[Guest]

As a much younger person, my sport of choice was small-bore rifle competition, at which, unlike math, I approached Olympic class. The equipment is a high precision .22 caliber rifle, firing a projectile with a muzzle velocity of 1080 feet per minute. Sights, either telescopic or ‘iron’, are mounted 2 inches above the barrel, centerline of sight to centerline of bore when parallel. Sighting adjustment is accomplished by turning either of two knobs, one for windage and one for elevation, which are graduated in one/sixth minute ‘clicks’. Common distances are 50 feet indoors and both 50 yards and 100 yards outdoors. If a rifle is zeroed perfectly for the 50 foot range, what adjustments, in clicks of elevation, must be made to score accurately at 50 yards and 100 yards? Assume both a no wind condition and constant projectile velocity.

[Guest]

As a much younger person, my sport of choice was small-bore rifle competition, at which, unlike math, I approached Olympic class. The equipment is a high precision .22 caliber rifle, firing a projectile with a muzzle velocity of 1080 feet per minute. Sights, either telescopic or ‘iron’, are mounted 2 inches above the barrel, centerline of sight to centerline of bore when parallel. Sighting adjustment is accomplished by turning either of two knobs, one for windage and one for elevation, which are graduated in one/sixth minute ‘clicks’. Common distances are 50 feet indoors and both 50 yards and 100 yards outdoors. If a rifle is zeroed perfectly for the 50 foot range, what adjustments, in clicks of elevation, must be made to score accurately at 50 yards and 100 yards? Assume both a no wind condition and constant projectile velocity.

Clarification please. Is this a simple projectile question assuming no air resistance? Incidentally, when I shot at Bisley (that's near Aldershot in the UK) one got two sighting shots before the scoring shots. Especially when one was on the 500 yards firing range, one had to gauge the wind by keeping half an eye on a flag which fluttered in the breeze close by, specifically located for the pupose. There was quite an art here. Our rifles' sights did not have a lateral deflection knob, so one had to judge everything with ones two "sighters."

[bonanova]

1 click = 1/6 x 1/60 x pi/180 = 48.48 microradians.

You need a little more than 40 clicks above horizontal to reach 50'.

Another 80-81 clicks gets you to 150'.

Another 120-121 clicks reaches 300'.

The three elevations are about 0.112^o, .335^o and .669^o.

I assumed 1080 feet per second muzzle velocity.

[Guest]

Yes. With a muzzle velocity of 1080 ft/min, namely 18 ft/s, the maximum range of this potent rifle would be, assuming a gravitational acceleration of 32 ft/s/s, a mind-boggling 10 and one eighth feet. This assumes an elevation of the rifle of 45º, and I doubt that gun sights are geared for such a correction (especially on a 25 yard or a 100 yard range!) Lateral corrections for wind would seem a bit superfluous too, whether one is dealing with still air or hurricane force winds.

Seriously though, I find that my own typos are embarrassing enough. Making arithmetic or algebraic errors are worse. Making a complete cockup of my approach to a problem is most definitely to be avoided.

[Guest]

I offer my profound apologies for my complete cockup, as it were, and I thank one respondent for his kind correction. I did indeed intend a muzzle velocity of 1080 feet per second.

[Guest]

It is a decent question, and we all knew what you meant.

[Guest]

no correct answers yet

[Guest]

1 click = 1/6 x 1/60 x pi/180 = 48.48 microradians.

You need a little more than 40 clicks above horizontal to reach 50'.

Another 80-81 clicks gets you to 150'.

Another 120-121 clicks reaches 300'.

The three elevations are about 0.112^o, .335^o and .669^o.

I assumed 1080 feet per second muzzle velocity.

Would be interested in seeing the math here, as I believe the solutions given are not correct. I calculate the 50 foot range to require

83 clicks. The angles given for 50 yards and 100 yards would score high and very high, respectively.

I am led to believe that I may have mis-stated the OP in a way that could give rise to these answers.

[Guest]

For 50'

83.88 clicks. Your sight is 2" above bore...you are basically computing the angle required to clear 2" from horizontal at 50'. That gave me a theta required of 0.233 degrees. There are 1/360 degrees per click, so that is 83.88 clicks.

1/6 = 50*v_0y/v_0x - 16*50²/v_0x²

v_0x = 1050*cos(theta)

v_0y = 1050*sin(theta)

etc..

Edit: my answer is slightly off because I misread the muzzle velocity to be 1050, instead of 1080...and I don't feel like redoing the calculations

Edited August 18, 2009 by tpaxatb

[Guest]

Something doesn't look right ... someone check my work???

