Problem with ellipse

Hey all,

I'm running into a problem with my LISP when creating an elliptical arc. I
first off can't seem to initiate the elliptical arc command, which is
probably why I can't get anywhere with this routine.

I have my routine set so that my pellipse variable is set to 1.

Here's the line that I'm having problems with;

(command "_ELLIPSE" "_A" p5 p3 p4 "0" "180")

It's pretty straight forward, but for some reason it will not envoke the arc
command within the ellipse command. Is there another way around this?

Thanks in advance,

Xolo

 

The ARC option is only available with PELLIPSE=1.
___

 

Sorry, I meant to say "The ARC option is NOT available with PELLIPSE=1."

It *is* available with PELLIPSE=0.
___

 

So, how do I get it to create an elliptical arc?

Do I have to create a full ellipse and trim it down somehow?

Xolo

Sorry, I meant to say "The ARC option is NOT available with PELLIPSE=1."

It *is* available with PELLIPSE=0.
___

 

Set PELLIPSE to 0. Then draw the elliptical arc.
___

 

Paul,

But I still want the ellipse to be drawn with pellipse set to 1. That way I
can join it to other pollylines and create a 3D model from it.

I really do need to be able to create it as a pollyline entity.

Xolo
Set PELLIPSE to 0. Then draw the elliptical arc.
___

 

In that case, you'll have to first draw the ellipse, then TRIM it.

However, joined polylines aren't the only way to create 3D objects. You
might try "joining" a "true" elliptical arc to other objects by creating a
Region, and using that instead of polylines.
___

 

Let me guess -- "Xolo" wants to make it a polyline, so it can be given
width, or so it can be offset without the result being a
gazillion-definition-point Spline. Maybe there are other possible reasons.

If so, making it a full ellipse and trimming it is one option, but hard to
do with something like a lisp routine, because you need the trimming
edge(s). You could probably come up with a lisp routine that would draw one
and then break it. It could use prompts similar to Ellipse's Arc option
when PELLIPSE = 0. But notice that there, the definition of the starting
and ending directions of the elliptical arc is relative to an axis of the
ellipse (it seems that zero degrees is at the opposite end of the major axis
from the side where you designated that axis), not relative to the
coordinate system. So to make it work similarly, you'd have to save some
directions in the process of making the polyline ellipse.

Kent Cooper, AIA

 

Paul,

I'm not sure that I follow you.....create a region from lines and elliptical
arcs (with pellipse set to 0) and then I can revolve the entity?

Just want to know if I'm following your information correctly.

Thanks,
Xolo

In that case, you'll have to first draw the ellipse, then TRIM it.

However, joined polylines aren't the only way to create 3D objects. You
might try "joining" a "true" elliptical arc to other objects by creating a
Region, and using that instead of polylines.
___

 

Hi Xolo,
I'm thinking we are working on the same kind of project. If you would be willing to post your mailing address I've got some routines that we could bang back and forth without clutter up the board here. I did have a routine that I was trimming the ellipse but abounded that if favor of just doing an arc ellipse. I've got another that I'm working on at the moment that does the outside, offsets it all for the inside and places either the skirt or a skirt and base plate. Just a routine in progress with a DCL...

Regards,
Scrutch

 

Yes.
___

 

Thanks for the nudge about regions even though it wasn't pointed at me... :~)

Thanks Paul,
Scrutch

 

I am working on something similar also. I want to prompt the user to select a start point (p1) and end point (p2). I want the quadrant of the ellipse to be .5" from the p1 and p2 points selected. How can this be done?

TIA

Collin

 

Collin,

Since no one else has taken a shot at this here's my attempt. Keep in mind that I'm not a hard core lisper and I'm sure there are a lot of different ways to do this. Probably some with a lot less code than I've used but I'm still learning the language.

