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bjyuen12-Amethyst
12-Amethyst
November 15, 2016
Solved

LISP for part rolling a angle by step

  • November 15, 2016
  • 13 replies
  • 6039 views

Hi~


I programed a LISP for a blank to roll one angle 0.9° step by step.( see the attachment LISP file)

Because position_pa command is a terminate action,so I use sd-call-cmds function in the LISP program.

When I ran the program to roll the part 0.9°，actually rolled angle was 51.566202°.

example-initial position.bmpexample-ran position.bmp

Does anybody could tell me why?

How can I  achieve my purpose to roll 0.9° step by step?


Thanks

Justin


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Best answer by ThomasKirkman

Justin

I am not 100% sure of the answer, I can give a best guess.  The calculations are done in lisp which requires radians.  Modeling takes a degree input and converts it to radians for calculations.

13 replies

F
FriedhelmK17-Peridot
17-Peridot
November 15, 2016

Hello Justin,

I do not know much about Lisp programming,

but I see the cause in radiant.

1 rad = 180 ° / Pi = 57.295779513082320876798154814105°

° 0.9 = 51.566201561774088789118339332695° 

It revolves around 0.9 radiants and not around 0.9°

Best regards

Friedhelm

B
bjyuen12-AmethystAuthor
12-Amethyst
November 16, 2016

Friedhelm,


I know that 0.9rad equal 51.566202degree.

In the LISP using position_pa action have to set the angle in degree. So I set a variable angle_factor_deg.

22.bmp

Ran the LISP，you could see variable angle_factor_deg value was 0.9.

11.bmp


Justin

T
ThomasKirkman16-Pearl
16-Pearl
November 16, 2016

A couple of tips  These will replace some of the work you did.

(sd-deg-to-rad angle)

(sd-rad-to-deg angle)

Lisp works in radians despite what units you have set.

So simply change the line from

:rotation_angle angle_factor_deg

to

:rotation_angle angle_factor_rad

regards

Tom

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