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Barrel Spacer Plate Dimensions

For those of you with Calipers

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#1 Lt Stefan

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Posted 18 October 2009 - 06:48 PM

I have some questions for those of you who have the means of measuring the distance from the center axis of the turret on a gun to the center of the barrel. I finally figured out how to design them with CAD, but do not have the means of getting proper dimensions. I'm ultimately trying to compile an open-source list of spacers for almost all turreted guns that are widely used. I have a list of guns whose dimensions I need below. Any accurate help you can provide will be much appreciated. If I have the dimension there I do not need it, but if you know it's wrong say so. UK means unknown. Suggest any guns you want.

Dimensions from center to barrel:

-AT2K: .5"
-AT3K: 11/16"
-AT4K: UK
-SM1500: 45/64"
-SM3000 (Yellow/Green): 11/16"
-DTG: UK
-LB: 0.90
-Firefly: 1.273"
-RFSG: UK
-Berserker: 139/64"
-Lanard Turreted Shotty: UK
-Arrowstorm: 2.222"

The following require more complicated measurements:
-Big Salvo: UK
-Hornet: UK
-Blastfire (New): UK
-RF20: UK
-Wildfire: UK

If you see anything wrong, let me know.
Thanks everyone.

Edited by Lt. Stefan, 19 October 2009 - 08:49 PM.

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#2 TantumBull

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Posted 18 October 2009 - 06:54 PM

Some of the turrets you mentioned don't have as nice dimensions as the easy 2k turret. Get some calipers and make some measurements. One of the most accurate ways I've found is finding a tubing of sorts that fits as snug as possible in between the barrels of a modded turret. Wrap e-tape till this fits snugly, then use the calipers to get the OD.

(R of OD of e-taped tubing) + (R of OD ofbarrel material) = (D from barrel centerpoint to turret centerpoint)
Edit: I have sm3k turret dimensions somewhere. I may dig them up at some point if you can't do it yourself.

Edited by TantumBull, 18 October 2009 - 06:56 PM.

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#3 cheesypiza001

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Posted 18 October 2009 - 07:10 PM

Okay I have a few of the blasters you want measurements for.

Arrowstorm: 2.222" (from turret axis to center of barrel)
Firefly: 1.273" (from turret axis to center of barrel)

More to come later....maybe.

Edited by cheesypiza001, 18 October 2009 - 07:15 PM.

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#4 Lt Stefan

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Posted 18 October 2009 - 08:49 PM

Thanks guys this really seems to be coming along. Anyone know if I can get calipers in your average nerf supply store or would I need to go online?
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#5 Samzilla

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Posted 18 October 2009 - 09:06 PM

It would be nice to have some people post templates for said turret spacers and turn it into a stickied thread.
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#6 Ryan201821

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Posted 18 October 2009 - 09:54 PM

It would be nice to have some people post templates for said turret spacers and turn it into a stickied thread.

Hey good idea.

http://nerfhaven.com...opic=17505&st=0
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#7 Lt Stefan

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Posted 18 October 2009 - 10:16 PM

I already made an at2k and sm1500 plate in circular fashion, but I'll wait until I male a substantial amount until I upload them to my website so you can download them. And forthe record, I got this idea from Ryan. However I wanted to make a greater variety of spacers than the ones he had so I decided to take matters (mostly) into my own hands.
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#8 Split

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Posted 18 October 2009 - 11:02 PM

I highly recommend making these in Google Sketchup as opposed to Cad, because seriously, who the fuck has cad. Stop being pretentious and use the damn freeware.
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#9 TantumBull

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Posted 19 October 2009 - 12:31 AM

Using sin laws:
sin 72 / X = sin 144 / R
sin 144 * X = sin 72 * R
R = (sin 144 * X) / sin 72
R = .6180 * X

X is the measurement you have now, R should be obvious.

Edit: Bad math, ignore. I'll update it when I feel like it.

Edited by TantumBull, 19 October 2009 - 07:11 PM.