(defun C:Elip (/ eco osm 9o p1 p2 px py pm pa p3)
(setq eco (getvar "CMDECHO");//save cmdecho
osm (getvar "OSMODE");//save osmode
);//end setq
(setvar "CMDECHO" 0);//set cmdecho
(setvar "OSMODE" 0);//set osmode
(setq 9o (* pi (/ 90 180.0) ) );//get 90 deg
(setq p1 (getpoint "\n Pick first point of ellipse: ") )
(setq p2 (getpoint "\n Pick first point of ellipse: ") )
(setq px (/ (+ (car p1) (car p2) ) 2) );//x midpoint
(setq py (/ (+ (cadr p1) (cadr p2) ) 2) );//y midpoint
(setq pm (list px py) );//midpoint of line
(setq pa (angle p1 p2) );//points angle
(setq ra (+ pa 9o) );//rotated angle
(setq p3 (polar pm ra 0.5) );//ellipse quadrant
(command "ELLIPSE" p1 p2 p3);//just do it...
(setvar "CMDECHO" eco);//reset cmdecho
(setvar "OSMODE" osm);//reset osmode
(princ);//quiet finish
);//end elip
(princ);//quiet load

Cheers,
Scrutch

 

Collin,

To be on the safe side you probably should insert the following in the above routine. This will assure that you are in Architectural mode while drawing the ellipse giving you the inches setting.

Insert in the beginning setq list...
lun (getvar "LUNITS");//save units
lup (getvar "LUPREC");//save angles
Insert in the upper setvar list...
(setvar "LUNITS" 4);//set units
(setvar "LUPREC" 4);//set angles
Insert in the lower setvar list...
(setvar "LUNITS" lun);//reset lunits
(setvar "LUPREC" lup);//reset angles

Don't forget to define the lun and lup in the defun variables list. Hope this is what you're looking for. I wasn't sure if you wanted the quadrant perpendicular to the midpoint or not.

Cheers,
Scrutch

 

Scruth,

Here's my e-mail address.



Yes, I think we're working on exactly the same type of LISPS too. I'm trying
to get an entire vessel drawn, but need to allow the user to make as many
choices as possible. I was toying with trying to create a 3D vessel too,
that's why I wanted to be able to revolve the ellipse after having drawn it.

Xolo

willing to post your mailing address I've got some routines that we could
bang back and forth without clutter up the board here. I did have a routine
that I was trimming the ellipse but abounded that if favor of just doing an
arc ellipse. I've got another that I'm working on at the moment that does
the outside, offsets it all for the inside and places either the skirt or a
skirt and base plate. Just a routine in progress with a DCL...

 

Files are coming at ya...<g>
I haven't been working on any of the 3D stuff with this project because I have a third party program that does all that stuffins for me. Just a bit hard to quickly produce a 2D drawing so that's why I'm into this routine.
Scrutch

 

There's ODDBALL STUFF GOING ON HERE! Even though I agree that the true
elliptical arc and region approach is probably better, I couldn't help
experimenting (in A2K4 if it makes any difference) with a way of duplicating
the effect with a polyline partial ellipse, which has to start with a full
ellipse and then be trimmed or broken to leave the elliptical arc. But
FASTEN YOUR SEAT BELTS.

The original posted problem line was:
(command "_ELLIPSE" "_A" p5 p3 p4 "0" "180")
which doesn't work with PELLIPSE set to 1, because the A option is not
allowed then.

Points p5, p3, p4 would obviously already be defined here. Whatever the
PELLIPSE setting, the defining of the starting ellipse is the same. So to
make a (complete) polyline ellipse with PELLIPSE set to 1:

(command "_ELLIPSE" p5 p3 p4)
gives you the same ellipse (within the polyline approximating limits) as the
true ellipse that it would be starting with when PELLIPSE is 0. Simple
enough.

Since the original post has a predetermined arc range making an exact
half-ellipse, we should be able to just break the polyline ellipse from one
quadrant point to the opposite one, right? It turns out to be REALLY HARD
to figure out how to do that with predictable results.

