Bullets in the Wind 17's and 22's all rimfire! Different reports!

ivanthehunter

Quick question!
I have heard some people claiming that 17s are better a cutting through the wind than their parents, the 22's.
Some people claim that larger 22 rounds suffer less from the wind due their extra mass.

Could both be part-true? especially when the range distance are taken into account.

Might it be that the faster 17's show more apparent resistance to the wind simply because of their super fast velocities and they are exposed to the wind for a shorter duration-- obviously their comes a point when the velocity becomes slower and wind has a longer time to push the bullet of course but this seems to only happen at ranges of 100-150m???

IMO i believe that the heavier 22 rounds show higher resistance to wind at longer ranges! But do they suffer from wind worse within the early stages of flight when compared to a 17cal??

clivej

I was at the ranges last week and a good cross wind was blowing. Using 22 subs my POI was 1 1/2" off zero.
But after I adjusted for this I was back with 12mm-15mm 10 shot groups @50m.:D

Don't know about the 17 rounds in the wind.

ejg

According to the Federal Data. The 17HMR has about 3-4" wind drift at 100yds and the 22 rimfire's have 4-7" wind drift at the same distance. (10mph wind at 90 deg.)
Not such a huge difference.
edi

ivanthehunter

ejg wrote: »

According to the Federal Data. The 17HMR has about 3-4" wind drift at 100yds and the 22 rimfire's have 4-7" wind drift at the same distance. (10mph wind at 90 deg.)
Not such a huge difference.
edi

All depend on the range.

kowloon

What's the weight of the heaviest common .17 in comparison to a velocitor or something along those lines?

Vegeta

ivanthehunter wrote: »

All depend on the range.

He gave the range as 100 yards

Mellor

ejg wrote: »

According to the Federal Data. The 17HMR has about 3-4" wind drift at 100yds and the 22 rimfire's have 4-7" wind drift at the same distance. (10mph wind at 90 deg.)
Not such a huge difference.
edi

Funny that, If I was given those numbers, i'd of said the difference was pretty big. .17 averaging at 3.5" and .22 at 5.5", almost 60% more drift

kowloon wrote: »

What's the weight of the heaviest common .17 in comparison to a velocitor or something along those lines?

The .17HMR comes commonly as a 17grain or 20grain round. Velocitors are 40 grain (if i remember correct)




Ivan, you are generally correct in your opening post.
Extra mass would reduce the affect wind has,
As would a higher speed, due to a greater force behind bullet (not suite the same as your time suggestion).

Also, the further the distance, wind (assuming wind speed is constant) will have a greater affect. This is due to the speed of the bullet is dropping and the force of the wind is greater in relation to the remainly bullet energy/force.

rrpc

The wind will have the greatest effect on the point of impact at the muzzle of the rifle, not towards the end of its travel.

This is because the wind doesn't shift the bullet in it's flight path, but changes it's angle.

So if the wind moves the point of impact on the target 2" to the left, it probably only shifted it millimetres at the muzzle, but the change in angle translated to 2" at (say) 100 yards, but this would be greater at 200 yards etc.

To be 9" off at 100 yards, the angular shift would only need to be .14 of a degree at the muzzle, at 50 yards the angular shift would have to be .29 of a degree and at 80 yards .72 of a degree.

firefly08

his is because the wind doesn't shift the bullet in it's flight path, but changes it's angle.

Are you sure? That doesn't make sense to me. The wind will begin pushing the bullet the moment it leaves the muzzle, and will keep pushing it until the bullet stops.

As far as I can make out, a constant wind pushing at the bullet in a constant direction will move it by a constant distance per unit time, throughout it's travel. I can't see how the wind will affect it more of less at either end. The reason it seems this way is because the bullet is slowing down (it has nothing to do with the amount of energy the bullet has).

Because it's going slower at the end of it's travel, the ratio of lateral movement to forward movement in a given time frame increases - i.e. the path of the bullet is a curve. If the speed of the bullet remained constant, the path would be a straight line (but would still be off target of course!).

Might it be that the faster 17's show more apparent resistance to the wind simply because of their super fast velocities and they are exposed to the wind for a shorter duration-- obviously their comes a point when the velocity becomes slower and wind has a longer time to push the bullet of course

This is exactly right. It's exactly the same as how gravity affects the bullet.

rrpc

firefly08 wrote: »

Are you sure? That doesn't make sense to me. The wind will begin pushing the bullet the moment it leaves the muzzle, and will keep pushing it until the bullet stops.

