By how much will I miss the target?

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I just started shooting with an air gun that I got. What I wonder is this; if I from 5 meters miss the bullseye with let's say 2 cm (0,02 m), does that automatically mean that I'll miss it by 4 cm from 10 meters? I thought I would miss it by much more in the latter case, but that's what I found playing around with the law of sines, cosines etc. Can that be right...?
 
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It depends.

If you miss because your barrel is not perfectly straight and aligned with the sight, because your pellet is not symmetrical, etc., then yes, 2 cm error from 5 meters means 4 cm error from 10 meters.

There are other possible sources of error. Earth's gravity makes the pellet drift downwards. As the pellet flies through the air, it slows down because of air resistance and, the slower it goes, the more it is susceptible to crosswinds.
 
In addition to the error sources mentioned by hamster, the Coriolis force can have a subtle effect.

This was experienced by British soldiers in the Falklands War; they had been trained to aim slightly off-target in order to compensate for the Coriolis effect, but going to the southern hemisphere, they had to re-adjust by aiming off-target to the OTHER side from which they were used to.
 
hamster143 said:
It depends.

If you miss because your barrel is not perfectly straight and aligned with the sight, because your pellet is not symmetrical, etc., then yes, 2 cm error from 5 meters means 4 cm error from 10 meters.

No, that's not necessarily true. The projectile (pellet, BB, etc.) loses speed as it continues to fly through the air, therefore, the farther you are from the target, the more speed the projectile loses and the farther it will drop from the point you are aiming at. Since the projectile's speed is not constant, and the effect of gravity is constant, the ratio of error to distance is not linear. That being said, there is a distance within which the projectile's speed remains fairly constant (which depends on the projectile, itself as well as the barrel design and means of acceleration); within that range, you can expect the results you quoted.
 
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