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#10 TantumBull

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Posted 19 October 2009 - 09:16 AM

Hmm, that's odd. I just did (Sin A / a) = (Sin B / b ). I must just be a little rusty, we did the bulk of this last year.

Edited by TantumBull, 19 October 2009 - 09:16 AM.

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#11 Darth Freyr

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Posted 19 October 2009 - 09:29 AM

I did a bit of math. I got about 1.376*x. To verify, your measurement is around 0.654. For all interested, the coefficient is the cotangent of 36, or tan(36)^-1, not to be confused with tan^-1(36), the inverse tangent.
Ignore the above failure. My original calculation lacked both correctness and common sense. Don't do math when tired. The actual diameter is 0.851*x or x/(2*sin(36)). This is what Just Some Bob originally said he did and what he said he was using regardless of the math.

Edited by Darth Freyr, 19 October 2009 - 03:53 PM.

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#12 TantumBull

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Posted 19 October 2009 - 09:49 AM

Wow, I'm retarded. My concept was right, I just assumed that there were 360 degrees in a triangle. I need to get off them circles.
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#13 Lt Stefan

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Posted 19 October 2009 - 06:37 PM

I highly recommend making these in Google Sketchup as opposed to Cad, because seriously, who the fuck has cad. Stop being pretentious and use the damn freeware.

Alibre Xpress is free.

And wow all this trig- next year maybe this will all make sense.

So am I going with .851 for the LB?
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#14 Zorns Lemma

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Posted 19 October 2009 - 06:46 PM

I highly recommend making these in Google Sketchup as opposed to Cad, because seriously, who the fuck has cad. Stop being pretentious and use the damn freeware.


This entire hobby is obviously about dickwaving and epeen. How are we supposed to do that if we can't be all pretentious with our super smarts and lazor cuttorz? We should only share if we can keep the high ground!
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#15 Darth Freyr

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Posted 19 October 2009 - 06:53 PM

Just Some Bob got 0.90" for the LB, and I believe his process was correct. The Berserker has an axis-barrel center measurement of 2.171875" or 139/64".
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#16 Lt Stefan

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Posted 19 October 2009 - 07:47 PM

Just Some Bob got 0.90" for the LB, and I believe his process was correct. The Berserker has an axis-barrel center measurement of 2.171875" or 139/64".

So .9 then- everyone agree?
And 139/64" nice. Don't take this personally but can anyone verify that dimension? It's just because there are a lot of barrels and I want to be ansolutely sure. Not like I don't trust that or anything...

EDIT

JSB I remember here you said at3k barrels were 11/16," is this right? It doesn't seem like the same size as an sm3k....

DOUBLE EDIT

Okay I have a few of the blasters you want measurements for.

Arrowstorm: 2.222" (from turret axis to center of barrel)
Firefly: 1.273" (from turret axis to center of barrel)

More to come later....maybe.


Are those rounded? Because I tried to get all the barrels evenly around it by using the dimension (for the firefly) as in the equation:

x=the linear length of one barrel to the next by forming a square (so every other barrel in the firefly)

2.564= Diameter

2.564^2=6.527944

6.527944=2x^2

3.263972=x^2

x=1.806646617

That dimension, when implemented, causes the barrels to be too far apart so that they can't fit on the R1.273 circle.

Any suggestions all you math people on how to go about doing this on things with other than 4 barrels (which is easy)?

Edited by Lt. Stefan, 19 October 2009 - 09:17 PM.

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#17 TantumBull

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Posted 19 October 2009 - 09:36 PM

4: R = .7071X
5: R = .8507X
6: R = X
8: R = 1.3066X
12: R = 1.9319X

X = distance between adjacent barrel center-points.

Edited by TantumBull, 19 October 2009 - 10:52 PM.

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#18 Lt Stefan

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Posted 19 October 2009 - 10:31 PM

So I divide the radius by those numbers, correct? And the number I am left with is the linear distance, not around the circumfrence (i forget what that's called), right? This is quite helpful, thanks.
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#19 TantumBull

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Posted 19 October 2009 - 10:51 PM

This is for solving R. Plug the distance between adjacent barrel center-points (X) into the appropriate equation to solve for the radius ( R ), which is the value you're going to want when making spacers. Of course CAD software could do the calculations for you, but this way is a little more straight forward.

