Dope shooting uphill - tell me why I'm wrong

[Baron23]

Hi guys – During the Frank/Marc show in PA a few weeks back, we all found that, at range, our true dope was less than what the weaponized math predicted and that the delta between dope at increasing range was not incrementing up as we would expect. The dope kind of flattened out at range.

Now this is in central (maybe a bit western) PA and the shooting line is in the valley and once past about 600 yards, the range started to go quite steeply up the side of the valley.

Frank explained what we were seeing as being a result of our shooting uphill and he used a diagram something like this which seemed to make a lot of sense at the time. Looking at this, it seemed clear to us at the time that yeah, the horizontal distance is less so that’s less of an horizontal distance for gravity to effect the drop of the bullet hence why our dope was coming out less than expected (and yeah, I believe we can discount any impact of a greater loss of velocity from going uphill as that would increase our dope and not reduce it from expected as we found and as Frank explained).

Now, I’m sure that Frank and Marc are right and I am not. And that this is why most good LRF will give you horizontal distance as well as slant range, if so chosen.

But as I thought about this more, I don’t understand why they are right (yeah, ole’ Baron23 asking why, why, why once again! haha).

The acceleration from gravity is defined as 9.8 m/s/s (9.8 meters per second per second, right). Note, there is nothing in here about horizontal distance. The absolute displacement resulting from this acceleration is purely a function of the time period that the bullet is exposed to gravity….that is, time of flight.

So, looking at the diagram above, I’d like to use a very simple example. 1,000 yard range target and a bullet going 3,000 fps (1,000 yds per second, right?). Also, for simplicity, let’s assume it’s a laser and there is no parabolic curve to the flight path and that the velocity remains constant across the time of flight.

So, a bullet shot purely in the horizontal will take 1 second to go 1,000 yards and the resulting drop will be 9.8 meters based on gravity accelerating the drop for 1 second.

Where I’m confused is that when shooting uphill the bullet also has a flight path of 1,000 yards to the target so it should also be exposed to the acceleration of gravity for 1 second and therefore the drop should be the same…i.e. 9.8 meters. But of course, it’s not.

Now, I’m really confident that Frank/Marc are correct and that there is a flaw in my thinking on this….but I don’t know what it is.

Any view on this…any explanation for why I’m wrong?

 

[pyrotechnic]

The acceleration due to gravity is a vector. When shooting uphill or downhill you have a component of that vector that is now parallel to your line of sight.

The "less horizontal distance" isn't technically correct, but it serves to illustrate the point.

 

[Matches Malone]

^which is a fancy way of saying a cosine or % of gravity is effecting the bullet flight. The more steep the angle, the less gravity has a full effect on the bullet.

 

[Baron23]

The acceleration due to gravity is a vector. When shooting uphill or downhill you have a component of that vector that is now parallel to your line of sight.

The "less horizontal distance" isn't technically correct, but it serves to illustrate the point.

Ah, thank you...need to think about this a bit. But is it really a vector component of gravity itself or that the bullet's flight path is effected by the gravity vector (always perpendicular to the earth and always 9.8 m/s/s) combined (in this case, offset to a degree) with the the upward vertical vector component of the bullet's path. So, the vector sum of gravity and the bullet path's vertical component results in a lower resulting downward influence.

But yes, I think I see what you are saying.

Wow, at 69, this stuff just doesn't come to me as quickly as it did when I was 22 and in college physics class! haha

Cheers

 

[pyrotechnic]

Our line of sight is our reference frame. The bullet doesn't really care where it is going, once it leaves the muzzle the forces acting on it are drag and gravity. We just have to figure out how these forces work in relation to our line of sight so that we can correct for it.

Hope that helps.

ETA: Litz does a fantastic job of putting everything in layman's terms in his book.

 

[pyrotechnic]

In your hypothetical in a vacuum, gravity would only act perpendicular to the line of sight with the magnitude of the green arrow and back towards the shooter with the magnitude of the blue arrow.
The blue and green components add together to form the red gravity vector.

Keep in mind there would still be a parabolic trajectory in a vacuum, so it's better to think of the lines you drew as your LOS.

 

[Baron23]

@pyrotechnic - thanks and....well, not to be argumentative but although I think we are talking about the same general principle, we seem to see it differently.

