Macro question

[Guest]

I am trying to work through a problem.

 

I am trying to come up with a Macro to cut a part but I think there is a piece of information missing.

 

Here's what I do have

 

I have a starting diameter on a face which is the staring point of a radius, the radius value is known.

 

What I can not figure is an endpoint, which I could use to come up with a an arc length value and then trig a location. this radius transitions off into another radius which is again known but I suffer from the same lack of a known point for it.

 

 

Any one have any equation ideas that might work for calculating one of these end points?

 

I can't come up with a way to get an end point or either RAD

[bigjohn]

Have a go atHere John. See if it helps

 

 

John

 

 

-------

[Guest]

Thanks John,

 

no help to me though, at then end of this I am missing a piece of Z information. The only thing I have is Z0 and a set length but nothing establishing a relationship between any other geometry.

 

I think they are going to need to define a Z location before I can figure it all out.

[Sambo]

I'm pretty sure you've already tried it but ... how badly does constraining it by tangents mess it up?

 

Sam

[Guest]

Sam,

 

Constraining it by tangents work perfectly fine.

 

However, it ONLY allows me to trig out the upper X position. There is nothing that allows me to calculate the Z position. I have Z zero at the face and the first radius is define as 4/5 of the diameter so as the diameter is changed the radius will change and without some Z point I can not calculate for it. I put the SW sketch up only to show the compound tangency of the 2 radii.

 

The only known numbers in the sketch are the .078 rad, the Z face and the 1.5° angle, everything else is a calculated value off of the diameter

 

But through a Fanuc Macro I need to be able to calculate and define those endpoints so I that I may cut it at any size.

[crazy^millman]

You may not know the Z but you can easily figure it out with the know information here.

 

1st figure out the triangle with the .8694 side and the 11.64 angle. Using trig of Z = .8694 x tan 11.64. So with that Z on the sweep of the R2.200 is .1764. Then we do the same for the next one which is Z = .0471 / .6773. So with that Z on the sweep of the R.078 is .0695. So if we add the 2 together we get .2459.

 

Now do you want that in a macro for a Fanuc format or does that help out?

 

Bigjohn that site is a spammers paradise.

 

HTH

[Thee Rickster ™]

Does this help

.

.

.

[bigjohn]

I didn't know Ron. What site should I use to up load. Can you suggest any. I have tried the cadcam's FTP Site before but it won't work for me.

 

Thanks Mate.

 

John

[gaijin_2007]

Wish I could figure out how to put pictures up!

 

Yes, this can be done from the info you have.

You have 2 triangles.

 

1. You have a triangle formed by the intersect of the 1.5 deg line and centre of the 0.078 rad. The hypotenuse is 0.078, the angle is 1.5 so the adjacent length (cos 1.5 x 0.078) is the distance in X from 2.75 to the 0.078 circle centre.

 

2. We know the centre of the 2.2 rad is X1.833 Z-2.2

 

3. Because the rads are tangent we now have another triangle with a length of 1 side being 2.2 (drawn from the centre of the 0.078 rad to the centre of the 2.2 rad) and another side being the X centre position (from 1) of the 0.078 rad minus 1.833 (drawn vertically from the centre of the 0.078 rad to X1.833)

 

4. Now we can calculate the angular sweep of the 2.2 rad and the rest is golden.

 

If I figure out how to put a pciture up I'll do, unless you can figure it out from the above ;-)

[bigjohn]

Upload your jpg file to http://www.photobucket.com and send the link by clicking on URL from Instant UBB Code. and pasting the url / kink between the codes or click on image and paste the url in window . Call it what you like in the second window.

 

 

Good luck

 

 

John

[Guest]

Ron,

 

Yes I can figure it when values are known but the only given values are diameter and length, the .078 RAD is constant but all other values are factors of the input dia.

 

Those shaded values are snapped off of Solidworks, they do not exist on the print

[gaijin_2007]

So you only get:

 

1. The start position of the 2.2 rad (variable, but in the example shown X1.833 Z0)

2. The size of the 2.2 rad (variable)

3. The size of the 0.078 rad (fixed)

4. The 1.5 degree angle (fixed)

 

Do you get the X value for the intersect of the 1.5 degree line and the 0.078 rad?

The X2.75 dimension shown in the example shown.

 

Or is that a factor of one of the dimensions given, maybe the X start position of the 2.2 rad?

 

If you have either the X dimension of the 1.5 degree/0.078 rsd intersect or the Z dimension of the vertical face at the end of the 1.5 degree line then you're good.

Otherwise you're missing one more piece.

[Guest]

gaijin,

 

You highlighted the info I get.

 

I am missing any locational information about the arcs.

 

Yes I can calculate the start position at the tangent point of the .078 and the 1.5° angle but I can not tie a Z to anypoint as the arc locations are not defined at all on the z.

 

and I believe that is the missing piece information.

 

I think even the relative height of the larger rad would allow me to them calculate it's position.

[Thee Rickster ™]

Hi John,

 

I sent you a file even though I probably have no Idea if it will help ya or that i even get it

 

sorry if i am no help

just trying to understand yer problem.

 

Cheers

Rick

[JDowe]

If the 2.200 radius is tangent to the face

location 1.833 - then the center of the 2.200

arc is 2.200 in the minus x direction from

the 1.833 point - if the arc ends at the

endpoint of the 11.64 chord then a triangle

can be constructed from the endpoint of the

arc center using 23.28 degrees - this shold allow

caculation of the first Z location

[Guest]

Thanks guys

 

 

Ron went over this with me and agrees, after looking at the print he sees there is not enough information to define this as a macro.

 

I do appreciate all of the ideas and effort

[Andy]

John,

I worked on this problem for an hour. I assumed this was lathe job. (interesting challenge).

It looks like the point common to the two radii will vary in x and z as the diameter of the part changes. It seems possible to quantify algebraically but would be very complicated and alot of algebra for a macro.

Good luck.

Andy

[Guest]

quote:

---

It seems possible to quantify algebraically but would be very complicated and alot of algebra for a macro.

---

Andy thanks for trying.

 

You're correct, it "seems" like is "might" be possible but as I solved through as I could I kept running into essentially the same dead end.

 

If they can define for me just "one" more piece of the puzzle I am confident I can get it all wrapped up.

 

I drew it in Solidworks looking to see if at different sizes I could come up with a ratio of change based on the size changes. As I built through and measure and recorded size changes. What I found was the ratio changed and was not a constant. I suppose mathematically I could capture that but then you would be spot on, way to complex for a lathe macro.

[Tim Johnson]

On the drawing is that 2.75 dia at the tangent point of the .078 radius or at the intersection point of the 2.200 radius"?

[crazy^millman]

Tim the drawing has no size only ratios of one to the other. What he put up was his own doing with his own number not ones on the print. There was nothing more.

[gaijin_2007]

I think I'm missing something because I can't see why this can't be solved from the information you have.

Have a look at ftp://mastercam:[email protected]/Ma...ro_question.png

 

I'm guessing I'm wrong on the info supplied somehow?

 

EDIT: I changed some of the dimensions to make it easier to see the triangles.

 

[ 05-19-2008, 11:52 PM: Message edited by: gaijin_2007 ]