Finding a Midpoint

[KEPelton]

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

 

[Jason Wilder]

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:

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

 

[ffejgreb]

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.

 

[kkosbau]

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

 

[mhr]

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

 

[Jason Wilder]

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

"mhr" <> wrote:

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

 

[Doug Broad]

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:

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

 

[Josh]

(defun midpnt (pt1 pt2)

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

"KEPelton" <> wrote:

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

 

[KEPelton]

That works! Thank you very much.

-K

 

[KEPelton]

Thank you everyone! It works!

-K.E.

 

[Joe Burke]

One more...

 

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

"KEPelton" <> wrote:

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

 

[John Uhden]

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. :]

 

[Doug Broad]

Hi John,

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

Regards,
Doug

 

[Jamie Duncan]

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

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. :]

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))

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

 

[Doug Broad]

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

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

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. :]

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))

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

 

[Jamie Duncan]

thanks for the clarification.

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

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

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

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. :]

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))

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