E.g. gravity is always perpendicular to the earth and I'm not sure how I understand how it could possibly be perpendicular to the flight path of the bullet (and yes, line of sight is prob more accurate but for simplicity let's discount a parabolic path of the bullet...maybe?) nor what force is exerted on the bullet backwards along the flight path toward the shooter.

I sort of see it like this.....again for simplicity, let's assume the angle is 45 deg, therefore the bullet's vector can be broken down into two vectors of equal magnitude...one horizontal toward the target and one vertical in the up direction. The vertical component added to the vector of gravity results in a smaller acceleration of gravity and hence the reason our dope is less that expected when shooting uphill.

And yes, I'm probably being annoyingly pedantic .....but I'm retired and somewhat bored at the moment and do very much appreciate your participation in this convo. Not quite to the point of how many angles can dance on the head of a pin....but I think I'm approaching that! hahaha ;-)

This illustrates how I see it....am I wrong in this? And yes, I should prob get Lintz book and read it. Absolutely.

 

[1moaoff]

Ummm the simple version is distance that gravity affects it

 

[pyrotechnic]

@Baron23 I don't know how to explain it differently, sorry

We are measuring deflection from our line of sight. That deflection due to gravity is maximum when our line of sight is level and gravity acts perpendicular to it. As soon as the angle of our line of sight changes the component of gravitational acceleration perpendicular to our line of sight gets smaller until it would become zero if we were to fire perfectly vertical.

Maybe try this as a thought experiment. If you were to drop a rock 200 yards away from you at point A and let it fall for one second to point B.
You then use your reticle to measure the distance in mils/MOA between A and B. That measured angle would be greatest if you were looking at it directly from the side (LOS perpendicular to gravity). As you changed the elevation of your point of view that measured angle would shrink until point A and B would look like the same point from directly above or below.


For practical purposes the rifleman's rule is just using the dope for the horizontal distance to target which works ok for shorter distances and smaller angles. For the improved rifleman's rule you multiply the cosine of your look angle times the dope of the actual distance.

Neither of these methods are entirely accurate but often get us close enough.

 

[DocRDS]

gravity is always 9.8 fps. shooting uphill or downhill.

Try this thought experiment instead.
Instead of thinking of aiming your rifle up or down--continue to think of shooting your rifle straight, but 'rotate the ground' and plot bullet impacts
That is the same as shooting uphill and downhill

We always think of shooting as a 1-D problem. I have to hit the target at 1000 yards.

Its actually a 2-D problem, I have to hit at 1000 yards and 0 elevation.
Shooting uphill and downhill changes the target. I no longer have to hit a 1000 yards and 0 elevation, I have to hit at 1000 yards at +10 Yards up or -15 yards down.

Wind makes it 3d....

 

[Baron23]

@Baron23 I don't know how to explain it differently, sorry

We are measuring deflection from our line of sight. That deflection due to gravity is maximum when our line of sight is level and gravity acts perpendicular to it. As soon as the angle of our line of sight changes the component of gravitational acceleration perpendicular to our line of sight gets smaller until it would become zero if we were to fire perfectly vertical.

Maybe try this as a thought experiment. If you were to drop a rock 200 yards away from you at point A and let it fall for one second to point B.
You then use your reticle to measure the distance in mils/MOA between A and B. That measured angle would be greatest if you were looking at it directly from the side (LOS perpendicular to gravity). As you changed the elevation of your point of view that measured angle would shrink until point A and B would look like the same point from directly above or below.


For practical purposes the rifleman's rule is just using the dope for the horizontal distance to target which works ok for shorter distances and smaller angles. For the improved rifleman's rule you multiply the cosine of your look angle times the dope of the actual distance.

Neither of these methods are entirely accurate but often get us close enough.

Thanks...I won't bother you again on this...but I do think we are talking about the same thing..more or less.

all good.

 

[KnowNothing256]

So, did you get your question answered? I’ll take a stab if not

 

[Baron23]

So, did you get your question answered? I’ll take a stab if not

Well, I think so but I want to be gentle here (not my forte' at all).

Pyrotechnic was kind enough to try to explain to me multiple times and I'm grateful for that....but, I respectfully don't agree with his analysis.

