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#176 | ||
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Join Date: January 13, 2018
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This may help the discussion.
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#177 |
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Join Date: September 28, 2013
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An other way to visualize the mechanism. In point mass model, wind deflection is proportional to lag time (Tlag), which is the difference between TOF and time (To) the projectile would have taken to travel the same distance in vacuum.
Tlag = TOF - To With higher MV, both TOF and To reduce. However if decrease in TOF can't catch up with decrease in To, which can happen with high BC, Tlag increases, and so does wind deflection. -TL Sent from my SM-N960U using Tapatalk |
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#178 | |
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I know, it is nit-picky and yes that force is effected by velocity and mathematically is calculated is almost the exact same way as drag/lift.....as a function of dynamic pressure to static. The adherence to the reference axis is how we keep the forces straight when doing the vector resolution. For example in a climb in an aircraft, Lift Force is decreased because a vector of thrust is now contributing to lift along the lift axis. |
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#179 |
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Join Date: September 28, 2013
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Unclenick has comprehensive explanations on the wind deflection and drag in post #70. The way I see it, the cross wind changes the angle of the apparent wind on the projectile. That minute angle change can be visualized as yaw. If the drag force vector is decomposed into 2 orthogonal components, a small force is pointing sideways and causes the deflection.
-TL Sent from my SM-N960U using Tapatalk |
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#180 |
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Excellent synopsis.
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#181 | |
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In terms of intuitive insight, the TOF-To difference relationship to POI change is actually even more problematic because it sort of implies that the projectile somehow "knows" its flight time in a vacuum. The point is that it's easy to see that if you have wind acting on a projectile, the longer it acts on the projectile, the more effect that will have on the POI. It's not at all easy to see why, in some cases that you can decrease the time the wind acts on the projectile without decreasing the effect on the POI. Pointing out that it's a function of the difference between the actual TOF and the theoretical TOF in a vacuum may explain how the model works, but it still doesn't really give someone a good "feel" for why the shorter TOF doesn't result in less POI error. Also, there's the issue of why this happens sometimes but not other times. Low BC is referenced, (airgun pellets, the referenced projectiles are not just low BC by bullet standards, they are off the charts low) but that implies that there's some sort of BC threshold. What is it? Pellets are usually shot at very low velocities, compared to rifle bullets, is there a limited velocity range where this occurs?
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#182 | |
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#183 |
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Post #70 and #175 may help. If it still doesn't, please allow me some more time to work out an mathematical equation. After that, I'm afraid I will be quite out of witts.
![]() High drag offsets the benefit of low TOF. That's what's going on. Low MV and super low BC for pellet. Probably super high MV and low BC of light varmint bullet would do the same. -TL Sent from my SM-N960U using Tapatalk |
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#184 | |
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"Everyone speaks gun."--Robert O'Neill I am NOT an expert--I do not have any formal experience or certification in firearms use or testing; use any information I post at your own risk! |
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#185 | |
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I'm freshly into airgun after buying my first PCP rifle due to some "bad influence". Airguners love to tune their guns for higher MV. That article pointed out that faster is not always better. There exists optimum MV for minimum wind deflection. For .177 cal pellets, 800fps ish is the optimum. -TL Sent from my SM-N960U using Tapatalk |
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#186 | |
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Parasitic Drag and Induced Drag. Induced Drag is a function of lift production. Parasitic Drag is everything else. Skin Friction Drag Form Drag Interference Drag Leakage Drag Profile Drag Low Time of Flight = Higher Velocity Velocity has a direct and squared relationship to Parasitic Drag. Mathematically, the relations of Parasitic Drag to velocity it is expressed as: Dp2/Dp1 = (V2/V1)^2 Dp1 = Parasitic Drag Force at beginning velocity Dp2 = Parasitic Drag Force and ending velocity V1 = Beginning velocity V2 = Ending velocity So doubling the speed is four times the Parasitic Drag forces. |
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#187 | |||
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Air Mass has a very specific definition: Quote:
Turbulence is characterized by changing Coefficients of Pressure resulting in torqueing moments about the CG. It is a Stability and Control problem NOT a Performance Problem. Different frame of reference and very different results. Torque is a rotational movement ABOUT an axis. It is not a positional change in the axis itself. Density will remain constant so the axis position does not change. That is why we say, "An object in flight does not feel the wind". Quote:
I don't pretend to know it all or understand everything about this particular problem. Nobody does and that is evidence by the report I posted earlier investigating the behavior of a bullet in flight. Fluid dynamics, Aerodynamic, Advanced Aerodynamics, and Stability and Control were all classes in college for me. I think it is useful to explain the behaviors of a bullet using aircraft analogy. Much more useful than arbitrary Ballistic Coefficients and previous physics expressions. Last edited by davidsog; January 21, 2024 at 12:49 PM. |
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#188 | |
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So, cutting to the chase--because I can't understand a flippin thing about what you're saying here--you are refuting that 1) shear is a form of turbulence; and 2) it can occur at or near the surface (0 agl)?
