RIGHT ANGLED SOLID CONE W CIRCULAR BASE

Can anyone help to create a solid cone with a circular base? I always tend to make a right angled solid cone with an elleptical base. Please do help me. Thanks

 

What you're looking for is the elusive eccentric reducer. So far it can only be approximated in plain AutoCAD by routines like xtornd-demo posted by jochen. I think verticals like MDT can produce native eccentrics.

 

As I said earlier, you first post was not explicit. I have xtornd so if you want to give me the dimensions I will try to construct one for you.

 

No - that's not what he wanted. Look at his last post on Saturday.

 

Sorry, i didnt know he wanted eccentric. stupid me , i should learn to read.
will look into it, I havent solid modeled in vanilla acad for a many years,
but will take a stab at it without MDT or Inventor.

Regards

 

You can forget it - won't work in vanilla autocad. Tried it and also tried
rndtornd. Right now I don't know how to make it a solid.

 

to make a right angled solid cone with an elleptical base. Please do help
me. Thanks

Good news.. it's possible

Bad news.. it's only just possible :-(

Method, construct an over length cone with an elliptical base, with the
ratio of the axes of the ellipse in the correct ratio dependant on the angle
of the cone.

Make a box or other object with a face at the correct angle to form the base
of the cone and subtract it to cut the cone into a right angle.


Assume you want a cone with height H and diameter D.

Draw a line from 0,0 to 0,H to D,0 then a perpendicular to this long enough
to intersect an extension of the first line segment.
Extend the first segment to intersect the line and trim off at the
intersection.
Rotate this triangle about 0,H to get the mid point of the base on the Y
axes.
Move the triangle until the mid point is at 0,0

Call the cone command.
use the elliptical option, and use the base of the triangle as one axis, and
0,D to 0,-D as the other.
use the height of the triangle as the cone height.

Move the cone tip to 0,0,H
Rotate the cone about 0,0,H in the XZ plane until one edge is at D,0 and
the opposite edge is aligned with the Z axis.

Use a suitable block with a face on the XY plane to SUBTRACT off the
unwanted lower part of the cone.


--
Jonathan

Barnes's theorem; for every foolproof device
there is a fool greater than the proof.

To reply remove AT

 

Hey what is "rngtornd" anyway? Is it a commang in autocad or just a system variable?

Up to now nobody has any bright idea making an eccentric "solid" cone. =)

 

SORRY

I got the details of making the cone wrong, the X Z plane is O.K. but the Y
needs more work.

The method outline is sound, and I have a method working... but to limit the
possibility of embarrassing myself again I will be taking a little more time
and checking it again before I post it.
--
Jonathan

Barnes's theorem; for every foolproof device
there is a fool greater than the proof.

To reply remove AT

 

----- Original Message -----
From: "Jonathan Barnes" <>
Newsgroups: autodesk.autocad.drafting
Sent: Tuesday, March 29, 2005 2:53 AM
Subject: Re: RIGHT ANGLED SOLID CONE W CIRCULAR BASE

What I have is a procedure that *should* draw the required 'cone', when I
ran it, there was an error of 1%... either my geometry is out, or stacking
up of small errors has made this rather large error.

( for practical use a slight tweak scaling line 8 could produce a corrected
result... )

O.K. here goes.

draw line 1 from 0,H to D,0.
draw circle 1 centre 0,H, rad length of line 1.
draw line 2 from D,0 to bottom quadrant of circle1.
draw line 3 from 0,H through the mid point of line 2 to circle 1.
Mirror line 3 about the X axis, to form line 4.
draw line 5 between the ends of lines 3 and 4 ( 3+4+5 = triangle 1 ).
Copy triangle 1 so triangle 2 has the centre of it's base at 0,0.
draw circle 2 centred at the tip of triangle 2, passing through the
intersections of
the base of triangle 2 and the sides of triangle 1.
draw line 6 1/2 D long, from the bottom quadrant of circle 2 parallel to the
x axis.
move line 6 up by the distance between the triangles.
draw construction line 1 from the top of triangle 2 through the end of line
6.
draw line 8 from 0,0 along the x axis to the intersection with the
construction line.
Rotate line 8 -90 degrees about 0,0
Start cone, elliptical,
use the ends of line 5 as one axes, and the end of line 8 to define the
other.
define the height as the distance from 0,0 to the tip of triangle 2.
rotate the UCS 90 about the X axes.
rotate the base of triangle 2 90 about it's tip
( orbit to get better view point )
rotate the cone about it's tip, use the top of the rotated base as a
reference.
return the UCS to it's original orientation.
move the cones tip to 0,0,H.
draw a large circle centred at 1/2D,0
extrude the circle down for H
subtract the cylinder from the cone.

WHEW...

check for any error on the y axis length, scale line 8 appropriately before
repeating the creation of a new cone.....

anyone want to program this ?

Jonathan

Barnes's theorem; for every foolproof device
there is a fool greater than the proof.

To reply remove AT

 

Hi Igor,

I can see Iggy that you are miffed a bit. (Can I call you Iggy?) No Iggy is short for Ignatius. Guess I'll jus' call you 'E'. You didn't like my RE (above) and OC has jabbed at you a couple of times still related to another thread, SO you feel you have to poke back. Human nature. Hence your snotty RE. Or is that snooty? I don't know. BUT;

1) it remains (for the time being at least) undoable.
2) what you and the OP produced were and STILL are not true right angled solid cones with a circular base.

Yours is a polygon mesh and when exploded, you have 50 triangular 3D Faces with no right angles. And as I've stated before, a very nice looking "SURFACE MODEL". You could have 360 triangular 3D Faces and it would appear even better than now. But, it is still a surface model, NOT a solid model.

Now I am not (obviously not) a scientist / mathematician to be able to explain mathematically why this is so. And I think, why can't I take a perfect cylinder say 6" in diameter and 3" in height (same proportions as your model (100 x 50) and start whittling away and finally arrive at that perfect right angle cone? Why can't I do it in Cad? Donno! Just can't! Not smart enough I guess.

But keep pluggin' away and keep us posted when you achieve (or believe you have) that outcome.

 

If you explode the figure you get at the end for the base 'circle':

LIST

ELLIPSE Layer: "0"
Space: Model space
Color: 4 (cyan) Linetype: "BYLAYER"
Handle = b0e
Area: 11.5018
Circumference: 12.0223
Center: X = 0.0000 , Y = 0.0000 , Z =
0.0000
Major Axis: X = 0.0000 , Y = -1.9134 , Z =
0.0000
Minor Axis: X = -1.9134 , Y = 0.0000 , Z =
0.0000
Radius Ratio: 1.0000


I already had that a long, long time ago...... : P

 

AH well, Kudos for trying. If I read anymore of this, You'll have me in this too.
Good luck!

 

The Following is a response from Co-Worker. Don't shoot the messenger.
Quote begins now

"It sounds to me like you are a real productive guy in your workplace, to be so perfect. Here in the real world, we use measurements and rounding to required precision. It doesn't matter what manufacturing Industry you're in, thats how its done, and no measurement is perect.
You also failed to note the cone is not perfectly round as asked for, rather its of eliptical properties but this is as good as it gets. Hey I'm not perfect, Just Highly paid. Have fun with your little conundrum."
End Quote

**runs out of the forum quickly** as to avoid any tomatoe throwing ~