doublegunshop.com - home Forums General Discussion DoubleGun BBS @ doublegunshop.com MOI machine calibration and demonstration

In Don's absence I will answer this for you. Yes, he does. The very first step in this process is to determine the balance point of the gun very closely. Then, this is the point that is centered on the wheel.

SRH

Good to see Don again - at the Southern. He's building quite a database.

Sorry to check in late, but here is is.

The teeter-totter balance point is the center of gravity. This is the point of lowest MOI for any object. If your hand placement and physique makes the gun pivot at a point other than the CG, the MOI will be higher. There is a simple formula to calculate MOI at any point once MOI at CG is known (it is called the parallel axis theorum). This is how I get the MOI at butt. The formula is weight divided by 3.22 times the length from the CG to the new pivot point multiplied times itself (this length must be in feet, not in inches)plus the MOI at CG. Keep in mind that I multiply true MOI by 10 to get whole numbers and decimals. Otherwise, we would need to divide the weight by 32.2. I'll work an example as soon as I have a bit more time (maybe later this week).

MOI at butt is typically about four times as high as MOI at CG. However, if the hand placement causes the pivot to move only an inch or two, very likely, then the change in MOI will be relatively low.

I'm very happy to see such astute questions and some very good answers. You guys are getting it!!

Dig shot my Ithica NID 4E at the Southern 5-stand and now demands hand sanitizer!! Hey, Dig good to see you again. Oh, did I mention he shot it well?!!!

DDA

That single trigger almost had me confused for a minute. No rash so far.

OK, here we have the promised sample problem:

Lets say we have a gun that weighs 6.44# (6# 7 oz), and has a MOI at CG of 1.50. The shooter checks for point of rotation while unmounted and finds that his hand placement causes the actual rotation point to be 1.20" from the CG. The formula then becomes ((6.44/3.22) X 1.2/12) X (1.2/12)) + 1.50 = (2 X 0.1 X 0.1) + 1.5 = (2 X 0.01) + 1.5 = 0.02 + 1.50 = 1.52. 0.02 is about a tenth of the amount of change in MOI to noticed by most shooters. Keeping in mind that doubling the distance causes a four fold increase in the change in MOI, 0.6" would be 1.505 and 2.4" would be 1.58. Around three inches would cause a noticable change for a gun of the stated weight. Heavier guns would require less change for a noticable MOI increase.

Same gun as above and having 4" balance in front of the front trigger and a 14" LOP. Mounted MOI is found as follows: (((6.44/3.22) X ((4" + 14")/12) X ((4" + 14") / 12)) + 1.50 = ((2 X ((18/12) X (18/12)) + 1.50 = (2 X 1.5 X 1.5) + 1.5 = (2 X 2.25) + 1.5 = 4.50 + 1.50 = 6.00 = MOI at butt.

I forgot to mention that balance tells the proportion of weight carried by each hand. A hand centered under the CG would carry 100% of the gun's weight. If the hands are spread equally from the CG each hand carries 50% of the weight. If the center of the trigger hand was three inches behind the trigger (4" CG to trigger + 3" behind trigger = 7") and the front hand was 14" in front of the CG, the front hand would carry 7/21 + 1/3 of the weight and the back hand would carry 14/21 = 2/3 of the weight. Some shooters are quite sensitive to the proportion of weight in each hand.

Do check my cipherin', I have recently mad some misteakz.

Questions?

Dig, it is possible to fire a gun twice by pulling the single trigger twice. It is not necessarily so, but sometimes.

DDA

Does the mounted gun actually rotate between the hands?

The shooter's entire body rotates and it does not rotate around a point on the gun.

Isn't the motion of a gun while mounted far more complex? It isn't planar and is not a simple rotatory motion.

Didn't you ask your hero Obama where he rotates when he is posing with a shotgun for propaganda pictures? I think he rotates to the extreme left in everything he does.