how I measured a 5 groove bore

[legi0n]

I wanted to know whether the chambers on my 357 GP100 are larger than the bore or not.
The thing is ... it's a 5 groove! And I'm not going to fork hard earned money for a one-off measurement tool (v-block micrometer).

Sooo, I took an old rod from a discarded printer. Said rod is made out of polished steel, and has a consistent diameter of 0.3124"

It so happens that the bore slug sits on the groves (bore groves) when placed in this contraption:

http://members.shaw.ca/le_r0umain/slug_measuring_device1.png

http://members.shaw.ca/le_r0umain/slug_measuring_device2.png

The device is made of 3 1.5" lengths of said rod tied snugly together.
Now I can measure the length between the lower rod and the top groove (barrel groove).

http://members.shaw.ca/le_r0umain/equation1.png

a is the measured length.
r1 is half the rod diameter.
r2 is half the slug diameter (what I needed measured)

Now I can't solve such equations but luckily there's a free program that can (google Microsoft Mathematics)

http://members.shaw.ca/le_r0umain/equation2.png

the diameter I needed to know is twice r2

given a = 0.9016"
r1 = 0.3124/2

I now know the groove diameter is 0.3573"

(the shim stock wrapped around the bullet only measured 0.355")

me happy

[leftiye]

If the grooves and lands are of the same width you can measure from the leading edge of one land to the trailing edge of the "opposite" land - they are diametrically opposite each other.

[legi0n]

This would work too if the land edges are not rounded. Which I cannot guarantee.

Today I tested the device with a sized hardcast bullet measuring 0.3597".

The calculus yielded 2*0.179879 which is... 0.359758"

Pretty accurate if you ask me :D

[leftiye]

Wrap it in a .002 or .003" shim, measure, deduct the shim thickness. Thas about all the calculus i know.

[legi0n]

Wrap it in a .002 or .003" shim, measure, deduct the shim thickness. Thas about all the calculus i know.

been there done that
didn't work
and I believe it usually doesn't otherwise the v-block micrometer wouldn't sell for so much money

[nanuk]

I can see it, but I don't get it

[legi0n]

I needed some contraption that would allow the slug to sit firmly on the grooves (barrel grooves - bullet outer diameter) and by measuring a linear dimension to be able to compute the slug diameter (barrel groove diameter).

Where I live a v-block micrometer sells for hundreds of dollars and extra v-blocks are about $80 a piece.
My device was almost free :D

Think of the 3 rod ensemble as a v-block. All the computations can be tabulated for fast measurement. I didn't bother since I don't own that many odd-number-of-grooves barrels.

[leftiye]

Look on eVil bAy - lotsa cheap machine toolz there LOL.

[nanuk]

I needed some contraption that would allow the slug to sit firmly on the grooves (barrel grooves - bullet outer diameter) and by measuring a linear dimension to be able to compute the slug diameter (barrel groove diameter).

Where I live a v-block micrometer sells for hundreds of dollars and extra v-blocks are about $80 a piece.
My device was almost free :D

Think of the 3 rod ensemble as a v-block. All the computations can be tabulated for fast measurement. I didn't bother since I don't own that many odd-number-of-grooves barrels.


I agree with the cost of a V-block mic.... $$$$$$$

no, my comment was that I can see how it works, I just don't get the math.

and to think, I graduated with high honours in algebra/geotrig, and my Dad was the teacher....

[legi0n]

PM me if you want to know.

[Gtek]

legiOn, May I ask? Are you not an engineer eh? Wrapping with shim stock works for me, I can count to twenty one if you give me a minute. Way cool on the math Dude! Gtek

[frnkeore]

I use pin gages and a blade mic (much cheaper than anvil mic's). You then measure the distance between the top of the land and bottom of the groove. Subtract the dia of the gage pin that fits the bore (if you have a lathe, you can make a pin to fit). That leaves you with the depth of the groove. Multiply that times two and add the pin gage dia and you have a groove diameter thats pretty close.

