Jump to content
henrynguyen

How to draw Cam lobe

Recommended Posts

henrynguyen

Hi everyone, when its come to autocad i'm a total noob. Thank to cadtutor.net and you guys i learn bits here and there. You guys are great sources when i have no where else to go. So i'm here today with this problem that have troubled me the past couple of days. I have included a picture of the drawing. I could draw the 45 radius Arc that goes from 0 degree to 90 degree. The other Arcs somehow i just can't. Hope you guys can help me on this

 

Many thanks

cam lb.jpg

Share this post


Link to post
Share on other sites
ReMark

Anyone able to help this member out? I tried but I always come up one arc off. Obviously I'm doing something wrong but I don't have the time to figure it out right now.

Share this post


Link to post
Share on other sites
easymike29
Anyone able to help this member out? I tried but I always come up one arc off. Obviously I'm doing something wrong but I don't have the time to figure it out right now.

 

I agree.

 

Gene

Share this post


Link to post
Share on other sites
ReMark

I'm going to *bump* this thread as I see JDM is back from his extended vacation. It must have been final exam week at the University of PepsiCola.:lol:

 

I'm willing to venture a guess he'll say it can be done using AutoCAD Inventor.:shock:

Share this post


Link to post
Share on other sites
Guest AARi

0000000000

Edited by AARi

Share this post


Link to post
Share on other sites
Bill Tillman

This looked like an interesting exercise and having a little time on my hands I gave it a shot. I'm not an ME but would be curious how close I came to getting this correct.

 

I started with the known center of the keyed hole and the 45R on the right. I then drew two circles, 61R and 80.6R respectively (red dashed circles) and just moved them using the quadrants and the centerlines of the keyed hole. Then I constructed the two lines which are 79° to each other, referenced to the 59° angular dimension given. I trimmed these red circles using the construction lines and then filleted the last part of it together using a radius of 47.3°. The ending cam shape I then joined as a polyline, it's shown in green. Like I said, I'd be interested to know if this is correct or not. I'm sure others may use different method to construct it. That's the nice thing about AutoCAD. There's always more than one way of doing something.

CamExercise.dwg

Share this post


Link to post
Share on other sites
Murph_map
I'm going to *bump* this thread as I see JDM is back from his extended vacation. It must have been final exam week at the University of PepsiCola.:lol:

 

I'm willing to venture a guess he'll say it can be done using AutoCAD Inventor.:shock:

 

Final exams :) I was with J.D. and ahhhhhhh yes it was final exams. ;)

If we had the cam profile along with the cam itself it's easy to create. But from what I know there's a lot of missing data to create it propery.

Share this post


Link to post
Share on other sites
Murph_map

Nothing from anyone else? Not even the OP? If anyone is interested I'll give it a shot and post the results.

Share this post


Link to post
Share on other sites
welldriller
Nothing from anyone else? Not even the OP? If anyone is interested I'll give it a shot and post the results.

 

i have been trying for the last 2 days off an on and i can not even come close with the information given. Wish i could look at Bill Tillmans drawing but LT 2000 will not open anything newer. Still going to work on it though.

Share this post


Link to post
Share on other sites
JD Mather

If we assume all of the arcs are tangent and they stop/start at the angles given - there is one solution. (none of the arc centers are on any of the construction lines, but it is fully defined with the given dimensions) (BTW - solution was found in less than 10 minutes.)

 

Tangent Arcs.PNG

Share this post


Link to post
Share on other sites
welldriller

CAM DRWG.jpgMr Mather Your drawing looks a heck of a sight better than mine but would like for some one to look at it for me.

 

i know that i am .01 off in the center but what else

Share this post


Link to post
Share on other sites
Guest AARi

]0000000000

Edited by AARi

Share this post


Link to post
Share on other sites
neophoible
If we assume all of the arcs are tangent and they stop/start at the angles given - there is one solution. (none of the arc centers are on any of the construction lines, but it is fully defined with the given dimensions)
Sounds right.
(BTW - solution was found in less than 10 minutes.)
In AutoCAD? Seems like someone guessed you would resort to Inventor. Hopefully, AutoCAD can do it, as well. Please post the DWG and explain. I can see how you can constrain the arcs to be coincident and tangent to each other. How did you determine the exact location of the shaft per the given angles? It looks like these are all interdependent and might require trial and error.

Share this post


Link to post
Share on other sites
neophoible
in microstation....
Are you saying you could not do it in AutoCAD? Where is the center of your shaft? Were you able to locate it per the problem?

Share this post


Link to post
Share on other sites
Guest AARi

0000000000

Edited by AARi

Share this post


Link to post
Share on other sites
JD Mather
Sounds right.

In AutoCAD? Seems like someone guessed you would resort to Inventor. Hopefully, AutoCAD can do it, as well. ..

 

The solution would be identical in AutoCAD - using parametric geometry constraints.

At some point I will try to get around to producing an AutoCAD solution - it is just much faster for me to work in a modern CAD program.

 

Here is the AutoCAD r2013 edu solution.

Put the hole for the shaft at the origin (0,0,0) I didn't bother with the trivial.

 

 

Cam Solution.dwg

Edited by JD Mather

Share this post


Link to post
Share on other sites
JD Mather
...seems somthing wrong ...

 

I don't see the specified angles in your "solutions" that were given in the original problem?

 

Can you turn off the grid in those programs to make it easier to visualize?

Share this post


Link to post
Share on other sites
neophoible
The solution would be identical in AutoCAD - using parametric geometry constraints.

At some point I will try to get around to producing an AutoCAD solution - it is just much faster for me to work in a modern CAD program.

Understood. Working out the solution is not quite identical, though, is it? If I ever get a modern computer, maybe I'll switch to a modern program!

 

Here is the AutoCAD r2013 edu solution.

Put the hole for the shaft at the origin (0,0,0) I didn't bother with the trivial.

Thanks, JD. I'll try to take a look when I get a chance to convert it to an earlier DWG format. If by trivial, you mean drawing the shaft hole, agreed. If you mean locating it to begin with, well, that seems to me to be a major part of the prob. In fact, it's crucial, as that is where all of the angles are being measured from, right? It seems to me that there would be no unique solution without it.

Share this post


Link to post
Share on other sites
JD Mather
Understood. Working out the solution is not quite identical, though, is it?

The solution is identical. Working out the solution is almost identical. AutoCAD doesn't automatically add coincident constraint to trimmed entities while Inventor does - cuts down on the amount of work a bit (AutoCAD solution took me 24 minutes, working on a laptop, would have been less on my desktop with dual screen larger monitors).

 

I'll try to take a look when I get a chance to convert it to an earlier DWG format.
I would have saved down myself, except that parametric constraints are critical to the way I found the solution.

 

 

 

If by trivial, you mean drawing the shaft hole, agreed. If you mean locating it to begin with, well, that seems to me to be a major part of the prob.

 

Rather obvious that it is a circle at the vertex of the angled lines. (keyway is also trivial)

 

I applied a Fixed constraint at the vertex (at the origin). If the fixed constraint is removed the entire cam can be dragged around the screen without distortion using any point as it is fully constrained (the position constraint being the only one defined by Fixed).

Share this post


Link to post
Share on other sites
Guest AARi

0000000000

Edited by AARi

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

×