TargetTalk

I'm trying to find out the lock & barrel time for an LP-10, for use in analysis of SCATT trace. Steyr have responded to email with a time of 2mS (0.002 second) for the time between the click of the action & pellet exiting the barrel. This is a calculation they have made based on average pellet velocity of 120m/S during time in barrel. 2mS seems to be so short that it should have negotiable effect on the final impact point on the target, this is contrary to what I would have read. Can anyone confirm the lock/ barrel time as being in this order?
Joe

joecon wrote:I'm trying to find out the lock & barrel time for an LP-10, for use in analysis of SCATT trace. Steyr have responded to email with a time of 2mS (0.002 second) for the time between the click of the action & pellet exiting the barrel. This is a calculation they have made based on average pellet velocity of 120m/S during time in barrel. 2mS seems to be so short that it should have negotiable effect on the final impact point on the target, this is contrary to what I would have read. Can anyone confirm the lock/ barrel time as being in this order?
Joe

Sounds about right. The pellet initial velocity is zero and the pellet muzzle velocity is typically 150m/s, so the Steyr people have estimated the average velocity to be 120 m/s. With a barrel length of 250mm, the transit time calculates to 250/120=2.08 ms.

Let's say the muzzle were to deflect at a rate of 1 m/s, not a particularly high rate of movement. Let's also assume that this deflection occurs because your arm has moved, not the sight alignment. A muzzle rotation of 2mm in 2 msec on a radius of ~ 1 meter (distance from shoulder to muzzle), will displace the shot 20mm on the target 10m away.

Fred Mannis wrote:

joecon wrote:I'm trying to find out the lock & barrel time for an LP-10, for use in analysis of SCATT trace. Steyr have responded to email with a time of 2mS (0.002 second) for the time between the click of the action & pellet exiting the barrel. This is a calculation they have made based on average pellet velocity of 120m/S during time in barrel. 2mS seems to be so short that it should have negotiable effect on the final impact point on the target, this is contrary to what I would have read. Can anyone confirm the lock/ barrel time as being in this order?
Joe

Sounds about right. The pellet initial velocity is zero and the pellet muzzle velocity is typically 150m/s, so the Steyr people have estimated the average velocity to be 120 m/s. With a barrel length of 250mm, the transit time calculates to 250/120=2.08 ms.

Let's say the muzzle were to deflect at a rate of 1 m/s, not a particularly high rate of movement. Let's also assume that this deflection occurs because your arm has moved, not the sight alignment. A muzzle rotation of 2mm in 2 msec on a radius of ~ 1 meter (distance from shoulder to muzzle), will displace the shot 20mm on the target 10m away.

Nice math skills Fred ;)

Thanks. At least some of the gray cells are still working :-)

Can be done. It is an approximation because there are many factors involved.

barreltime = (barrel length [meters] * 4) / (pi * V0)

250mm barrel length minus pellet length (5.5mm) and a V0 gives precisely 0.00208 second - I assume that Steyr has used this formula for their answer. Average velocity: 0.25m barrellength / 0.00208 sec. = 120.19 m/s

Albert B
(The Netherlands)

Pellet weight and size are also a small factor in the overall calculation.

Peeps

My Math isn't great but where does the 4 and Pi come into the calulation? Pi is only used for angular measurements. Here we are talking a straight line - its simply length divided by average velocity.

Cheers

David

higginsdj wrote:My Math isn't great but where does the 4 and Pi come into the calulation? Pi is only used for angular measurements. Here we are talking a straight line - its simply length divided by average velocity.

Cheers

David

I believe Albert is offering a formula for estimating barrel time from measurement of the barrel length and the muzzle velocity. Then, having calculated the barrel time, you can calculate the average pellet velocity in the barrel. The average pellet velocity in the barrel cannot be directly measured.

Does anyone know the force exerted on the pellet? If we do then:

dTime = force/(muzzel velocity * pellet weight)

Interestingly if we use dTime = 0.00208seconds then we can calculate the force and have a complete picture to determine velocity or dTime for any pellet weight (if we assume the original calculations were based on a known pellet weight)

BUT as I said my Math skills aren't great so I could be talking a whole heap of rubbish :)

higginsdj wrote:Does anyone know the force exerted on the pellet? If we do then:

dTime = force/(muzzel velocity * pellet weight)

Interestingly if we use dTime = 0.00208seconds then we can calculate the force and have a complete picture to determine velocity or dTime for any pellet weight (if we assume the original calculations were based on a known pellet weight)

BUT as I said my Math skills aren't great so I could be talking a whole heap of rubbish :)

force=0.25*pi*(d^2)*P, where d=pellet diameter, P=air pressure

d=4.5mm and p=80 psi. Don't have all psi to metric conversion handy, so won't bother with the calculation

http://www.unitconversion.org/pressure/ ... rsion.html

But I don't know which one to use.......

joecon wrote:I'm trying to find out the lock & barrel time for an LP-10, for use in analysis of SCATT trace. Steyr have responded to email with a time of 2mS (0.002 second) for the time between the click of the action & pellet exiting the barrel. This is a calculation they have made based on average pellet velocity of 120m/S during time in barrel. 2mS seems to be so short that it should have negotiable effect on the final impact point on the target, this is contrary to what I would have read. Can anyone confirm the lock/ barrel time as being in this order?
Joe

This appears to not include the 'lock' time; i.e. for an air gun, the time for the mechanicals to operate and the pressure to build up to overcome the various factors before the pellet begins to move.

Which has started me on a seach for definitions of 'lock time'. A quick google returns acceptable definitions for cartridge guns (release of sear to impact on primer), but not for air guns

Spencer

Fred -
I just knew those years at MIT would eventually yield a benefit.

Your pal - Dennis

Fred.Mannis wrote:
.... and p=80 psi. Don't have all psi to metric conversion handy, so won't bother with the calculation

I think this should be 80 bars/~1200 psi

OzzieM

You can calculate the acceleration of the pellet using the simple f=ma equation if you know its mass and the nett force acting on it. This nett force is the forwards force from air pressure behind, less friction, acting in the opposite direction. Because this type of air gun uses a metered small volume of air to propel the pellet, the force drops off as the pellet moves along the barrel (as the gas expands, its pressure and temperature go down). If the force were constant, the acceleration would be linear and the average in-barrel velocity would be half the exit velocity. The fact that the factory quote an average velocity so close to the exit velocity suggests the pellet is accelerated at a high rate initially, then, as the propelling force reduces, acceleration falls off.

This begs the question why have such a long barrel? If you want a short barrel time, shorten the barrel! You won't lose much muzzle velocity.

Anyway, if acceleration were constant, v=at, where v=muzzle velocity and t=barrel time. Since v=s/t, (s=barrel length) substituting this in the previous equation and rearranging gives t^2=a/s, so you have the relationship between all the parameters under consideration, but this schoolboy calculation is invalid for the reasons given above. Real life aint so simple.

I'd have thought instrumentation of the rifle and direct measurement of barrel and lock times had been done by the factory - not too difficult these days.