Only in relation to the ground to hold a constant heading or position. Otherwise the aircraft does not feel the wind.
Great explanation in this book:
https://www.kevinkoperski.com/post/stick-rudder/
Physics of shooting a rifle
- Thread starter tangolima
- Start date
- Status
Only in relation to the ground to hold a constant heading or position. Otherwise the aircraft does not feel the wind.
Great explanation in this book:
https://www.kevinkoperski.com/post/stick-rudder/
Anyone who's had to hold hard rudder & aileron landing into a strong crosswind . . . .
https://www.youtube.com/watch?v=NglxhkfP1ds&t=409s
https://www.youtube.com/watch?v=NglxhkfP1ds&t=409s
stagpanther
New member
For all airspeeds there is an associated angle of attack to the apparent wind at which an aircraft can be stalled. In the past few months commercial airliners have been able to exceed 700 or more mph indicated groundspeed over the US and when transiting the pond; nobody is breaking the sound barrier. Friend of mine just posted this yesterday doing a regional flight over the US.
Attachments
Last edited:
stagpanther
New member
stagpanther
New member
What? You don't like my drawing? I am an artist you know.
Bingo.
There are quite a few terms confused and convoluted to obscure a very simple truth.
"the aircraft does not feel the wind". Consequently, a bullet does not feel it either.
However, the airmass is moving and we will perceive the "wind drift" when our frame of reference is two stationary points on the ground that are NOT moving in the airmass.
Convoluting separate concepts of turbulence at airmass shear does not help illustrate that fact of physics.
Even in turbulence, an airfoil maintains a constant air density as it only produces the lift required. We perceive that maintenance of constant air density as movement about the CG and because our frame of reference is seperate in not being attached to the aircraft...we feel the acceleration.
Last edited:
stagpanther
New member
Errr...what??? I'm taking a wild guess and assuming what you mean is that due to camber/variable airfoil the Bernoulli effect (some people have argued that a sheet of flat plywood can fly quite well under the right circumstances) creates a net positive lift pressure under the wing when splitting the airflow over the wing?
I'm not arguing the drift vs push thing--what I'm pointing out is that in general where there's wind, there's turbulence, especially near the ground. Of the three sources of disturbed airflow I show in my crude drawing above (there are others, like downdrafts/catabatic winds) chances are very high you're going to be exposed to at least one of them while shooting from the ground.
Last edited:
When sideways (off centerline axis) pressure is applied, aircraft and boats "skid". They get moved off course. SO do bullets. Aircraft and water craft have control systems to compensate for that. Bullets do not.
Can we agree on that?
Car's tires have traction, which "anchors" them to the ground more firmly than boats in the water or airplanes in flight, they don't drift as much, but you can feel your car being rocked by a gust of sideways wind, so its obvious the pressure is there.
A bullet cannot compensate for sideways force the way a steerable vehicle can. So only the shape, surface area and speed are factors in how much drift from a given "push" the bullet undergoes.
It was stated that as the bullet goes faster the effect of "wind drift" goes down, (meaning the amount of off axis movement is reduced) and that makes sense to me.
However, it was also stated that when a bullet goes faster than the point where minimum drift happens, the amount of drift goes back up, and that I cannot figure out. Still hoping someone more skilled in this matter can explain it to me in a way I can grasp.
Thanks.
Can we agree on that?
Car's tires have traction, which "anchors" them to the ground more firmly than boats in the water or airplanes in flight, they don't drift as much, but you can feel your car being rocked by a gust of sideways wind, so its obvious the pressure is there.
A bullet cannot compensate for sideways force the way a steerable vehicle can. So only the shape, surface area and speed are factors in how much drift from a given "push" the bullet undergoes.
It was stated that as the bullet goes faster the effect of "wind drift" goes down, (meaning the amount of off axis movement is reduced) and that makes sense to me.
However, it was also stated that when a bullet goes faster than the point where minimum drift happens, the amount of drift goes back up, and that I cannot figure out. Still hoping someone more skilled in this matter can explain it to me in a way I can grasp.
Thanks.
stagpanther
New member
I would add time (duration) of flight exposed to the wind to that.
CL = L/qS
p = Density
q = 1/2 * p * V^2
L = Lift = weight in steady state flight = constant
S = Lifting Surface Area = constant
When an aircraft moves thru the airmass it must maintain that constant density to remain in flight. When that density changes due to a change in air mass velocity at the shear.....
The density remains constant from the aircrafts FRAME OF REFERENCE as the aircraft will remain at a constant density. YOUR Frame of Reference is not the aircrafts and because you are not in that same Frame of Reference you will feel it as motion as the aircraft keeps that same density.
https://physics.info/frames/
That sideforce is called spindrift. It is a factor in long range precision shots.
I don't think that's an explanation that provides any sort of intuitive insight. I suspect he's looking for answers to these questions that will provide an intuitive grasp of the situation.
Drag always goes up with velocity, so it's not simply a matter of drag increasing. WHY does the increased drag, even with lower TOF result in more POI change due to wind?
Intuitively, with a shorter time for the wind to operate on the bullet, one thinks it should move less. What effect is resulting in more POI change due to wind even though the wind is the same and has less time to operate on the projectile?
You mention that this doesn't happen all the time. What are the specific circumstances under which increasing velocity also increases POI change due to wind and how do those circumstances result in this effect?
Drag always goes up with velocity, so it's not simply a matter of drag increasing. WHY does the increased drag, even with lower TOF result in more POI change due to wind?
Intuitively, with a shorter time for the wind to operate on the bullet, one thinks it should move less. What effect is resulting in more POI change due to wind even though the wind is the same and has less time to operate on the projectile?
You mention that this doesn't happen all the time. What are the specific circumstances under which increasing velocity also increases POI change due to wind and how do those circumstances result in this effect?
There is a fundamental misunderstanding in the term "Wind Shear". There seems to be some belief that turbulence is wind shear. It is not. Wind shear is a very isolated phenomenon that occurs between airmass in only four situations. Turbulence occurs in all airmasses. We feel turbulence because we are not physically a part of the aircraft. That is why they tell you to remain seated with your seat belt buckled when experiencing it.
Wind shear is a very different animal. In order for your bullet to experience Wind Shear effects it would have cross gust front of a Thunderstorm and the wind shift line between airmasses. They convergence of those two things MIGHT create a shear. Mostly it will just make turbulence.
The other conditions for Wind Shear are simply not something you would ever experience on a rifle range as they occur at altitude.
Wind deflection and drag, see unclenick's explanation in post #70. Basically the force that makes the bullet move sideways is syphoned off the drag. The magnitude of the cross wind determines how much of the drag force got diverted.
With high speed and low BC, the drag force increase, so does the force that move the projectile sideways. If the reduced TOF can't offset the effect of the increased force, wind deflection comes back up.
I will dig up the McCoy book to get a analytical solution when I have a chance. For now just try this on a ballistics calculator. Keep increasing MV and observing the change in wind deflection.
-TL
Sent from my SM-N960U using Tapatalk
This may help the discussion.
https://www.mdpi.com/2226-4310/9/12/816/pdf?version=1670847097
Yes you did and a very good concise explanation too.
https://www.mdpi.com/2226-4310/9/12/816/pdf?version=1670847097
Yes you did and a very good concise explanation too.
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
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
The only thing I would change in your explanation is the term "drag". Drag is opposed by thrust and works only in that axis. What you are talking about is the sideforce created by the yaw of the bullet in the direction of airmass movement.
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.
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
-TL
Sent from my SM-N960U using Tapatalk
- Status