Finding a Midpoint

Hello. I am fairly new to LISP so I apologize in advance if this has already been answered.

I am trying to write a rountine that will ask the user for two points, determine the midpoint between these points and insert a block at that midpoint. Is there a way to do that? I have:

(setq sptd (getpoint "\nFirst Point?"))
  (setq eptd (getpoint "\nSecond Point?"))

How can I determine the midpoint between sptd and eptd? Thank you very much.

K.E. Pelton
SDG

 

Well, I hope someone could provide you a better answer, but two methods come to mind.



 



One - draw a temporary line, get the midpoint of the line, place the block, erase the line.



 



Two - mathematically calculate the midpoint.



"KEPelton" <> wrote in message news:...

Hello. I am fairly new to LISP so I apologize in advance if this has already been answered.

I am trying to write a rountine that will ask the user for two points, determine the midpoint between these points and insert a block at that midpoint. Is there a way to do that? I have:

(setq sptd (getpoint "\nFirst Point?"))
  (setq eptd (getpoint "\nSecond Point?"))

How can I determine the midpoint between sptd and eptd? Thank you very much.

K.E. Pelton
SDG

 

Rather than write a routine, why not use acad's internal calculator?  When prompted for an insertion point, type 'cal - mee.  Works perfectly and been using for years.

 

You do a little math to get the mid point.
(setq x1 (car eptd))
(setq x2 (car sptd))
(setq y1 (cadr eptd))
(setq y2 (cadr sptd))

then

(setq midx (/ (+ x1 x2) 2))
(setq midy (/ (+ y1 y2) 2))
(setq midpt (list midx midy))

I think that will work

 

(Setq pnt (Polar sptd (Angle sptd eptd) (/ (Distance sptd eptd) 2.0))

 

That's the ticket, forgot about that one.  Good call!



"mhr" <> wrote in message news:...

(Setq pnt (Polar sptd (Angle sptd eptd) (/ (Distance sptd eptd) 2.0))

 

Caveat:



POLAR is the wrong function to use unless you are doing 2D math.  It should



almost never be used for such a function.



 



If you're not going to use the built in calculator, then:



 



(defun mp (p1 p2) (mapcar '(lambda(a b)(/(+ a b 0.0) 2.0)) p1 p2))



 



 



"mhr" <> wrote in message news:...

(Setq pnt (Polar sptd (Angle sptd eptd) (/ (Distance sptd eptd) 2.0))

 

(defun midpnt (pt1 pt2)



 (mapcar '+  pt1 (mapcar '* (mapcar '- pt2 pt1) '(0.5 0.5 0.5)))
)




"KEPelton" <> wrote in message news:...

Hello. I am fairly new to LISP so I apologize in advance if this has already been answered.

I am trying to write a rountine that will ask the user for two points, determine the midpoint between these points and insert a block at that midpoint. Is there a way to do that? I have:

(setq sptd (getpoint "\nFirst Point?"))
  (setq eptd (getpoint "\nSecond Point?"))

How can I determine the midpoint between sptd and eptd? Thank you very much.

K.E. Pelton
SDG

 

That works! Thank you very much.

-K

 

Thank you everyone! It works!

-K.E.

 

One more...



 



(defun MidPoint (p1 p2)
 (mapcar '* (mapcar '+ p1 p2) '(0.5 0.5 0.5))
)



"KEPelton" <> wrote in message news:...

Hello. I am fairly new to LISP so I apologize in advance if this has already been answered.

I am trying to write a rountine that will ask the user for two points, determine the midpoint between these points and insert a block at that midpoint. Is there a way to do that? I have:

(setq sptd (getpoint "\nFirst Point?"))
  (setq eptd (getpoint "\nSecond Point?"))

How can I determine the midpoint between sptd and eptd? Thank you very much.

K.E. Pelton
SDG

 

That IS the best, BTW. Actually, it's just shy of the best. Use * instead of
/. But you knew that and was just checking to see if anyone was paying
attention. :]

 

Hi John,

Catching up on your NG reading after a nice vacation?
Ya got me.

Regards,
Doug

 

Why is that John/Doug? - to avoid possible divide by zero error if user
picks same point twice?

also, you use lambda and mapcar 'cause points are lists? why the 0.0?

Jamie

 

0.0 probably isn't necessary. The concepts are a holdover from
my experiences from earlier versions of LISP where adding two
large integers had unpredictable results or overflow. The 0.0
ensures that the result of the addition is a real rather than an integer.
The type conversion is automatic and does not require float.

If there had been a function called "average", then it could
have been written:
(defun midpt (p1 p2) (mapcar 'average p1 p2))

But since there is no buit in function "average", the choice is
define one or use lambda to create a nameless version.

Multiplication by 0.5 has been demonstrated by some in
the newsgroup to be faster than dividing by 2.

So this is probably enough:
(defun midpt (p1 p2)(mapcar '(lambda (a b) (* 0.5 (+ a b))) p1 p2))

Does that help?

The main reason I posted was to warn folks not to use
polar unless the z-coordinate value was unimportant.

Regards,
Doug

 

thanks for the clarification.

the 0.5 verus /2 seems insignificant unless one was performing this on
thousands of points.


--
Jamie Duncan

"Don't walk in front of me, I may not follow. Don't walk behind me, I may
not lead. Walk beside me and be my friend. "