There is no horizontal component of gravity. Gravity is purely a force downward toward the center of mass (of the earth and for our purposes its perpendicular to a flat plane and we can discount earth's curvature). Nor does gravity care about the attitude of the bullet or the angle of its flight. It is a force directly downward on the bullet that is equivalent to its weight (weight really is sort of the definition of the force of gravity on a mass). The 9.8 m/s/s is the acceleration, but weight is how we measure the force of gravity applied to an object.

I sort of...without being presumptuous or arrogant...think that my analysis is correct. That is, gravity is a constant but when shooting up hill its opposed to some degree by the vertical component of the bullet's upward flight path and so drop is reduced....and drop is also by definition (I think) completely in the vertical.

The result is less drop than a horizontal flight patch for the same range (horizontal range or slant range....in my example drwg its 1k yards in both cases).

And of course there is a lot of simplification in my little model just to facilitate the discussion. E.g.....no aerodyamic drag or parabolic flight path. I simplified it to be basically a laser in a vacuum.

Now, I could be quite wrong but that's how I see it presently.

Please feel free to school me, however.

 

[Carlos Danger]

Hi guys – During the Frank/Marc show in PA a few weeks back, we all found that, at range, our true dope was less than what the weaponized math predicted and that the delta between dope at increasing range was not incrementing up as we would expect. The dope kind of flattened out at range.

Now this is in central (maybe a bit western) PA and the shooting line is in the valley and once past about 600 yards, the range started to go quite steeply up the side of the valley.

Frank explained what we were seeing as being a result of our shooting uphill and he used a diagram something like this which seemed to make a lot of sense at the time. Looking at this, it seemed clear to us at the time that yeah, the horizontal distance is less so that’s less of an horizontal distance for gravity to effect the drop of the bullet hence why our dope was coming out less than expected (and yeah, I believe we can discount any impact of a greater loss of velocity from going uphill as that would increase our dope and not reduce it from expected as we found and as Frank explained).

View attachment 7722965

Now, I’m sure that Frank and Marc are right and I am not. And that this is why most good LRF will give you horizontal distance as well as slant range, if so chosen.

But as I thought about this more, I don’t understand why they are right (yeah, ole’ Baron23 asking why, why, why once again! haha).

The acceleration from gravity is defined as 9.8 m/s/s (9.8 meters per second per second, right). Note, there is nothing in here about horizontal distance. The absolute displacement resulting from this acceleration is purely a function of the time period that the bullet is exposed to gravity….that is, time of flight.

So, looking at the diagram above, I’d like to use a very simple example. 1,000 yard range target and a bullet going 3,000 fps (1,000 yds per second, right?). Also, for simplicity, let’s assume it’s a laser and there is no parabolic curve to the flight path and that the velocity remains constant across the time of flight.

So, a bullet shot purely in the horizontal will take 1 second to go 1,000 yards and the resulting drop will be 9.8 meters based on gravity accelerating the drop for 1 second.

Where I’m confused is that when shooting uphill the bullet also has a flight path of 1,000 yards to the target so it should also be exposed to the acceleration of gravity for 1 second and therefore the drop should be the same…i.e. 9.8 meters. But of course, it’s not.

Now, I’m really confident that Frank/Marc are correct and that there is a flaw in my thinking on this….but I don’t know what it is.

Any view on this…any explanation for why I’m wrong?

Was this the range of which you speak?

 

[pyrotechnic]

There is no horizontal component of gravity. Gravity is purely a force downward toward the center of mass (of the earth and for our purposes its perpendicular to a flat plane and we can discount earth's curvature).

You're correct, and I probably have done a poor job of explaining. Gravity always points the same way, however when we are looking at an angle other than purely horizonal gravity points slightly towards or away from us. That component (towards or away) is really small with comparison to drag forces on the bullet. However, the portion of gravity that points perpendicular to our line of sight is reduced (as compared to when we are looking purely horizontal), and that is why we perceive less drop when shooting uphill or downhill.

 

[Baron23]

Was this the range of which you speak?

View attachment 7723636View attachment 7723637

Nope, but that is a damn fine looking range. We were at Mifflin Sportsman in Reedsville, PA and the hill the long range targets were on was a good bit steeper that it appears is the case for this range.

Where is this, may I ask?

