Author Topic: [ non-lisp ] Rotary die distortion equation  (Read 5536 times)

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Strucmad

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #15 on: March 24, 2009, 07:09:45 PM »
How can I figure out the distortion between a rotary die and a flat surface? 

Ahhh. I think I get it... yes the die will be distorted (only in one direction) to compensate for the for the arc of the roller, guessing it should be elliptical..

Just to make sure we are on the same page,  are we talking about the difference of A & B

T.Willey

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #16 on: March 24, 2009, 07:14:29 PM »
I understand the process you illustrate, but having no actual experience, I can't think of any reason there would be variance due to principle or an "ideal situation" and any problems that occur seem, to me, that they would be coming from natural deviance in machine operation and the general chaos of happenstance.

How good are your operators?  That makes all the difference in the world that no engineer or designer can plan for.
Ummmmm.... operators.... ( don't want to get fired here )... Good!!

I just know what people that clam to know more tell me.  I can see why certain circumstances would affect the outcome of the pattern; mainly speed comes to mind.  If they are moving at the same speed, then I don't see why there should be distortion ( in a perfect environment, when no metals expand / contract when heated / cooled ), but those who are in the know, say there is.  I have no data to defunct their statement.  This is a big learning idea for me, so anything someone can find, stating one way or the other, I'm willing to read.

My thought was that maybe your tolerance is too tight for the equipment.

What is the tolerance?


Not sure, but I don't think they are too tight right now.  What I've heard is that, say for example the circle, will be out of spec by as much as half the circle at one point, and then at others, totally out of the rectangle pattern, until it comes all the way around, and will be back in spec for a little while.
Tim

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T.Willey

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #17 on: March 24, 2009, 07:15:33 PM »
How can I figure out the distortion between a rotary die and a flat surface? 

Ahhh. I think I get it... yes the die will be distorted (only in one direction) to compensate for the for the arc of the roller, guessing it should be elliptical..

Just to make sure we are on the same page,  are we talking about the difference of A & B

Yes.
Tim

I don't want to ' end-up ', I want to ' become '. - Me

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Strucmad

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #18 on: March 24, 2009, 07:41:47 PM »
O.k - Circle maths.

As per the pic. inprevious post.

The relationship between A & B..

You know 'A'

so using the cosine rule (A2 = 2.52+2.52 - 2x2.5x2.5 cos0) you can calculate 0 in terms of `A'

Then simply (B=0 x 2.5) will give you an equation of B in terms of A. - note 0 must be in Radians.

So you can plug in any `A' and get the `B' dimension

T.Willey

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #19 on: March 24, 2009, 07:44:01 PM »
O.k - Circle maths.

As per the pic. inprevious post.

The relationship between A & B..

You know 'A'

so using the cosine rule (A2 = 2.52+2.52 - 2x2.5x2.5 cos0) you can calculate 0 in terms of `A'

Then simply (B=0 x 2.5) will give you an equation of B in terms of A. - note 0 must be in Radians.

So you can plug in any `A' and get the `B' dimension

Time to go, but I will try this out on some real life prints, and see what comes up.  Thanks for taking the time.
Tim

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gile

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #20 on: March 25, 2009, 06:26:37 AM »
Hi,

I'm not sure having understood the request, but for calculating B (arc length) knowing A (chord length) and R (arc radius), you can do this (A have to be smaller or equal to 2*R):

(setq B (* R 2 (asin (/ A 2 R))))

With asin define as:
Code: [Select]
;;; ASIN Returns the arcsinus of a number in radians
(defun ASIN (num)
  (cond
    ((equal num 1 1e-9) (/ pi 2))
    ((equal num -1 1e-9) (/ pi -2))
    ((< -1 num 1)
     (atan num (sqrt (- 1 (expt num 2))))
    )
  )
)
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ElpanovEvgeniy

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Strucmad

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #22 on: March 25, 2009, 07:07:45 AM »
I was having fun with the math..Well sad but true..

