circle's radius) = C * SIN( ACOS(A / C))

 

 

This is the type of stuff I do which makes me identify with my signature quote.

 

 

--HC

Yeah, it's simple. Draw a triangle on the origin plane and you will know the radius of the circle you need to draw. No 3D sketches or projections needed.

Oh, and it's not the yellow one, I don't know why that's there.

ANYWAY, back to the question, once I do get the circles on the face of the ball, I want to trim out those small eye-shaped areas where the circles intersect.

Attach your file here.

I didn't answer your original question.

To trim projected geometry you have to Break Link.

Here's the file. Thanks in advance. BALL.ipt

 

-matt

Also, I really need to find the answer as to why the circle I project onto one side of the ball also appears on the other! Look at this pic:

 

p.s. I considered what you said about there not being a need to create, but I couldn't solve the triangle, I knew only one angle (90) and one side (the radius of the circle i want to make on the ball)

I tried to "break link" between the parent 2d and its 3d sketch by right clicking on the 2d sketch. It's not an option on my right click menu.

The circle will be projected to any part of your selected "project to" face that intersects that circle in a direction normal to the plane on which the circle lies. Your sphere only has one face, therefore the face crosses that projected circle in two places ... so you get two projected circles.

 

If you don't want to have the circles projected to both sides of the sphere, you need to split the surface of the sphere into multiple faces.

Here's the file. Thanks in advance.

-matt

It will be interesting to see this part when you get it finished as I am pretty certain you are doing an astonishing amount of extra work to create the geometry. But I am not sure where you are going with it yet to remodel it from scratch for you to show you how I would do the model.

I'll try to keep involved in working to final solution, but I am a bit busy with work for now.

 

Are you reproducing an existing part? If so, can you attach picture of actual part?

Interesting indeed. Once I trim out the eye shaped portions where the circles intersect I will attempt to sweep a profile along it for a concave sort of groove. It is a ball for a valve.

Check it out. The lines in red need to go. The profile marked by the green lines have to stay. I know you know what to do. Help please pretty please.

-Matt

 

Once installed in the valve body, the ball will rotate 90 degrees about the x axis

Hello everyone,

 

I'm new to this forum, and relatively new to Inventor. I have never needed to use 3d sketches much, but I thought I would try this challenge. After breaking the link of the projected circles I went into the 2nd 3d sketch and included the circle of the first 3d sketch, trimmed them off and drew a 2d circle on each path and then swept the groove.

 

 

Unfortunately I am using 2014. Sorry, but perhaps the above info will help.

 

Dave

 

How did you break the link? (see my above posts i mentioned something about it)

How did you include the 3rd one in the 2nd one? How did you trim?? More screenshots, please please thankyou thankyou

I believe I was able to break it by right clicking on the projected loop in the browser. I expanded the 3d sketch until it exposed the loop and then right clicked. Of course I am using 2014, so it may be different.