Frank

[legi0n]

legiOn, May I ask? Are you not an engineer eh? Wrapping with shim stock works for me, I can count to twenty one if you give me a minute. Way cool on the math Dude! Gtek

Used to be an engineer. Switched to computers. And so I lost my equation solving skills... :mrgreen:

[legi0n]

I use pin gages and a blade mic (much cheaper than anvil mic's). You then measure the distance between the top of the land and bottom of the groove.

How do you do that with a flat mic?

[Gtek]

If you take a shim stock , say - .002" about 3" long. Bend it around a pencil 180 degrees. Tweek bends to run parallel to each other along side of boolit, stand Boolit vertical ( of coarse in little window outside land grooves). Using vernier dial or C type mic, Total = <.004". Not a slam on engineer thing, I have been around them my whole life, I had one that was called DAD. I have an understanding of that thought process!
Gtek

[sffar]

Clever solution for getting that measurement, and seems very accurate. I frequent a used tool store where lot's of micrometers pass through, but the V block micrometers only rarely.

[Pepe Ray]

V-blok guages for 5 land/grove slugs are simple, rugged, easy to use and available from this site. Much less expensive than Mikes.
Contact 45Nut.

[frnkeore]

For flat anvil mic's, they used to sell a assortment of tips that have a rubber sleeve to hold them in place on the moveable anvil such as a .200 ball. I haven't seen them in years, mostly because I haven't looked and don't need it. Try ENCO and MSC catalogs to find one.

Frank

[frnkeore]

LegiOn, if your asking about how it's done with a blade mic, the blades are about .03 wide and .250 long. You measure by putting the blade inline with the groove. You then twist it in the groove so it picks up the center and take that measurement.

Frank

[Tom Myers]

I wanted to know whether the chambers on my 357 GP100 are larger than the bore or not.
The thing is ... it's a 5 groove! And I'm not going to fork hard earned money for a one-off measurement tool (v-block micrometer).

Sooo, I took an old rod from a discarded printer. Said rod is made out of polished steel, and has a consistent diameter of 0.3124"

It so happens that the bore slug sits on the groves (bore groves) when placed in this contraption:

The device is made of 3 1.5" lengths of said rod tied snugly together.
Now I can measure the length between the lower rod and the top groove (barrel groove).

http://members.shaw.ca/le_r0umain/equation1.png

a is the measured length.
r1 is half the rod diameter.
r2 is half the slug diameter (what I needed measured)

Now I can't solve such equations but luckily there's a free program that can (google Microsoft Mathematics)

the diameter I needed to know is twice r2

given a = 0.9016"
r1 = 0.3124/2

I now know the groove diameter is 0.3573"

(the shim stock wrapped around the bullet only measured 0.355")

me happy

I know that this is an old post but I just happend to stumble upon it and wondered if I could work out the math and reduce the equation to a fairly simple algorithm that could be run on a calculator or even use a pencil and paper with simple arithmetic and would return the diameter of the sample.

After a lot of Algebra to reduce the equation, this works and is not really too complex.

Let r = the diameter divided by 2 of the rods that are taped together into a stack.
let a = the distance from the top of the slug to the bottom of the stack.
let z = the square root of 3 = ( 1.732 )
let d = the diameter of the slug.

Then

d= ( a - r - r*z )² / ( a - r*z)

substituting 1.732 back in for z, we get

d = ( a - 2.732*r )² / ( a - 1.732*r )

Hope this helps.

[wallenba]

My brain hurts.

I would like to see a go-no-go gage plate for us sluggers of five grooves offered. Kind of like a drill bit gage plate.

[felix]

Good job, Tom! A job done right for the measurement desired. ... felix

[gwpercle]

After seeing all this I remembered why I flunked calculis every time I took it.... it don't make any sense whatsoever. I own a 5 groove 303 British Enfield and managed to find a site on the net with simple instructions for doing it for persons with a degree from MIT in advanced mathmatics and it didn't require the use of a v-block either. If anyone needs it I'm sure I can find it again.

But I guess if you are blessed with mathmatical skills you may as well us them to calculate this sort of thing. Once while trying to get thru some college math course the professor kept telling me not to give up , that one day the light will go on and I will under stand it all. Well the lights are still out, I never got it , He gave me a D just because I tried and maybe he didn't want to see me next year

gary

[Tom Myers]

I was looking at this result and discovered I could make it a little more simple.