The nominal zero-degree direction in a TRUE elliptical arc is at one end of
the major axis, but WHICH end depends on the relationship between the three
points. It seems that if p5-to-p3 is the major axis, the zero-degree
direction is toward p5 (the first point given). But if p4 is farther from
the p5-to-p3 midpoint (the center of the ellipse) than p5 & p3 are, then p4
is what defines the major axis, but zero degrees is determined somehow not
by p4 alone, but by how it relates to p5 & p3: it's hard to describe except
to say that it seems the 90-degree direction is toward p3. The zero-degree
direction can be at or near P4, or it can be at the opposite end.

So depending on the relationship between those three points, if you were
making a partial true-elliptical arc, you might want to do it between p5 &
p3, or you might want to do it between the at-or-near-p4 quadrant point and
opposite that. For the same end-result ellipse, zero degrees will be at one
of the major-axis quadrant points, but which one will depend on the sequence
of the three points and their relative distances from each other.

That all determines where zero degrees is in relation to starting and ending
angles for an elliptical arc when PELLIPSE is 0. But if you BREAK a true
ellipse, it removes part of it from the first point you give it, in the
counterclockwise direction around to the second, as it does with circles, no
matter how it was defined.

It's WEIRDER, but different, in breaking POLYLINE ellipses (made with
PELLIPSE = 1). It took some experimenting and comparisons, but I think I
figured it out. It seems to consider the first point given as sort of like
zero degrees (similar to the way true ellipses work IF the first two points
make the major axis), whether or not that ends up being on the major or
minor axis (which is UNlike true ellipses), and no matter in what kind of
relationship the other defining points fall. And it DOESN'T CARE in which
order you give it two break points, or in which direction they're closer to
each other -- it breaks out the same part no matter what. It seems to get
both break points before it decides anything, then works its way around from
its nominal zero-degree first point, counterclockwise until it finds either
of the break points (including if either of them happens to be ON that first
point), and then it breaks out the portion of the polyline ellipse
continuing counterclockwise from there until it finds the other break point.
If either break point is AT that first defining point, it breaks
counterclockwise from there. If neither is at that point, the portion of
the ellipse containing that point is what stays, and the rest is broken out,
AND that first point becomes the common endpoint of TWO SEPARATE
partial-ellipse polylines.

Suppose you want to be left with the top half of an ellipse with its major
axis horizontal. Suppose its major axis is 4 units and its minor axis 2 (or
the half-ellipse is 1 unit high in the middle), and its left end is at 0,0.
Suppose that p4 is ON a quadrant point (equidistant from p5 & p3).

If p5 is (0,0), p3 is (4,0) and p4 is (2,1), you can do either
(command "_ELLIPSE" p5 p3 p4 "_BREAK" p5 p3)
or
(command "_ELLIPSE" p5 p3 p4 "_BREAK" p3 p5)
[The same is true if p4 is (2,-1).]

But if p5 & p3 are reversed, either of those would leave you with the BOTTOM
half instead. Apparently the only way to have the top half remain would
then depend on whether p4 is (2,1) or (2,-1). If the former, you'd need to
calculate the bottom quadrant point, which in this case would be (2,-1) and
call it, say, p6; if the latter, you'd have that location already. Then you
can Break from p3 to that bottom quadrant point. After that, the remaining
lower right quadrant of the original ellipse is now a different polyline
from the upper half (the two meet at p5), so you'd have to erase it.

If conditions are the same but the first two points define the vertical
axis, it's different. If p5 is (2,1), p3 is (2,-1), and p4 is (0,0) or
(4,0), you could break between p4 and its opposite quadrant point (which
you'd have to calculate). If p5 & p3 were reversed, you have to do a
similar breaking out of a quarter of it, from p5 to the right quadrant
point. But in this case it isn't like the one with p5-to-p3 horizontal --
since the break started at the first-point nominal zero-degree direction,
the remainder is now ONE 3/4-ellipse polyline, so instead of erasing a
leftover quadrant, you'd break between p5 and the left quadrant point.

