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Bubble level for rifle

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Unread 03-07-2004, 11:00 AM
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Join Date: Dec 2002
Posts: 374
Re: Bubble level for rifle

<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR> Well dude, your trying to tell me that you couldn't see the mill table<HR></BLOCKQUOTE>

Gee, that's right! I'm leaning into (tward) the mill with my elbow on the mill table and my head sideways, turned to my side looking at an index head that obscures the rest of the table. I'm looking at a part with 3 intersecting radii in a 3 jaw chuck, and I'm close enough to this small part that I have probably not more peripheral vision than you have while looking in a scope. I have my left arm running the chuck key and my right turning the part. Doesn't seem so tough to hit a degree.


From where I'm sittin, and I haven't scrutinized the math and don't care. I thought quickly about a statement you said about sight height adding, not reducing and I think you're choosing a different datum than the RSI software and I'll describe.

Back to fantasy land:

We have a real target at 100 yards with a nice aiming point. We drill a hole in the middle of the aim spot and put an immaginary string through. Tie the string to a real washer so It cant come through and take the other end of the string that has no weight (so it won't sag) back to the bench.

An immaginary bench, I'd like one of these. Has the ability to hold the gun perfectly in 2 axes of lateral/vertical movement (up/dn - sd/sd), then also in 3 axes of pitch, yaw, and roll, then ALSO, it can orbit the gun upon any point within those points.

Now we have an immiganary gun, and I don't want one of these because it has a scope level on on it that I have to pay extra for [img]images/icons/smile.gif[/img] The gun is a hologram so we can wave our hand through it, move our string through it etc. Yet, somehow, it still shoots lead/copper bullets. Maybe I do want one.

If we take the string and hold the end of it behind the gun so it passes through it, we establish a datum upon which to rotate the gun.

I think most of your math is based upon the barrel being the datum. If we start with a gun sitting level on the bench, scope aimed at the aim spot, yours with a scope level, mine plenty close enough without, and we hold the string so it passes through the bore, it will pass closer to the bottom of the bore at the muzzle, and closer to the top of the bore, near the throat. The amount closer we will say here, is your hypothetical 3moa drop and sight height @100.

Now, we freeze the gun and rotate it as you suggest around this string. First remove the action, stock, scope and level. The barrel is NOT dynamically balanced on the string. Kinda like a bent drill sticking out of a chuck. Thus, that will contribute to an additional error which you are not accounting for because as you rotate the gun and the scope orbits this string at a radius of 1.75", you need to constantly adjust the point of aim by this additional error which is only pitch now, but becomes pitch and yaw at angles other than 0, 90, 180, 270. Our datum is only a straight line while the bullet is in the perfect bore on our perfect gun while aimed 3MOA below the aim point.

Lets say now we put the gun back and take the string and move it up. Someone may contend that there are an infinite number of points where we can rotate the gun but I think you and I can agree (no way!) that there are only 2 really reasonable points from which to do the math.

If the string passes through the scope, dead center, and we freeze it there, we now have a datum that is along a straight line. This is much easier to deal with because when we remove the barreled action, stock and level, the scope now has a dynamic balance along this line. (The turrets have no weight on this perfect immaginary scope).

Now, if we allow the barrel to orbit the string, all of the numbers are different. This is what I think RSI is doing though I have never seen the program other than the screenshots you provided. Also, I think you may be causing some other error with a sight height of 0 (impossible) causing a division by 0 error in thier code. It may make the math easier, but try 10" and it'll make more sence to the code I bet.

The 3moa now, is not going to make a 6moa circle because the origin point of the bullet is in an orbital pattern around the sight line, and will REDUCE the "cone" because the bullet passes the sight line (at least when PERFECTLY level, with a level) and when perfectly upside down (now your cooked, but I can see that). The cone is now 2 cones with the points intersecting. Yes I'm aware there will be no 2 perfect points.

