Topic: Description on a distance

[Coder]

Hello everybody .

I have seen the below codes posted in this forum and I can't understand the logic thing of it .

Can anyone of you masters describe it for me please ?

Code - Auto/Visual Lisp: [Select]

1. (distance '(0. 0. 0.) (mapcar '- pt1 pt2))
2.  

Thanks in advance . 

[Lee Mac]

Assuming the variables pt1 & pt2 represent 3D points, the expression (mapcar '- pt1 pt2) is returning the vector from pt2 to pt1. Since a vector is simply a displacement, the vector can also be viewed as a 3D point relative to the origin, as shown by the following diagram:



Now (distance '(0.0 0.0 0.0) (mapcar '- pt1 pt2)) will return the distance from the origin to this point, i.e. the length of the vector from pt2 to pt1 (or indeed pt1 to pt2).

[irneb]

The question does beg why such is used though. Would the exact same result not be obtainable from:

Code - Auto/Visual Lisp: [Select]

1. (distance pt1 pt2)

[Lee Mac]

The question does beg why such is used though. Would the exact same result not be obtainable from:

Code - Auto/Visual Lisp: [Select]

1. (distance pt1 pt2)

Exactly. 

[Coder]

Thank you so much Lee for that clear explanations 

Regards .

[Lee Mac]

Thank you so much Lee for that clear explanations 

You're welcome 

[CAB]

It also returns the x & y delta values (Vector Difference).
(mapcar '- ' p2 p1)
(mapcar '- '(30 20) '(10 10))
(20 10)

p2 is 30 units greator on the x axis
p2 is 20 units greator on the y axis


and this can be used to get the distance

_$ (mapcar '- '(30 20) '(10 10))
(20 10)
_$ (defun sqr (S) (* S S)) ; needed subroutine
SQR
_$ (mapcar 'sqr (mapcar '- '(30 20) '(10 10)))
(400 100)
_$ (apply '+ (mapcar 'sqr (mapcar '- '(30 20) '(10 10))))
500
_$ (sqrt (apply '+ (mapcar 'sqr (mapcar '- '(30 20) '(10 10)))))
22.3607
_$ (distance '(30 20) '(10 10))
22.3607