Edit: Those value are also rounded to the ten-thousandths place

Edited by TantumBull, 19 October 2009 - 10:53 PM.

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#20 Zorns Lemma

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Posted 20 October 2009 - 04:26 PM

4: R = .7071X
5: R = .8507X
6: R = X
8: R = 1.3066X
12: R = 1.9319X

X = distance between adjacent barrel center-points.


Instead of all this fancy math, for even-barrel blasters, why not just measure the distance from two opposite barrels, known to most people as "along the diagonal," and divide by two?

As for odd-barrel blasters, the only one I can think of right now that would need barrel spacers is the lightning blitz, but the barrels on that aren't perfectly pentatonic so there's no point in doing that either.
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#21 Lt Stefan

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Posted 20 October 2009 - 06:09 PM

This is for solving R. Plug the distance between adjacent barrel center-points (X) into the appropriate equation to solve for the radius ( R ), which is the value you're going to want when making spacers. Of course CAD software could do the calculations for you, but this way is a little more straight forward.

Edit: Those value are also rounded to the ten-thousandths place


So those dimensions listed are the distances between centers if adjacent barrels? Wouldn't they vary, though, depending on the gun? Because I'd already know the radius, unless you're saying to isolate x, which I think you are. And do you think you could not round them?

On a side note, how'd you figure that out?
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#22 TantumBull

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Posted 20 October 2009 - 07:28 PM

Instead of all this fancy math, for even-barrel blasters, why not just measure the distance from two opposite barrels, known to most people as "along the diagonal," and divide by two?

As for odd-barrel blasters, the only one I can think of right now that would need barrel spacers is the lightning blitz, but the barrels on that aren't perfectly pentatonic so there's no point in doing that either.

A very valid point. Maybe use this if you're using a turret that your calipers won't reach far enough with?


So those dimensions listed are the distances between centers if adjacent barrels? Wouldn't they vary, though, depending on the gun? Because I'd already know the radius, unless you're saying to isolate x, which I think you are. And do you think you could not round them?

On a side note, how'd you figure that out?

Holy crap, dude. This is for IF YOU DON'T HAVE R. If you already have the radius then there isn't any other value you should be caring about for spacing the holes. I wouldn't have set up the equations like that if I wanted you to solve for X. The point of these equations is to solve for the radius if all you have is X and you are capable of counting the number of barrels. Also, the only variance comes in the number of barrels. The equations will work for any symmetrical turrets with the given number of barrels.

I used the law that sin a / A = sin b / B = sin c / C

Edited by TantumBull, 20 October 2009 - 07:29 PM.

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#23 Lt Stefan

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Posted 20 October 2009 - 10:11 PM

I need to learn trig......

And I just didn't understand what they were for, I guess I didn't ask my question clearly enough. The reason I want to know how far apart the are with a given radius is so I can space them along the circumfrence of the barrel centers evenly. Unless anyone has a sugestion on how to do that.
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#24 TantumBull

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Posted 20 October 2009 - 10:28 PM

I need to learn trig......

And I just didn't understand what they were for, I guess I didn't ask my question clearly enough. The reason I want to know how far apart the are with a given radius is so I can space them along the circumfrence of the barrel centers evenly. Unless anyone has a sugestion on how to do that.

Ah, sorry. I thought you had said you were using CAD. If you're doing it by hand then maybe that would be the way to go, though I think using a protractor would be much easier and more accurate.
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#25 Zorns Lemma

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Posted 21 October 2009 - 03:10 PM

Maybe use this if you're using a turret that your calipers won't reach far enough with?


Oh god I don't have my turret machined down to the micrometer whatever shall I do? How I can mount my manta on my crossbow now?
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"In short, the same knowledge that underlies the ability to produce correct judgement is also the knowledge that underlies the ability to recognize correct judgement. To lack the former is to be deficient in the latter."
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