There may still be a bug, and I really don't give a damn if there is. It is in the sight height to a majority degree, and could be all in there yet. Who knows what else they have attempted to account for. I'm surprised they didn't ask for the twist so they could calculate the precession at the same time. Then again, maybe you should try a -90 deg rotation to check for that.

While all these numbers are well and good for someone who "needs" them. I have to ask. If you go to one of these sniper shoots, and the target is at say 250 yards and the shot will be at a cant angle. Do you know what the angle is ahead of time? Is it specified what angle you must hold and who determines if you are holding it? How can you possibly hold for a particular angle at random? Lets say I'm the director of the shoot and I say, hold at 32.5 degrees of cant. What's the tolerance you get? Who checks? Do you have a degree wheel on the gun too? It would make more sence.

[ 03-07-2004: Message edited by: 4mesh063 ]

[ 03-07-2004: Message edited by: 4mesh063 ]
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Unread 03-07-2004, 01:23 PM
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Join Date: Dec 2002
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Re: Bubble level for rifle

<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR>Not let's figure the same for a zero of 600 yards. I'll guess you use a magnum and that your 600 yard zero is about 11 MOA (sea level) ON TOP of your 100 yard zero (a hidden ~3 MOA). You can see that if you perform the 100 yard "round robin" test again your "cone base" circle is going to be at least ~22 MOA plus the "hidden" 100 yard zero correction of ~3 MOA (but 6 MOA of "cone base"). Overall I'd say you would have a circle of about 28 MOA.


28 moa, And you may well, but,
You keep refering to bullet path as drop. I realise that different software uses different terms. I call drop, drop. How much does the bullet fall from the bore line. I call bullet path, how much does it fall from the point of aim.

Your calculations will be infinitly easier to do and for others to understand if you use the drop value instead of path. This way, a zero at 600 yards is a zero at 600 yards. Zero is zero. Not a recalculation from some 100 yard zero. Take the drop total, multiply by 2 for when you rotate the gun 180 degrees, and subtract twice the sight height becuase the scope is now the datum and the barrel is now 3.5" higher from where it was. (This excluding the few thousanths of scope incline we have)

When I made my scope mounts on my gun, I put the action in the mill vise and put a .030 shim under one side. Then milled the ring channels. It's a Kurt vise and is roughly 5" wide on the bed. I get .030 of incline for every 5" of barrel length. (less the sag). Now, at the end of a 25" barrel(average of 24"/26" factory barrels) , there would be 5x.030 or .150 of additional bore height to subtract from the sight height. I wonder if they are not doing something with that as well. There are simply too many different ways to come up with an answer without establishing some standards for where to measure from. Some are nearly impossible to visualize as they rotate around because so many different errors affect the result in differnt directions. Even in my example above of 25", where are we measuring from. The middle of the scope tube between the rings, or at the objective lens. Where do we consider the muzzle to be, at the muzzle, or the bolt raceway area under the 1.75" sight height because the sight height at the true muzzle is less than it is where we measured it at the action.

You picked on me for not looking at the numbers 80"-78" to 3 decimals when I only pointed out the location of the variation... Go ahead and now calculate the sight height to the millionth. You can see where this is going, I have no idea what they are using as a baseline reference for thier data, but there is probably some good reason for the change in numbers.
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Unread 03-07-2004, 02:41 PM
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Join Date: Jun 2001
Location: Palmer, Alaska
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Re: Bubble level for rifle

You misunderstand the numbers in Dave's sreen shots at 90 degrees cant I think.

If you look back to the ones I posted, you'll see where your misunderstanding is.

The first number in Dave's example above is "0" because the LOS and BL (bore line) is on the same plane, they both are level. His scope height is not zero, but 2.6" or 1.83" depending on which one above you look at. Both are canted 90 degrees and on the same plane now, so effective scope hight in the vertical plane is "zero inches", the number to the left ouside the parenthesis. At 6 degree cant, my example resulted in 1.99" in the vertical plane. Scope height is (2.0" @ 6 deg). Make more sense now?