Well you can use the analogy of a speeding car as an example. If a speeding car is hit on the side, it won't shift x feet to the other side and continue moving in the same direction, but will deviate from its path and continue to travel in a different direction.

firefly08

Well, I don't think the analogy of a speeding car is appropriate - do you mean a car being propelled by it's own energy source, as cars normally are? And by "hit on the side", do you mean hit once, by a transient force(i.e another car, which then stops), or continuously pushed from the side by a constant force?

The bullet is slowing down from the moment it leaves the muzzle. The force is aplied only once, at at the start.The wind, on the other hand, is constantly pushing it throughout it's journey. As the bullet slows, the ratio of lateral to forward movement in a given time frame increases; hence the curve. Gravity has exactly the same effect.

Aside form that difference, I think you are mistaken about how the force affects an object's direction anyway. I could be wrong about this, but here's what I think:

In theory, if you applied a small transient force, from the side, to a bullet fired from a gun, the bullet would of course start to move in that direction. But if left to travel for long enough (here the "in theroy" bit kicks in !) the momentum of that sideways force would eventually run out. Now, there is no longer any force pushing the bullet in that sideways direction - it will therefore cease to move in that direction, but will continue in in the original direction (parallel to the original path).

This would be the same with a car (since the wheels dictate the direction it can move in), nor would it be the same in a vacuum (since the momentum would never run out, it would always keep traveling in the new direction, imparted by the sideways force.) But for a bullet in the air, that energy must eventually run out.

rrpc

It's important to know what is happening when wind strikes a bullet. It doesn't actually blow the bullet off course; what happens is if the wind strikes the left hand side of the bullet, the air pressure on that side is higher and the bullet will veer to the lower pressure side. A sailing boat being hit by the wind is another good example as it won't jump to a position opposite the wind direction and continue on course, but will change course.

THere's also the fact that the bullet is spinning clockwise at around 50,000 rpm and so the high pressure area will creep upwards on the left side of the bullet and down on the right side, which will cause the point of impact to be to the right and low. The windchart below illustrates this.



The numbers are the wind direction on the clock face with 12 being straight at you and 6 being behind you. You can see that the 9 o'clock wind is striking right and low and the three o'clock wind is left and high.

Once you accept that it's an angular deviation that you're dealing with and not a shift, it's obvious that the further along the bullet's travel the wind strikes, the less the point of impact will change.

clivej

rrpc wrote: »

The wind will have the greatest effect on the point of impact at the muzzle of the rifle, not towards the end of its travel.

But then what about if you shoot a bullit in a sheltered enclosure where there is NO wind at the muzzeland for the next 20m. The bullit travels say 20m before a side wind hits it. Will it still get blown off course for the rest of it travel say the next 30m to the same extent as it would if there was no shelter?

rrpc

clivej wrote: »

But then what about if you shoot a bullit in a sheltered enclosure where there is NO wind at the muzzeland for the next 20m. The bullit travels say 20m before a side wind hits it. Will it still get blown off course for the rest of it travel say the next 30m to the same extent as it would if there was no shelter?

No, that's the point. Look at the bullet travel as a triangle with the point of aim and the muzzle as a straight line and the point of impact to the muzzle as the hypotenuse of that triangle (the deviation at the target being the base).

If the bullet travels along the line of sight for the first 20m, and then gets an angular shift, then the base of the triangle is shorter meaning that your point of impact is closer to your point of aim. Assuming it's a constant wind of course.

Mellor

rrpc wrote: »

This is because the wind doesn't shift the bullet in it's flight path, but changes it's angle.

rrpc wrote: »

Well you can use the analogy of a speeding car as an example. If a speeding car is hit on the side, it won't shift x feet to the other side and continue moving in the same direction, but will deviate from its path and continue to travel in a different direction.

Sorry rrpc, this is wrong.
The huge difference with the car analogy is that it is struck only once. Its addition of two vector forces, which gives a linear resultant force from the point of impact. ie changes it's course but remains travelling straight.

A bullet (or any object travelling through the air) and the wind differ because the wind is a constant force. The bullet path is an arc, just like the effects of gravity.

rrpc wrote: »

It's important to know what is happening when wind strikes a bullet. It doesn't actually blow the bullet off course; what happens is if the wind strikes the left hand side of the bullet, the air pressure on that side is higher and the bullet will veer to the lower pressure side.

You are right that its the change in air pressure that causes the bullet to shift but thats exactly what blowing is. From blowing a small ball to pieces of paper, its all changes in pressure. they are the same thing.