 

[Baron23]

You're correct, and I probably have done a poor job of explaining. Gravity always points the same way, however when we are looking at an angle other than purely horizonal gravity points slightly towards or away from us. That component (towards or away) is really small with comparison to drag forces on the bullet. However, the portion of gravity that points perpendicular to our line of sight is reduced (as compared to when we are looking purely horizontal), and that is why we perceive less drop when shooting uphill or downhill.

haha....I still don't see how there can be any component of gravity in any direction other that absolutely downward toward the center of the earth. If your reference is the chord line of the bullet (line from tip to center of the tail) then yes, gravity will be exerted at an angle to that chord line.....but we could be launching perfectly round shot at this angle and still have the same effect of reduced drop and there is no chord line reference.

BUT, I'm afraid I'm starting to get argumentative when seeking help (which is really annoying) so I'm going to stop here and take your advise and buy Litz's book and see if I can grasp what he says about.

Also, I have a Phd Physics (ret) friend at the club....maybe I'll let him shatter my mind also! haha

Thanks again for your patience.....I still don't see it....but, you sure have tried hard to explain your view of it and I do appreciate that very much.

Cheers

 

[Nik H]

@Baron23

Go to Wikipedia and search for Rifleman's rule...it is all there. Simple physics...you are overthinking it

 

[KnowNothing256]

Well, I think so but I want to be gentle here (not my forte' at all).

Pyrotechnic was kind enough to try to explain to me multiple times and I'm grateful for that....but, I respectfully don't agree with his analysis.

There is no horizontal component of gravity. Gravity is purely a force downward toward the center of mass (of the earth and for our purposes its perpendicular to a flat plane and we can discount earth's curvature). Nor does gravity care about the attitude of the bullet or the angle of its flight. It is a force directly downward on the bullet that is equivalent to its weight (weight really is sort of the definition of the force of gravity on a mass). The 9.8 m/s/s is the acceleration, but weight is how we measure the force of gravity applied to an object.

I sort of...without being presumptuous or arrogant...think that my analysis is correct. That is, gravity is a constant but when shooting up hill its opposed to some degree by the vertical component of the bullet's upward flight path and so drop is reduced....and drop is also by definition (I think) completely in the vertical.

The result is less drop than a horizontal flight patch for the same range (horizontal range or slant range....in my example drwg its 1k yards in both cases).

And of course there is a lot of simplification in my little model just to facilitate the discussion. E.g.....no aerodyamic drag or parabolic flight path. I simplified it to be basically a laser in a vacuum.

Now, I could be quite wrong but that's how I see it presently.

Please feel free to school me, however.

So, some of what you've laid out I agree with, but I'd suggest a couple changes if it's not too argumentative

No horizontal component of gravity: Agreed. But for that to be true, we need to set a reference frame where that is true. So let's do that, the reference frame is perfectly horizontal with respect to gravity (so gravity is only down). This works on the relatively short distances where small-arms ballistics occur.

Weight vs. acceleration: Agreed here too, I think you spelled it out well. However, I think it's more useful to think about the acceleration on the bullet in this discussion, since that makes it independent of bullet weight and is any case what really matters when speaking (roughly) about DOPE.

"Drop is also by definition completely in the vertical": This is where I fork a bit. This statement is true if you're firing perpendicular to gravity, but it's NOT true if you are firing at an angle above or below the level. Drop, as we talk about it, is essentially drop from line of sight, as @pyrotechnic stated. I think this might be where the confusion is happening, and the best way to "feel it out" in your mind, so to speak, is the extreme example he gave in the same post of the rock, but this time use a bullet. So, say you point your rifle perfectly vertical, and there's no wind or Coriolis or spin drift or any of that. And let's also say that you've mounted your scope perfectly level with the bore, but off to the side instead of on top; so as the bullet travels, it'll appear to move in the left/right dimension but not in the up/down dimension. I think you'll agree that if you fired the bullet in this idealized scenario, it would stay perfectly level with your "horizontal" stadia, and would appear to drift towards the center starting from whichever side the bore is on.

If that part makes sense, then see what happens in the mental scenario if you tip the rifle juuuuuust slightly down from perfectly vertical. Now as the bullet flies, it'll appear to drift very, very slowly "down" in the scope as the gravity pulls it off the line of bore (you can visualize this more dramatically by tipping the rifle down more). So the trajectory in this scenario would look kinda like this:

At the risk of being frustrating (many apologies if so), this pull "down" from your line of sight is because in your line of sight isn't the same as the reference frame of gravity, and if you use the reference frame of your line of sight (which is what we all instinctively do), then gravity does have a horizontal component. Sorry to switch reference frames there right at the end, lemme know if I can explain that better.