Tim,

here is the formula for your situation (5 diameter)

cos-1 is the inverse cosine, probably called arccosine in your world..

equation craps out at A=5 anything under is cool
« Last Edit: March 25, 2009, 07:33:56 AM by PS_Port »

Strucmad

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #23 on: March 25, 2009, 07:09:07 AM »
You other guys had to go and ruin it with Lisp and Autocad... :lmao: :lmao: just muckin guys  :-)

gile

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #24 on: March 25, 2009, 09:05:30 AM »
PS_Port,

converting degrees to radians isn't necessary. LISP deals with radians.


Tim,

To use PS_Port formula (which returns the same as the one I posted), you can use this acos sub:

Code: [Select]
(defun ACOS (num)
  (cond
    ((equal num 1 1e-9) 0.0)
    ((equal num -1 1e-9) pi)
    ((< -1 num 1)
     (atan (sqrt (- 1 (expt num 2))) num)
    )
  )
)

(setq B (* R (acos (/ (- (* 2 R R) (* A A)) 2 R R))))


Evgeniy,

Here're 2 more routines to complete your arc stuff.
They return the arc angle knowing tha arc length and respectivly the chord length and the sagitta length.
These routines use the Newton's method to solve equations as, respectivly:
chord / arc = sin(ang/2) / (ang/2)
2*sagitta / arc = (1 - cos(ang/2)) / ang/2)

Code: [Select]
;;; ArcChord->Angle (Newton's method)
;;; Return the arc angle knowing the arc length and chord

(defun ArcChord->Angle (arc chord / k x)
  (setq k (/ chord arc)
x (sqrt (- 6 (* 6 k)))
  )
  (repeat 6
    (setq x (- x (/ (- (sin x) (* k x)) (- (cos x) k))))
  )
  (* 2 x)
)

;;; ArcSagitta->Angle (Newton's method)
;;; Return the arc angle knowing the arc length and sagitta

(defun ArcSagitta->Angle (arc sagitta / k x)
  (setq k (/ (* 2 sagitta ) arc)
x (* 2 k)
  )
  (repeat 6
    (setq x (- x (/ (+ (cos x) (* k x) (- 1)) (- k (sin x)))))
  )
  (* 2 x)
)
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Strucmad

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #25 on: March 25, 2009, 09:26:19 AM »
PS_Port,
converting degrees to radians isn't necessary. LISP deals with radians.

Thanks gile, I understand that Lisp deals with Radians..

I am not very fluent in LISP  :oops:, and would not have attempted to answer this question if it was in LISP. I am always amazed at the programming talent here, and someone (like yourself) would easily answer this.

Tim's title "[Non-Lisp] -- equation] just gave me enough incentive to give it a go..  :-)

Hope we helped Tim  :-)

James Cannon

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #26 on: March 25, 2009, 09:32:49 AM »
I love seeing manufacturing application topics like this come up.  Even though I'm out of the manufacturing environment, I still find the machinery and design concepts so interesting.  I love being able to observe and soak up what I can about it.

T.Willey

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #27 on: March 25, 2009, 11:00:40 AM »
Thanks for all the help guys.  I'm not sure if I will get to it today, but I will get to it this week, and then I will let you know what I find out.
Tim

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SEANT

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #28 on: March 26, 2009, 05:37:43 AM »
It is likely just a problem with this example but: Would the 16.25+ pattern actually fit on the 5 dia. roller?
Sean Tessier
AutoCAD 2016 Mechanical

T.Willey

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Re: [ non-lisp ] Rotary die distortion equation
« Reply #29 on: March 26, 2009, 11:02:49 AM »
It is likely just a problem with this example but: Would the 16.25+ pattern actually fit on the 5 dia. roller?

Don't know, as I just wanted to show an example.  How would one find out?  I guess if we find the needed space for one pattern, then we could see if it would fit on said roll.
Tim

I don't want to ' end-up ', I want to ' become '. - Me

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