THere's also the fact that the bullet is spinning clockwise at around 50,000 rpm and so the high pressure area will creep upwards on the left side of the bullet and down on the right side, which will cause the point of impact to be to the right and low. The windchart below illustrates this.

This bit is correct. The spinning causes the bullet to shift up or down.

Once you accept that it's an angular deviation that you're dealing with and not a shift, it's obvious that the further along the bullet's travel the wind strikes, the less the point of impact will change.

No, the reason the bullet shifts less when the bullet is struct by wind further along its travel, is because that the time that it is exposed to wind is less and/or the distance that the wind affects it is left.

No, that's the point. Look at the bullet travel as a triangle with the point of aim and the muzzle as a straight line and the point of impact to the muzzle as the hypotenuse of that triangle (the deviation at the target being the base).

If the bullet travels along the line of sight for the first 20m, and then gets an angular shift, then the base of the triangle is shorter meaning that your point of impact is closer to your point of aim. Assuming it's a constant wind of course.

I don't see how this backs up your idea that the path is linear.
Obviously, when affected by wind over shorter distance, the shift is less, no matter if the path is a straight line at an angle or an arc.

rrpc

Sorry Mellor, I was simplifying the situation. It's obvious that the effect is an arc (dependant on the wind conditions of course).

The car analogy is fine though because if you hit the car again, it will shift direction again, but the difference is that we are measuring that shift against a common endpoint (the target). The shift in angle at the start of the travel will have a far greater impact on the endpoint of it's travel then it would halfway along its path (everything else being equal).

The wind hitting a bullet can be looked at as a series of 'impacts' along it's route, each minutely changing it's angle but each one having less and less of an effect on it's point of impact as it gets closer to the target because the adjacent side of the triangle is getting shorter.

Mellor wrote:

The huge difference with the car analogy is that it is struck only once. Its addition of two vector forces, which gives a linear resultant force from the point of impact. ie changes it's course but remains travelling straight.

We are agreeing here. What firefly was implying is that the wind shifts the bullet laterally, but it's angle of travel doesn't change, i.e. it's still travelling in a straight but parallel line to the line of sight. In other words, if it was a single gust of wind at the start of its travel, the bullet would only deviate from the point of impact by the amount it was offset at the start.

kowloon

Not quite on topic, but: if I were to get a dedicated bunny rifle, would you go for the .17 or the .22? all things being equal.
I shoot in quite a windy area and the bunnies can see me coming a mile away because of the terrain.

Suggestions?

rrpc

Mellor wrote: »

I don't see how this backs up your idea that the path is linear.
Obviously, when affected by wind over shorter distance, the shift is less, no matter if the path is a straight line at an angle or an arc.

I'm not saying the path is linear, I'm saying that the deviation is angular. With an angular deviation you will get an arc because the angle keeps changing.

firefly08

In other words, if it was a single gust of wind at the start of its travel, the bullet would only deviate from the point of impact by the amount it was offset at the start.

Yes, that's exactly what I meant. That and the fact that the bullet travels in a curve - I didn't realize you agreed with that (that's what I was going to ask next!)

I'm open to correction about the point referenced above - but I can't shake the idea that the momentum pushing the bullet sideways must run out eventually, if the force is momentary. I'm thinking of it as two vector forces working perpendicular to each other -, I think that would certainly apply if the bullet was self-propelled - but maybe I'm wrong to treat them as separate components in this case.

rrpc

kowloon wrote: »

Not quite on topic, but: if I were to get a dedicated bunny rifle, would you go for the .17 or the .22? all things being equal.
I shoot in quite a windy area and the bunnies can see me coming a mile away because of the terrain.

Suggestions?

Depends on the kinds of distances you're shooting at. If it's over a hundred yards, the .17 would be better. Under 100 yards, the .22 would be fine, but if the distances vary a lot especially beyond 100 yards, you'd want to be pretty good at reading distances and settingyour sights correctly.

Mellor

rrpc wrote: »

Sorry Mellor, I was simplifying the situation. It's obvious that the effect is an arc (dependant on the wind conditions of course).

The car analogy is fine though because if you hit the car again, it will shift direction again, but the difference is that we are measuring that shift against a common endpoint (the target). The shift in angle at the start of the travel will have a far greater impact on the endpoint of it's travel then it would halfway along its path (everything else being equal).

The wind hitting a bullet can be looked at as a series of 'impacts' along it's route, each minutely changing it's angle but each one having less and less of an effect on it's point of impact as it gets closer to the target because the adjacent side of the triangle is getting shorter.