Clear as mud?

 

[pyrotechnic]

No worries! And again, you're right in that there is no component of gravity that acts other than directly down to the center of the earth. The only thing that changes is how we are looking at that vector. Like you said gravity doesn't care how we launched the bullet or even that it is moving. It gets accelerated downward all the same.

One last attempt with an illustration. You can think of the below as a free body diagram of the bullet (black circle) in a vacuum, for correctness we also assume that this is the instant it leaves the barrel. The green line is our line of sight and our velocity vector (this ignores any sight height). On the left we are looking perfectly horizontal and all of the (red) gravitational force acts perpendicular to it.
On the right we have some arbitrary upward angle for our line of sight and launch angle, Now while the gravitational force still has its full value straight down to the center of the earth, there are now components, one of which (the blue one) which will decrease the magnitude of our velocity the other which will accelerate the projectile away from our line of sight. Both of these (the blue and orange) components are less individually than the red vector. But If you add these two together you get the red gravity vector which is still pointed straight down towards the center of the earth.

 

[KnowNothing256]

@pyrotechnic A+

ETA: OP, you can see in that free body diagram that in the angled scenario, the component of gravity that is pulling perpendicular to the line of sight (our normal frame of reference) is smaller that in the horizontal scenario; this is why dope is impacted less by distance on steeply angled shots, because the vector of gravity that is pulling the bullet away from LOS is smaller and smaller the more steeply angled it is, and the rest of gravity is either speeding up the bullet along LOS (downward shot) or slowing it down (upward shot).

 

[Nik H]

 

[metalicat]

Lowlight will be along shortly to lay there with @Baron23 to explain it again

 

[KnowNothing256]

View attachment 7723667

Hmm, I would suspect the faster/slower part is less impactful than the reduced vector component of gravity. A vector portion of 9.8 m/s^2 doesn't add up to much velocity change during a typical bullet's flight (at 2 seconds of flight, straight down to milk 100% of the acceleration, that's only a change of 64 fps, which isn't much, and most TOF values are WAY below that, and also you're only getting a portion of the "available" acceleration since you aren't shooting straight down or up).

 

[DocRDS]

stop saying reduced vector component of gravity.

Gravity is constant for ballistics. Period. End of story.

What is changing is the initial velocity in the vertical direction. this gives the bullet its 'time of flight' if you will before it hits something. I have to teach, but after class, ill show the equations and how they affect motion.

 

[KnowNothing256]

stop saying reduced vector component of gravity.

Gravity is constant for ballistics. Period. End of story.

What is changing is the initial velocity in the vertical direction. this gives the bullet its 'time of flight' if you will before it hits something. I have to teach, but after class, ill show the equations and how they affect motion.

Actually, no, I'll disagree here. Most of the TOF arises from MV and distance to target. If that target is straight up, at 1000 yds, the TOF is quite similar to perfectly level at the same distance.

I tried to game this out with the Hornady 4DOF, but it caps at 65 degrees and my back-of-the-napkin (actually a literal take-out box) underestimated drag too much, so I clearly can't swag it well enough to stand by my numbers. If there's a solver that'll take a 90-degree inclination as an input, we can crunch the numbers.

 

[pyrotechnic]

I think we are getting hung up on frames of reference.

And yes, the (de)acceleration of the projectile due to gravity when fired up or down pales in comparison to the effects of drag.
My .260 with whatever my last environmentals were goes from 2790fps to 1480fps over 1000yds with TOF of 1.48s. This is an average acceleration of -885 ft/s^2 compared to the 32.2 ft/s^2 acceleration due to gravity.

 

[DocRDS]

solve for t

this is the time of flight (I used -10 instead of -9.8 because its easy). Why time of flight? because it is the only thing shared between x and y motion--the time.

Lets say we shoot straight level--perpendicular to the ground, that v_y0 is 0. All MV is in x. You may say "hey genius your y has to be negative" and you would be right. I factored in the initial y or vertical position and called it 0. ANything below me is negative, anything above me is positive. So if I fire level, gravity pulls the bullet down and I can only hit targets below me.