(edit: add assumption implied in calculations - scope horizontal to target point)

We need t (theta) the angle to fire.

g = 32ft/sec²

y = x v_0y/v_0x - gx²/2v_0x²

y = x tan(t) - gx²/2v₀²cos²(t)

y = x tan(t) - gx²sec²/2v₀²

y = x tan(t) - (gx²/2v₀²)*(1+tan²(t))

let a = tan(t)

g = 32ft/sec²

v₀² = 1080² = 1,166,400

y=1/6 (2 inches)

x = horizontal distance (50', 50yd, 100yd)

a²16x²/1166400 - ax + 16x²/1166400 + 1/6 = 0

so that gives

One click is 1/360 of a degree

(a = tan(t))

at 50 ft:

(40000/1166400)a² - 50a + 40000/1166400 + 1/6 = 0

tan(t) = a = -(sqrt(53143514)-7290)/10, (sqrt(53143514)+7290)/10

t = 89.960702381, 0.230282843

At 50 feet, required elevation is 82.90182348 clicks above horizontal

at 150 feet (50 yards):

(360000/1166400)a² - 150a + 360000/1166400 + 1/6 = 0

tan(t) = a = -(sqrt(5904746)-2430)/10, (sqrt(5904746)+2430)/10

t = 0.181555104, 89.882106847

required elevation is 65.35983744 clicks above horizontal (you have to DOWN click from the 50' mark)

at 300 feet (100 yards):

(1440000/1166400)a² - 300a + 1440000/1166400 + 1/6 = 0

tan(t) = a = -(sqrt(5904446)-2430)/20, (sqrt(5904446)+2430)/20

t = 0.267619287, 89.764211698

96.34294332 clicks above horizontal.

Edited August 18, 2009 by tpaxatb

[Guest]

Something doesn't look right ... someone check my work???

(edit: add assumption implied in calculations - scope horizontal to target point)

We need t (theta) the angle to fire.

g = 32ft/sec²

y = x v_0y/v_0x - gx²/2v_0x²

y = x tan(t) - gx²/2v₀²cos²(t)

y = x tan(t) - gx²sec²/2v₀²

y = x tan(t) - (gx²/2v₀²)*(1+tan²(t))

let a = tan(t)

g = 32ft/sec²

v₀² = 1080² = 1,166,400

y=1/6 (2 inches)

x = horizontal distance (50', 50yd, 100yd)

a²16x²/1166400 - ax + 16x²/1166400 + 1/6 = 0

so that gives

One click is 1/360 of a degree

(a = tan(t))

at 50 ft:

(40000/1166400)a² - 50a + 40000/1166400 + 1/6 = 0

tan(t) = a = -(sqrt(53143514)-7290)/10, (sqrt(53143514)+7290)/10

t = 89.960702381, 0.230282843

At 50 feet, required elevation is 82.90182348 clicks above horizontal

at 150 feet (50 yards):

(360000/1166400)a² - 150a + 360000/1166400 + 1/6 = 0

tan(t) = a = -(sqrt(5904746)-2430)/10, (sqrt(5904746)+2430)/10

t = 0.181555104, 89.882106847

required elevation is 65.35983744 clicks above horizontal (you have to DOWN click from the 50' mark)

at 300 feet (100 yards):

(1440000/1166400)a² - 300a + 1440000/1166400 + 1/6 = 0

tan(t) = a = -(sqrt(5904446)-2430)/20, (sqrt(5904446)+2430)/20

t = 0.267619287, 89.764211698

96.34294332 clicks above horizontal.

Yes, my error. It is essential to add that the sight is horizontal.

Will chec the rest later, as my wife has the idea that we must now go out to dinner for our thirtieth. Yay us!

[bonanova]

..

Would be interested in seeing the math here, as I believe the solutions given are not correct. I calculate the 50 foot range to require

83 clicks. The angles given for 50 yards and 100 yards would score high and very high, respectively.

I am led to believe that I may have mis-stated the OP in a way that could give rise to these answers.

There was an error in calculating time of flight of the bullet.

Here are the new results.

Assume the barrel and target are at the same height.

Assume the sight and the barrel axes are parallel at "zero" clicks.

Adjusting the sight can be thought of as two steps.

.

1. Lower the sight by (1/6)/Range radians to intersect the barrel axis at the target.
2. Raise the barrel enough for the bullet to return to its starting height at the target.

.

Call these angles a1 and a2.

Their sum is the total sight correction from zero clicks at a given range.

At 50' a1 = 68.75; a2 = 14.22. Sum = 82.98 clicks

At 150' a1 = 22.92; a2 = 42.67. Sum = 65.59 clicks

At 300' a1 = 11.46; a2 = 85.34. Sum = 96.80 clicks

Correction from 50' to 150' = -17.39 clicks

Correction from 50' to 300' = +31.21 clicks

One click is pi/(180x360) radians = 48.48 x 10^-6 radians.

Assume that at 0 clicks the barrel is horizontal.

Find the range for a given number of clicks.

.

1. a₂ = 48.48 x clicks x 10^-6 radians.
2. v_y = 1080 sin(a2)
3. v_x = 1080 cos(a2)
   .
   h = v_y²/2g - highest point in the bullet's trajectory.
   t_fall = 2 sqrt(2h/g) = v_y/g - time for an object to fall a distance h.
   .
4. t = 2t_fall = 2vy/g - time of flight of the bullet - rising to h, then falling to target.
5. Range = v_xt

.

Adjust clicks until Range = 50, 150 and 300.

50 feet: 14.22 clicks

150 feet: 42.67 clicks

300 feet: 85.34 clicks

[Guest]

Interesting. Unless my calculations were off (which is why i asked for a check)...raising the barrel 2" (from the scope being on the horizontal to the target to the barrel being horizontal) doesn't affect the required clicks as much as I would have thought...