Trajectory pop quiz

[Ultrastu]

Ok so the image below relates to the reticule of a rifle  -airgun  , rim fire ,centrefire  even .

The small ret on the left is the scope on gun in the usual up right position  .

Lets say this is an airgun zeroed at 30 yds .177 . When fired at the 30 yd  target the pellet hits dead centre of the crosshairs .

now the quiz .

if i rotate the gun 90 degree sideways to the left so the gun is laying on its side .and take aim at my target using the centre crosshairs as my aim point  where does my pellet impact the target relative in the ret .

Position A, B ,C  etc on the large ret image 

Thats the challenge .

Its possible to work it out before shooting .

 So how well do you know the relation ship between scope and projectile trajectory  ? 

Answers please 

 

 

[strimmer_13]

I guess 'D'. Its a 1/9 chance 😂

 

[mad1]

Should still be "e"......same zero too....

Tin hat and popcorn time I think....🙃

[old'un]

G..

[Shufti]

I'm thinking E.

[Ttfjlc]

H, 1/9 chance, spent all week doing fractions with my son 

Edit: so it doesn't look just like a 'guess' the reason I went for H was due to gravitational forces at work. 

Edited May 2, 2020 by Ttfjlc

[JKD]

F

[Ultrastu]

Interesting to see how there are so many different guesses. 

I will divulge when we have a few more people guess. 

[GingerCat]

B.

[Pistol p]

E

[JKD]

Thought about it for a few more minutes and I think it's 'I'

Barrel is to the right of cross, pellet dropping when fired.

Edited May 2, 2020 by JKD

[Shufti]

Gravity will have an effect?

Maybe H

My head hurts!

[WalkedUp]

I do not believe it is necessarily computable given the information provided. The scenario relates to a 90° counterclockwise rotation from vertically aligned. As the projectile is always falling under gravity and the sight plane is horizontal, that excludes A-F. As the sight plane and barrel/trajectory were convergent to 30 yards with the barrel tilted upwards to the plane of the sight that excludes I.
 

The exact position between G&H is dependent upon projectile speed and the offset distance (height in the original orientation) between the centreline of the scope and centreline of the barrel, thus original drop, as the effect of gravity is now perpendicular to the original set up the absence of its depression of the rise of the projectile will shift POI to a position “higher” in the original, or leftwards in the rotated set up. Therefore I would guess at a smidgeon to the left of H assuming that scope was originally zeroed at the first zero point, a fast projectile and a minimal scope mount height.
 

However, as a parabola (the projectile’s path) and a straight line (the path of light into the reticle) always have two intersections (unless the line is tangential which is highly improbable in a real world scenario) there are always two zero points. If the original zero was to the second zero, which is unlikely to be what would be intended in a real world set up as 30 yards is not near the limit of the rifle’s effective range for target shooting, the projectile is slow, or the mount height is large then the impact would be closer to G.

[Ultrastu]

Im pleased to see some real grey matter going into working this out .

Yes your right there isnt a computor program i know of that will give you it exactly. But with the help of chairgun or strelock and some left field thinking a very good approximation can be made .certainly enough to hit a 25 mm spinner at said 30 yds first time  .😉

The scope hight  would be a normal 1.6 - 1.8 inches high and pellet speed around the usual 770 fps  .

But to be fair if we were talking .22 lr zeroed at 50 yds the correct  poi would be very similar ..

 

[johnphilip]

E

[Ultrastu]

15 minutes ago, WalkedUp said:

I do not believe it is necessarily computable given the information provided. The scenario relates to a 90° counterclockwise rotation from vertically aligned. As the projectile is always falling under gravity and the sight plane is horizontal, that excludes A-F. As the sight plane and barrel/trajectory were convergent to 30 yards with the barrel tilted upwards to the plane of the sight that excludes I.
 

The exact position between G&H is dependent upon projectile speed and the offset distance (height in the original orientation) between the centreline of the scope and centreline of the barrel, thus original drop, as the effect of gravity is now perpendicular to the original set up the absence of its depression of the rise of the projectile will shift POI to a position “higher” in the original, or leftwards in the rotated set up. Therefore I would guess at a smidgeon to the left of H assuming that scope was originally zeroed at the first zero point, a fast projectile and a minimal scope mount height.
 

However, as a parabola (the projectile’s path) and a straight line (the path of light into the reticle) always have two intersections (unless the line is tangential which is highly improbable in a real world scenario) there are always two zero points. If the original zero was to the second zero, which is unlikely to be what would be intended in a real world set up as 30 yards is not near the limit of the rifle’s effective range for target shooting, the projectile is slow, or the mount height is large then the impact would be closer to G.

Whats your guess then ? 

[Mice!]

Late to the party but thinking F as that's where the barrel is. 

[bluesj]

g

[figgy]

I'd go with G due to scope height from barrel with rise preset for gravity to its zero and drop from a level barrel will end up below the crosshairs.

I'm late to this,wonder how many went out and tried it 😂

Edited May 2, 2020 by figgy

[old'un]

so come on then whats the answer..zzzzzzzzzzzz

[Ultrastu]

Ok .the honours go to .

Old'un

Walkedup

Bluesj

And figgy 

The correct answer is 

G.

its late now will try and explain why in the morning .well done guys there is a lot going on in this puzzle and its way more complicated than it may first appear .

 

[enfieldspares]

C. Why. The pellet is still rising when fired. Therefore if the gun is turned its strike will be above the line of the barrel. But the 90 degree rotation will have the effect of it striking to the right of the point at which the crosshairs have been placed.

Edited May 2, 2020 by enfieldspares

[WalkedUp]

9 hours ago, Ultrastu said:

Yes your right there isnt a computor program i know of that will give you it exactly. But with the help of chairgun or strelock and some left field thinking a very good approximation can be made .certainly enough to hit a 25 mm spinner at said 30 yds

Oh I only meant computable in the old cerebral matter! I would be interested to do it for real today, simple enough to test! 

[bluesj]

Its all about gravity!

If you can do it without upsetting people at the moment have a go at loophole shooting.

[Shufti]

Gangsta shooting!