We are agreeing here. What firefly was implying is that the wind shifts the bullet laterally, but it's angle of travel doesn't change, i.e. it's still travelling in a straight but parallel line to the line of sight. In other words, if it was a single gust of wind at the start of its travel, the bullet would only deviate from the point of impact by the amount it was offset at the start.

rrpc wrote: »

I'm not saying the path is linear, I'm saying that the deviation is angular. With an angular deviation you will get an arc because the angle keeps changing.

Yeah, we are saying the same thing, the angle keeps changing = an arc. I completely misunderstood you at first.

I no what you mean now about the greatest shift being at the start, a gust at the start cause a bigger shift due to the longer baseline. A gust at the end will cause a greater change in angle, but the shift in inches is less.

rrpc

firefly08 wrote: »

Yes, that's exactly what I meant. That and the fact that the bullet travels in a curve - I didn't realize you agreed with that (that's what I was going to ask next!)

Sorry firefly, but I was just explaining what you were saying, not agreeing with it.

I'm probably not explaining it very well and it's a tough concept to get across without diagrams, but I suppose the best way is to try a test.

If you could set up two targets in your line of sight: one at 25 yards and one at 100 yards, making sure that the centre of each were in line with your muzzle (vertical deviation doesn't really matter here) and fired a shot on a gust of wind. You would then see that the point of impact on the 25 yard target was less offset than the one at 100 yards.

If you shoot with wind flags, you could watch the closest ones to you and if they are not showing much wind, even if the farther ones were, the point of impact won't be that affected as if the situation was reversed.

rrpc

Mellor wrote: »

Yeah, we are saying the same thing, the angle keeps changing = an arc. I completely misunderstood you at first.

I no what you mean now about the greatest shift being at the start, a gust at the start cause a bigger shift due to the longer baseline. A gust at the end will cause a greater change in angle, but the shift in inches is less.

That's it exactly, I was beginning to despair if anyone would understand me

Sparks

firefly08 wrote: »

I can't shake the idea that the momentum pushing the bullet sideways must run out eventually, if the force is momentary.

It does, it runs out instantly if the force is momentary. Thing is, it almost never is, it operated continously over the flight time of the bullet.
Doesn't matter really though, you're not really interested in the flight path (that's what range safety is for), you're interested in the point of impact some N metres away. So you can integrate the force over the flight path and treat the whole thing as a simple vector sum (the integration isn't quite so trivial though, you're integrating a function over a path).

Of course, in the end the maths doesn't really matter too much, you can learn to aim off or correct with sights without needing to be able to solve fluid mechanics' stochastic differential equations in your head!

firefly08

Sorry firefly, but I was just explaining what you were saying, not agreeing with it.

Haha, don't worry, I knew that - I just meant we were agreeing about the curved trajectory of the bullet:

It's obvious that the effect is an arc

In fairness, you are explaining it well. I just thought you were wrong! But I concede now; you're right.

Vegeta

rrpc wrote: »

This is because the wind doesn't shift the bullet in it's flight path, but changes it's angle.

Ok I agree with you guys (Mellor and rrpc) but just want to ask rrpc about this statement

Can you explain what you mean by "changing its angle"?

rrpc

Vegeta wrote: »

Ok I agree with you guys (Mellor and rrpc) but just want to ask rrpc about this statement

Can you explain what you mean by "changing its angle"?

Angular deviation of the bullet path from the line of aim.

Vegeta

rrpc wrote: »

Angular deviation of the bullet path from the line of aim.

Cool beans

For some reason I thought you were referring to the bullet* as opposed to its path.

* before someone points it out I know it does go nose to the wind and it doesn't fly perfectly straight either

rrpc

Vegeta wrote: »

Cool beans

For some reason I thought you were referring to the bullet* as opposed to its path.

* before someone points it out I know it does go nose to the wind and it doesn't fly perfectly straight either

Cheers. For a minute there, I thought I'd misquoted myself

For anyone who wants to learn the basics of shooting in the wind, you can get a CD by Lones Wigger and Lanny Basham (great names - great shooters too), entitled 'Winning in the Wind'. It's published by askachampion.com

Lots of really useful stuff in there including the wind target I posted earlier in this thread.

Sparks

'The Wind Book' is pretty good as well. It's more aimed (pardon the pun) at fullbore shooters, especially TR, but wind's wind, pretty much.

kowloon

A thread with good shooting literature, videos and links might not go astray if we don't already have one.