I edited the post and added the last two lines. That, I think, is about as simple as it can be.



I know that this is an old post but I just happend to stumble upon it and wondered if I could work out the math and reduce the equation to a fairly simple algorithm that could be run on a calculator or even use a pencil and paper with simple arithmetic and would return the diameter of the sample.

After a lot of Algebra to reduce the equation, this works and is not really too complex.

Let r = the diameter divided by 2 of the rods that are taped together into a stack.
let a = the distance from the top of the slug to the bottom of the stack.
let z = the square root of 3 = ( 1.732 )
let d = the diameter of the slug.

Then

d= ( a - r - r*z )² / ( a - r*z)

substituting 1.732 back in for z, we get

d = ( a - 2.732*r )² / ( a - 1.732*r )

Hope this helps.

[markinalpine]

...measure from one side of the bore slug to the opposite side of one of the known diameter rods, using a common anvil micrometer, and subtract the known diameter?

Looking at the drawing, the "contraption" is three known diameter rods arranged as an equilateral triangle standing on its apex. If you take one of the rods on either the left or right side, I'm pretty sure you could just measure from the side of that rod across the slug and get one good measurement without going through the math.

Mark :coffeecom

[Tom Myers]

...measure from one side of the bore slug to the opposite side of one of the known diameter rods, using a common anvil micrometer, and subtract the known diameter?

Looking at the drawing, the "contraption" is three known diameter rods arranged as an equilateral triangle standing on its apex. If you take one of the rods on either the left or right side, I'm pretty sure you could just measure from the side of that rod across the slug and get one good measurement without going through the math.

Mark :coffeecom

I don't believe that would work. to obtain the measure of the full diameter of one rod and the slug, the micrometer would still close on the center of a groove on the slug.

http://www.tmtpages.com/LinkSkyImages/forum_images/5%20groove%20dia.jpg

[markinalpine]

I don't believe that would work. to obtain the measure of the full diameter of one rod and the slug, the micrometer would still close on the center of a groove on the slug.

http://www.tmtpages.com/LinkSkyImages/forum_images/5%20groove%20dia.jpg

First, your illustration perfectly shows what I meant.
Secondly, the barrels grooves and lands, being spiral, or more properly helical, will generate helical mirror images on the slug, so the measuring tool could be oriented to measure the upraised portion of the slug produced by the barrel grooves, as opposed to the indentation made by the lands.

Great picture, though [smilie=w:

Mark [smilie=s:

[legi0n]

I was looking at this result and discovered I could make it a little more simple.

I edited the post and added the last two lines. That, I think, is about as simple as it can be.

Tom, your math-fu is greater than mine! :drinks:

[Tom Myers]

Tom, your math-fu is greater than mine!
Legi0n
Your Trigonometry skills must be way beond me to come up with the original formula. I just used Algebra to reduce the formula to a simple algorithm.


............ the barrels grooves and lands, being spiral, or more properly helical, will generate helical mirror images on the slug, so the measuring tool could be oriented to measure the upraised portion of the slug produced by the barrel grooves, as opposed to the indentation made by the lands.
Mark [smilie=s:

Mark,

Now I get it, I was just thinking in a two dimensional cross section. I'm thinking that if the lands were somewhat narrow then the slug would need to be somewhat longer to offer a measureing spot that would offer full diameter.

My previous explanations and examples were somewhat vague.
I recently acquired a new screen image measuring tool so I worked up another drawing that might make things a little easier to understand.

For what it's worth, this is it.

http://www.tmtpages.com/LinkSkyImages/forum_images/new-5%20groove%20dia.jpg

[Roger Ronas]

Hello Tom,
I have worked your equation several times and it works each time.
But I don't know how you got the number 2.732. What would that number be for rods of a diameter of .375?
Thanks
Roger




I know that this is an old post but I just happend to stumble upon it and wondered if I could work out the math and reduce the equation to a fairly simple algorithm that could be run on a calculator or even use a pencil and paper with simple arithmetic and would return the diameter of the sample.