The way I see this, or imagine it is, the LOS is straight throught the center of the bull, the BL drop is straight through the bull as well (dialed in at 100 yards for example), and this draws a circle under the bull when the rifle is rotated around the LOS, with 6 O'clock being 6 MOA low, 3 and 9 O'clock being diagonal to 12 and 6 O'clock. Picture a bent wire in the shape of the trajectory ending at 100 yards, with a weight hanging on the "end" of it so gravity keeps muzzle exit and the 100 yard point and all points in between aligned in the vertical plane when you rotate the rifle around LOS axis, and while the wire rotates in the bore. Follow?

Canting to the right (LOS maintained on bull remember) rotates the bore and wire to the left and begins to lower it at the same time until you hit 180 degrees, then it moves to the right and back up again. The closer to 12 and 6 O'clock you are, the higher the rate of deflection per degree is realized, but less rate in drop. The closer to 3 and 9 O'clock you are the opposite is true, more drop per degree of rotation and less deflection on the horizontal plane there is. So horizontal deflection is greatest per degree where we normally hold at 12 O'clock. At 90 degrees, vertical deflection from cant is what you will see more than anything, and horizontal deflection would be little at that point.

Consider the imaginary circle it draws, horizontal deflection is more pronounced in the first 45 degrees of cant than it is in the second 45 degrees where vertical deflection is now greatly being increased.

The same thing goes for wind deflection at near 12 and 6 O'clock directions. Each degree of wind direction change will prove "more" significant to horizontal deflection than it would had the wind direction been near 3 or 9 O'clock. Wind speed changes will however have less efects at 12 and 6 O'clock and more effect at 3 and 9 O'clock.

I missed a couple things in Dave's screen shots (read too fast) I'm now seeing, namely the inches of deflection being very oddly different, and the 100 yard deflection.

I will get Jim on here hopfully to explain what we're seeing, and if there is a bug causing this or what. I talked to him last night, but just now see this problem. I'll send him a link to the thread.

Heading to the range...

[ 03-07-2004: Message edited by: Brent ]
Brent Moffitt
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Unread 03-07-2004, 05:09 PM
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Join Date: May 2001
Posts: 2,369
Re: Bubble level for rifle

Maybe a graphic will help clear up any ambiquities.

The red circle represents a line extending
through the center of the bore and
intersecting the target at 100 yards. As
we rotate the rifle this line is formed.

The grey circle is that of where our bullets
will strike as they are effected by gravity
and fall toward the center of the earth no
matter the orientation of the rifle.

The green circle is where we place the cross
hairs of the scope no matter the orientation
of the rifle.

Again I'll state. I could be wrong but
it'll take a proper explanation to convice me of that.

Late edit

Those 3 units of whatever to get the
bullet to "fall" into the POA with
zero -0- cant will always exist when
shooting on the horizontal.
Notice that the diameter of the circle(s)
is six (6) whatever units, not three (3).

The more adjustment (elevation) on the
sights, the larger the circles will become.

[ 03-07-2004: Message edited by: Dave King ]
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Unread 03-07-2004, 09:53 PM
dwm dwm is offline
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Join Date: Feb 2003
Location: Texas
Posts: 738
Re: Bubble level for rifle

I propose that the rotation is about the centerline of the scope, IE the whole rifle rotates about the line of sight.

When the cross hair and rifle are at 0 degree cant, the windage is corrected to 0 error and the elevation is adjusted such that the curved trajectory causes the POI to intesect the POA at the desired range.

When rotated at an angle, the error in the windage is no longer 0, a component of the elevation is now being applied to the windage.

There is also error in the elevation, the resulting elevation will be less than the original elevation.

Rotate counter-clockwise and the POI will be low and to the left of POA.

A 90 degree rotation (cant) causes the entire amount of elevation to be applied to windage and the windage (0) to be applied to the elevation. (You could click the elevation into the windage knob and the windage into the elevation knob and shoot from a 90 degree cant!)