It could get even more complicated if p4 isn't actually ON the ellipse
(which it doesn't have to be), because calculating its opposite quadrant
would be much more involved, and p4's effect as a break point could be
different.

I think LOTS of comparisons and analysis would be required to make all of
that act predictably under any possible combination of point sequence and
relationship. If the earlier routine (or part of one) that defines the
three points always does them in the SAME kind of relationship, it would be
simple enough to make something that would work with that particular
combination.

But maybe the better way would be to draw the ellipse, draw a temporary line
through the middle, and use that as a trim edge to cut half of the ellipse
off (then erase the line), because you can do that without worrying about
what direction the ellipse considers to be like zero degrees.

Kent Cooper, AIA

 

Kent,

Thanks for taking the time to investigate this. I think I also understand
most of it too.

Anyways, to create the ellipse, I only need 1/4 of it to remain once it's
created, so breaking it at p5 and p3 should give me what I want. I can then
try erasing the remainder from p4 and see if it works out.

Again, thanks a lot of all of this information,

Xolo

There's ODDBALL STUFF GOING ON HERE! Even though I agree that the true
elliptical arc and region approach is probably better, I couldn't help
experimenting (in A2K4 if it makes any difference) with a way of duplicating
the effect with a polyline partial ellipse, which has to start with a full
ellipse and then be trimmed or broken to leave the elliptical arc. But
FASTEN YOUR SEAT BELTS.

The original posted problem line was:
(command "_ELLIPSE" "_A" p5 p3 p4 "0" "180")
which doesn't work with PELLIPSE set to 1, because the A option is not
allowed then.

Points p5, p3, p4 would obviously already be defined here. Whatever the
PELLIPSE setting, the defining of the starting ellipse is the same. So to
make a (complete) polyline ellipse with PELLIPSE set to 1:

(command "_ELLIPSE" p5 p3 p4)
gives you the same ellipse (within the polyline approximating limits) as the
true ellipse that it would be starting with when PELLIPSE is 0. Simple
enough.

Since the original post has a predetermined arc range making an exact
half-ellipse, we should be able to just break the polyline ellipse from one
quadrant point to the opposite one, right? It turns out to be REALLY HARD
to figure out how to do that with predictable results.

The nominal zero-degree direction in a TRUE elliptical arc is at one end of
the major axis, but WHICH end depends on the relationship between the three
points. It seems that if p5-to-p3 is the major axis, the zero-degree
direction is toward p5 (the first point given). But if p4 is farther from
the p5-to-p3 midpoint (the center of the ellipse) than p5 & p3 are, then p4
is what defines the major axis, but zero degrees is determined somehow not
by p4 alone, but by how it relates to p5 & p3: it's hard to describe except
to say that it seems the 90-degree direction is toward p3. The zero-degree
direction can be at or near P4, or it can be at the opposite end.

So depending on the relationship between those three points, if you were
making a partial true-elliptical arc, you might want to do it between p5 &
p3, or you might want to do it between the at-or-near-p4 quadrant point and
opposite that. For the same end-result ellipse, zero degrees will be at one
of the major-axis quadrant points, but which one will depend on the sequence
of the three points and their relative distances from each other.

That all determines where zero degrees is in relation to starting and ending
angles for an elliptical arc when PELLIPSE is 0. But if you BREAK a true
ellipse, it removes part of it from the first point you give it, in the
counterclockwise direction around to the second, as it does with circles, no
matter how it was defined.