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"Everyone speaks gun."--Robert O'Neill I am NOT an expert--I do not have any formal experience or certification in firearms use or testing; use any information I post at your own risk! Last edited by stagpanther; January 21, 2024 at 01:06 PM. |
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#189 | ||
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Turbulence is created as an airmass moves across the ground due to trees, buildings, terrain, and other obstacles. Turbulence is not shear. Turbulence is defined by creation of torque forces around the CG. Turbulence is a stability and control problem. Shear is defined by large changes in density and velocity across an airmass. Shear is a performance problem as because of the magnitude of density and velocity changes. Does that help? |
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#190 |
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Stagpanther,
Maybe this will help. You know what a Center of Gravity is right? NASA defines it as the "the average location of the weight of an object". It is just a single point. I can rotate that point any direction I want without changing the position in space of that point. That is the effect of turbulence. If I move that point in space to a different location, that is the effect of Shear. |
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#191 | |
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#192 | |
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Join Date: March 2, 2014
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"Everyone speaks gun."--Robert O'Neill I am NOT an expert--I do not have any formal experience or certification in firearms use or testing; use any information I post at your own risk! |
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#193 | |
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#194 | |||
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All wind shear will create turbulence but all turbulence is not wind shear. In fact, turbulence created by wind shear is a rare occurrence. Quote:
Nobody ever said anything about this being strictly horizontal or vertical. I don't know where you got that from. Last edited by davidsog; January 21, 2024 at 02:05 PM. |
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#195 |
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Join Date: March 2, 2014
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I agree with your definitions for the most part, I think the differences between your interpretations and mine are due mostly due to nit-picking interpretations. For example, an upper air inversion--the bane of thermal soaring pilots--I would describe as more of a stabilizing body of air than one associated with turbulence since it's warmer than the previous layers of air underneath it it tends to repress or stop altogether adiabatic lapse activity from rising convectively heated air. You can easily observe this when flying and see a clearly-defined boundary between "clean air" and murky, polluted air trapped below it. Inversions also can aid as a "bounce barrier" in setting up wave amplitude propagation to great heights and distances; enabling soaring unpowered aircraft to reach altitudes that are attained by airliners and fly well over a thousand miles.
What does that have to do with us shooting on the ground? Not much, except that any time the sun is out we're likely going to be experiencing convective heating and there are periods during the day when greater equilibrium between ground and near above surface air temperature will be reached. By the way, when I clicked on the link for your linked article, this is what popped up on my browser; I just love those people who cherish our privacy. ![]() ![]() ![]() ![]()
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"Everyone speaks gun."--Robert O'Neill I am NOT an expert--I do not have any formal experience or certification in firearms use or testing; use any information I post at your own risk! |
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#196 |
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Join Date: March 11, 2006
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I guess this is one of those physics problems I just don't get. Like the one where a guy gets into the pool, swims to the far end, swims back and climbs out where he went in.
The instructions for that one said to use physics to prove he went nowhere. I don't think he went nowhere. Aircraft terminology isn't very helpful to me, when looking at bullets in flight all I know is that a cross wind moves the point of impact of my bullet, proportional to the force and direction of the wind.
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All else being equal (and it almost never is) bigger bullets tend to work better. |
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#197 |
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He (the swimmer) in fact "went" nowhere.
Vector A + vector B = 0 No physics involved. Looking backwards though... the bullet does feel the full force of the crosswind upon muzzle exit. But as it gradually succumbs to the force over time, and increasingly drifts in the direction of that cross force, it "feels it less and less |
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#198 | |
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All else being equal (and it almost never is) bigger bullets tend to work better. |
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#199 | |
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"Everyone speaks gun."--Robert O'Neill I am NOT an expert--I do not have any formal experience or certification in firearms use or testing; use any information I post at your own risk! |
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#200 | ||||||||||||
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What happens is the bullet rotates about its CG until those torque moments equalize the force and the bullet achieves equilibrium just like an airplane. Yes, it is decelerating from the moment it is fired as a finite amount of thrust force has been applied by the powder charge but does not negate the principle of it achieving equilibrium in the side force yaw axis. That is an almost instantaneous process as the force is applied. While it is interesting to examine what is going on in that tiny moment in time, it is a Tree and not the Forest. The POSITION of the CG remains the same despite it rotating about its axis. Once equilibrium is achieved, the bullet no longer feels the wind but moves as any other object suspended in fluid. The whole mass of fluid is moving so the bullet moves just like a passenger on a train. Quote:
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Last edited by davidsog; January 22, 2024 at 11:22 AM. |
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