How about:

[ 03-07-2004: Message edited by: dwm ]
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Unread 03-07-2004, 11:15 PM
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Join Date: Dec 2002
Posts: 374
Re: Bubble level for rifle


I just typed a lengthy explaination for this, edited it twice, and will now do it a third time. I think I have it. There are another 2 circles you need to add.

Your graphic is nice. But it does not show the origin from where the bullet came. Since this is confusing, I will break it into small parts to digest.

Your gun shoots a bullet drop of 3 units. Lets just say that your sight height is 1 unit for this example.

We need to establish where the barrel and scope are in the diagram. One of them is NOT stationary and that movement must be accounted for. The easy way is: (in my mind)

We will assume for this example, that the line of sight is allways absolutely dead level so that our scope and our point of aim are allways the same point as viewed from behind the gun. (both, allways the green dot)

If you place another circle around the green aim spot that is 1 unit in radius This is the starting location of the barrel for each shot. This takes into account the distance the barrel is below the scope.

The center of that circle is the actual centerline location of the scope exactly as we look at the target, our eye is always in the green circle, the scope, the POA, and our eye.

The Red circle needs to be 1 unit in radius smaller.

For shot # 1, (upright level fire) (the only easy one) The barrel is down at the bottom of our new circle, 1 unit below the scope and point of aim. The Bore LOS is now, 2 units above the green dot. The bullet impact is in the green circle as you show.

The trouble is, it only impacts 2 units below the bore los even though it has 3 units of drop.

Shot # 2. We rotate the gun clockwise 90 to the right but we rotate it around the sight line, so the barrel is what moves. This eliminates a lot of math. The barrel is now on the 1 unit radius circle you have added.
LEFT of the sight line by the 1 unit of sight height.

The Barrel is also 1 unit of elevation higher than it was for shot 1, so the bullet starts out, 1 unit vertically higher.

The Bore LOS is now, 2 units right of the point of aim. That establishes the lateral movement of the shot. The drop, is still the same and is 3 total units. Our impact is 3 units below the point of aim and 2 units right.

Unless I'm mistaken, it is forming an eliptical impact circle. But it gets worse later. Is it a teardrop?

Shot 3, has the barrel above the scope when firing at 180. It is now 2 units higher in starting elevation than it was for shot 1. Above the scope instead of below it. The bore LOS now is 2 units below the green circle. That plus 3 units of drop gives you an impact 5 units below the LOS (green dot).

I have read this statement to myself about a dozen times and can't help but beleive, something is wrong with it. If you have any ideas, I'm all ears. In shot 1, we deduct 1 unit of difference from Blos to POA/POI. In this shot, we don't. ????? Seem right.

With 3 units total drop, it arrives 5 units below the sight line. It is 5 units away from shot # 1. But only 2 units lower than shot 2. Shot 2 is 3 units lower than shot 1. Again, something seems wrong with this thinking.

Shot 4 is with the gun canted 270, or 90 to the left. The barrel is now 1 unit to the right of the scope. Again, the barrel starts off 1 unit higher in elevation than shot 1 and is 1 unit right of there.

This shot, impacts 3 units below the point of aim and 2 units to the left.

I think at this point, even if RSI is wrong, I would rather take thier word for it and miss, than try to think this through any more. Life is too short.

[ 03-07-2004: Message edited by: 4mesh063 ]
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Unread 03-07-2004, 11:31 PM
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Re: Bubble level for rifle

<BLOCKQUOTE><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><HR> A 90 degree rotation (cant) causes the entire amount of elevation to be applied to windage and the windage (0) to be applied to the elevation. (You could click the elevation into the windage knob and the windage into the elevation knob and shoot from a 90 degree cant!)

You would be close, but no cigar. While typing that mess I just posted, I see you have added this. Unless I am mistaken, your impact on the 90 deg shot would be off windage by the amount of the sight height and closer to the barrel, and your drop would miss by the same amount, low.
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