Subject:
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Re: SSClagorpion - Compressor
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Newsgroups:
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lugnet.technic
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Date:
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Wed, 16 Jun 2004 19:57:15 GMT
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Viewed:
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23428 times
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In lugnet.technic, Steve Hassenplug wrote:
> In lugnet.technic, Kevin L. Clague wrote:
> > Sometimes it is not easy to get perfect geometry out of LEGO.
> >
> > I've been wrestling with geometries to get the pumps just the right distance
> > apart, and also having the compression strokes force the pumps through thier
> > full range.
>
> This sort of work is much easier, if you're working with larger distances.
> Getting a circle between 2 and 3 studs wide to spin is NOT easy. Good luck.
Thanks..... Like I said using two 1x2's in an L shape got me the square root of
two (1.414) or diameter of 2.828..... All the holes in 1x2's are axle holes, so
the make the center connector of the radial compressor free to turn I had to use
a special part. I tried to find it on peeron, but no luck.
It has size pin holes around a central pin hole, same geometry as wedge belt
wheel except that the center hole is a pin hole, not an axle hold.
This morning I came up with the combination of a distance of 1.5 using a technic
CAM, combined with a 30 degree cutback (back towards the center) using this:
http://www.peeron.com/inv/parts/44374
It looks ugly, but it works quite well and provides very good diameter circle..
I'll get a picture up soon.
>
> >
> > I ended up using a little geometry and trignometry, and a C program to help
> > run through the possible geometric combinations...... Never done that
> > before.
>
> You should post a screen-shot or two. I'm interested to see what kind of
> data you came up with. I'm sure it's not written in a "presentable" format,
> but personally, I really don't care. I'd just like to see it. (Actually,
> I'm not even interested in the exact data, I just want have a look at your
> thought process...)
I was trying to study geometries using the technic liftarm rotor 3 blades....
http://www.peeron.com/inv/parts/44374
Assuming we create a hexagonal shale using six of them, and we can put spacers
between adjoining rotors, what kind of distances are there in the result.
Assuming an orientation like this, where A represents the spacers between the
rotors:
B
\__@ D
/
C @
A __/
\
What is the distance horizontal distance between the center of A to the center
of the spacer @D (center of the hexagon)? This one always turns out to be nice
integral multiples of 1/2 a stud. This was annoying at first because 1/2 stud
multiples were not helping me get the geometry I wanted. The assumption was I
was going to hook pistons at the center of A and B and have them point to the
center of the hex formed by the six rotors.
What is the vertical distance from the center of @D to the horizontal line
projected by rotor A, presuming we hook the pistons at the center of @C and @D
pointing toward the center of the circle. The distance from @C to the center of
the hex is the same as @D to the center of the hex.
These distance do not turn out to be nice multiples of 1/2 stud.
Here is the program I wrote to do the calculation of @D to center (+/- multiples
of 1/2 stud to see if we could get the distance we want).
#include "stdio.h"
#include "math.h"
main(int argc, char *argv[])
{
/* calculate pi and sin(60 degrees) */
double pi = -4*atan(-1);
double sin60 = sin(2*pi*60/360);
int L,M;
double h,x,y,l,m;
printf("pi = %g\n",pi);
printf("sin(60) = %g\n",sin60);
/* l is the amount of spacing between two rotors */
for (L = 0; L < 16; L++) {
l = L*0.5;
/* the hypotneus is two rotor arm lengths plus spacing */
h = 3 + 3 + l;
/* the vertical component of the hypoteneus */
/* soh, cah, toa */
/* in this case we have all the angles and the hypoteneus */
/* so I used soh = sin(a) = opposite over hyp */
/* a is 60 degrees, opposite is unknown. */
/* sin(60) = opposite / hyp */
/* opposite = hyp * sin(60) */
y = h * sin60;
/* the horizontal component */
x = sqrt(h*h - y*y);
/* print out spacing, hypoteneus and horizontal distance */
printf("L %3.2g h %3.2g x %3.2g y ",l,h,x);
/* for a range of stud distances going toward the center and
* away from the center staring at X,Y, print the distances */
for (M = -4; M < 5; M++) {
m = M * 0.5;
printf("(%3.2g) %6.4g ",m, y-m);
}
printf("\n");
}
}
I would have paid a lot more attention in trigonometry class if they'd given me
something interesting like this to solve..... Now I think trig is great!
I obviously added comments for presentation purposes. ;^)
In the end, I didn't need this information, because the 1/2 stud spacing that I
get from the center of A to the center of the hex is good. I just needed a
better radius for circle in the center.
On a personal note, I'm traveling out to Silicon Valley from the 26th through
the 1st. I have to be moved out of my current residence before then.....
bummer. Plus all my LEGO is unsorted (if your curious about *that*, email me).
Bottom line is I won't be making much more forward progress until July.
LEGO geometry sure can be challenging at times.
Kevin
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Message has 2 Replies:
 | | Re: SSClagorpion - Compressor
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| In lugnet.technic, Kevin L. Clague wrote: <snip> (...) Here is the picture. (URL) got the whole compressor back together and it is mechanically sound. Whether this provides a superior compressor or not, I've got to try to use the same geometry to (...) (20 years ago, 17-Jun-04, to lugnet.technic)
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Message is in Reply To:
| | Re: SSClagorpion - Compressor
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| (...) This sort of work is much easier, if you're working with larger distances. Getting a circle between 2 and 3 studs wide to spin is NOT easy. Good luck. (...) You should post a screen-shot or two. I'm interested to see what kind of data you came (...) (20 years ago, 16-Jun-04, to lugnet.technic, FTX)
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