It's WEIRDER, but different, in breaking POLYLINE ellipses (made with
PELLIPSE = 1). It took some experimenting and comparisons, but I think I
figured it out. It seems to consider the first point given as sort of like
zero degrees (similar to the way true ellipses work IF the first two points
make the major axis), whether or not that ends up being on the major or
minor axis (which is UNlike true ellipses), and no matter in what kind of
relationship the other defining points fall. And it DOESN'T CARE in which
order you give it two break points, or in which direction they're closer to
each other -- it breaks out the same part no matter what. It seems to get
both break points before it decides anything, then works its way around from
its nominal zero-degree first point, counterclockwise until it finds either
of the break points (including if either of them happens to be ON that first
point), and then it breaks out the portion of the polyline ellipse
continuing counterclockwise from there until it finds the other break point.
If either break point is AT that first defining point, it breaks
counterclockwise from there. If neither is at that point, the portion of
the ellipse containing that point is what stays, and the rest is broken out,
AND that first point becomes the common endpoint of TWO SEPARATE
partial-ellipse polylines.

Suppose you want to be left with the top half of an ellipse with its major
axis horizontal. Suppose its major axis is 4 units and its minor axis 2 (or
the half-ellipse is 1 unit high in the middle), and its left end is at 0,0.
Suppose that p4 is ON a quadrant point (equidistant from p5 & p3).

If p5 is (0,0), p3 is (4,0) and p4 is (2,1), you can do either
(command "_ELLIPSE" p5 p3 p4 "_BREAK" p5 p3)
or
(command "_ELLIPSE" p5 p3 p4 "_BREAK" p3 p5)
[The same is true if p4 is (2,-1).]

But if p5 & p3 are reversed, either of those would leave you with the BOTTOM
half instead. Apparently the only way to have the top half remain would
then depend on whether p4 is (2,1) or (2,-1). If the former, you'd need to
calculate the bottom quadrant point, which in this case would be (2,-1) and
call it, say, p6; if the latter, you'd have that location already. Then you
can Break from p3 to that bottom quadrant point. After that, the remaining
lower right quadrant of the original ellipse is now a different polyline
from the upper half (the two meet at p5), so you'd have to erase it.

If conditions are the same but the first two points define the vertical
axis, it's different. If p5 is (2,1), p3 is (2,-1), and p4 is (0,0) or
(4,0), you could break between p4 and its opposite quadrant point (which
you'd have to calculate). If p5 & p3 were reversed, you have to do a
similar breaking out of a quarter of it, from p5 to the right quadrant
point. But in this case it isn't like the one with p5-to-p3 horizontal --
since the break started at the first-point nominal zero-degree direction,
the remainder is now ONE 3/4-ellipse polyline, so instead of erasing a
leftover quadrant, you'd break between p5 and the left quadrant point.

It could get even more complicated if p4 isn't actually ON the ellipse
(which it doesn't have to be), because calculating its opposite quadrant
would be much more involved, and p4's effect as a break point could be
different.

I think LOTS of comparisons and analysis would be required to make all of
that act predictably under any possible combination of point sequence and
relationship. If the earlier routine (or part of one) that defines the
three points always does them in the SAME kind of relationship, it would be
simple enough to make something that would work with that particular
combination.

But maybe the better way would be to draw the ellipse, draw a temporary line
through the middle, and use that as a trim edge to cut half of the ellipse
off (then erase the line), because you can do that without worrying about
what direction the ellipse considers to be like zero degrees.

Kent Cooper, AIA

 

Given the point sequence in your original code, breaking between p5 and p3
would appear to be cutting the ellipse in half, rather than leaving a
quarter of it, since those two points will be the opposite ends of one of
the axes. Point p4 may or may not be on the remaining part, anyway,
depending on the relationship. I assumed a half-ellipse because your
original code would make an arc from 0 to 180 degrees (if it were making a
true elliptical arc with PELLIPSE = 0).

If p4 is actually ON a quadrant point, you can break between p4 and one of
the others, but I think you'll still have to be careful about how you define
that, depending on the relationship of the points and which quadrant you
want to have left. Some varieties of break between p4 and another point
will leave you with a quarter-ellipse, but some will leave you with
three-quarters; some of the latter will leave you two partial ellipses, and
some of them one. That's why I hope the earlier part of your routine
defines those three points always in the same relationship, so you can
tailor the breaking (and erasing, if appropriate) to that particular way of
building the ellipse.

Kent Cooper, AIA


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