After a lot of Algebra to reduce the equation, this works and is not really too complex.

Let r = the diameter divided by 2 of the rods that are taped together into a stack.
let a = the distance from the top of the slug to the bottom of the stack.
let z = the square root of 3 = ( 1.732 )
let d = the diameter of the slug.

Then

d= ( a - r - r*z )² / ( a - r*z)

substituting 1.732 back in for z, we get

d = ( a - 2.732*r )² / ( a - 1.732*r )

Hope this helps.

[Tom Myers]

Hello Tom,
I have worked your equation several times and it works each time.
But I don't know how you got the number 2.732. What would that number be for rods of a diameter of .375?
Thanks
Roger

Roger,

The number 2.732 in the equation will be the same no matter what the diameter of the rods are.

Just substitute the radius of the rods (diameter of a rod divided by 2) for " r " in the equation.
Then substitute the height of the stack and slug for " a " in the equation.
Then do the arithmetic.

The number 2.732 is simply 1 + the square root of 3
The square root of three is 1.732050809

1 + 1.732 = 2.732

The reduction of the equation that Legi0n supplied is

d= ( a - r - r * z )² / ( a - r * z)
where z = the square root of 3

factoring [ a - r - r * z ] gives [ a - ( r * ( 1 + z) ] or [ a - r * ( 1 + 1.732 ) ] or ( a - r * 2.732 )

[ - r -r * z ] is the same as ( - r * ( 1 + z ) ) or [ - r * 2.732 ]

finally
d = ( a - [ r * 2.732 ] )² / ( a - [ r * 1.732 ] )
or
d = ( a - [ r * 2.732 ] ) * ( a - [ r * 2.732 ] ) / ( a - [ r * 1.732 ] )

Hope this helps

[Roger Ronas]

Yes it does, Thank You,
Roger




Roger,

The number 2.732 in the equation will be the same no matter what the diameter of the rods are.

Just substitute the radius of the rods (diameter of a rod divided by 2) for " r " in the equation.
Then substitute the height of the stack and slug for " a " in the equation.
Then do the arithmetic.

The number 2.732 is simply 1 + the square root of 3
The square root of three is 1.732050809

1 + 1.732 = 2.732

The reduction of the equation that Legi0n supplied is

d= ( a - r - r * z )² / ( a - r * z)
where z = the square root of 3

factoring [ a - r - r * z ] gives [ a - ( r * ( 1 + z) ] or [ a - r * ( 1 + 1.732 ) ] or ( a - r * 2.732 )

[ - r -r * z ] is the same as ( - r * ( 1 + z ) ) or [ - r * 2.732 ]

finally
d = ( a - [ r * 2.732 ] )² / ( a - [ r * 1.732 ] )
or
d = ( a - [ r * 2.732 ] ) * ( a - [ r * 2.732 ] ) / ( a - [ r * 1.732 ] )

Hope this helps

[machanic]

Ever try a set of small hole gages, takes a little practice to get the right feel but if you have a 1in. mic. you are good to go!

[Tom Myers]

Ever try a set of small hole gages, takes a little practice to get the right feel but if you have a 1in. mic. you are good to go!

Talk about coincidence! About an hour ago I received a spam email from a tool company advertizing that very same set of small hole guages and was just now trying to decide whether to get the set or not.

I think that I just might go ahead and order the set

[machanic]

If you look around you can find deephole sets, allows you to get 6 to 8 in. down a bore.

[Roger Ronas]

Those work for bore diameter but I don't see how they would work for groove diameter.

Roger

[machanic]

Groove depth can be measured with a vernier caliper, measure between the groove and the outside of the barrel and the bore and the outside of the barrel, the differance is the groove depth, times 2 plus the bore dia. is groove bore.
Of course measure carefully, when you multiply by two you double your mistake!!

[Roger Ronas]

I understand that, but I think with 3-4 measurements, the chance of getting an error is more. Also a caliper measures in thousands and a mic can go down to tenths.

[Michael J. Spangler]

Tagged for info.
Now I need to go home to measure my slug from my s&w shield.

Thanks guys.