Lets make y = -1/5 (1/5 meter down) firing level. Then 20y is 4 and sqrt(4) is 2 so in roughly 0.2 seconds my bullets drops -0.20 meters (that is a coincidence).

How far did it go?

x_0 is 0 so it went a distance vt--the same "t" I got from above. (no drag here). at 3000 FPS it went 600 feet. So I ended up a y = 0.2m and x = 600 feet.

Now Frank wants me shooting uphill 10 degrees. v_y0 is now 915 m/s (thats 3000 FPS to me and you) * sin (10) or about 159 m/s

Now the problem is my target is ranged at 1000m, but part of that distance is vertical, and part is horizontal.

The horizontal distance is 1000 * cos (10) or 985 meters. so I need a time of flight of 985/901 or 1.09 seconds. (only PART of my MV is in X now!)

but I can get that same answer from the equation for y, as i need the final y to be 1000 sin (10) or 174m.

the sqrt part will be 147.5 so my two answers are either (159-147.5)/10 or (159+147.5)/10 (this would be a super included suborbital trajectory--think about it--as I inclince farther and farther up--my bullet goes farther and farther, but at 45 degrees it has more 'up' so now my x velocity is so low, its not travling far until I shoot straight up. Obviously if I show straight up the bullet goes 0 in the x, but at some time between 45 and 90, the bullet hit my target again).

The first answer (cause i have been rounding like crazy) happens to be equal to 1.09 seconds had I done it properly without rounding like a lazy bitch..

Point being you are hitting a point in 2-d space, but those coordinates are connected by time

So the challenge of angled shooting is a significant portion of your velocity is now in the vertical direction, as well as we 'dope' for y=0 while in fact, in angled shooting y (final elevation) can be postive (above us) or negative (below us).

But notice the important part: All i really need was the horizontal distance to the target..

And then drag comes along and crits you for 1000.
You die.

 

[357Max]

Actually, no, I'll disagree here. Most of the TOF arises from MV and distance to target. If that target is straight up, at 1000 yds, the TOF is quite similar to perfectly level at the same distance.

I tried to game this out with the Hornady 4DOF, but it caps at 65 degrees and my back-of-the-napkin (actually a literal take-out box) underestimated drag too much, so I clearly can't swag it well enough to stand by my numbers. If there's a solver that'll take a 90-degree inclination as an input, we can crunch the numbers.

On the Kestral set to 90 degrees up I'd have to dial down 1.2 mil to hit 1050 with 6GT due to 2.3" scope height zeroed at 100y

In reality the 1050 target at Miflen is +6 degrees & requires 7.99 U mil vs 8.04 @ zero angle.

White spec top right is 1050y from the conex.

@Baron23 - I'm going to re-frame you perspective.

Assuming you understand the wind rose?

Now flip the wind rose on its side.

Wind = force
Gravity = force

With the wind rose on its side & shooting 12:00 gravity is acting @ 3:00 full value.

+15 degrees up hill or 3:30 on wind rose = 96% wind or cosign correction of .966

+30 degrees up hill or 4:00 wind = 86% wind value or cosign correction of .866

+45 degrees up hill or 4:30 wind = 70% wind value or cosign correction of .707

See the pattern emerging?

Wind and gravity are both force.

For a more practical comparison gravity is constant and easy compared to wind.

I shot a match at Miflen this past Saturday and had my turn on the stage (below) during a down poor as a storm front pushed through. Winds gusty & switching. Rooftop was next to conex so couldn't really feel the wind at position. Up hill +6 degrees 1st target 6" wide vertical bar @ 701y. 1st shot missed didn't see splash, 2nd shot missed way right elev good, 3rd shot held additional .8 mil L still missed right, 4th shot added another .4 L & missed left about .6 Grrr! Next target horizontal 6" bar hit 3 of 4, next target diagonal bastard hit 1 of 4.

So from the above I went 0 for 4 thanks to gusty wind & inability to spot my first shot in the down poor.
The horizontal target (elev dependent) 6" tall I went 3 for 4 & the diagonal is just pure evil in those conditions.

Now if you really want to get your panties twisted think about this gravity is constant.
Wind not only changes speed and direction, but the force imparted on the bullet changes with air mass/density.
The literal perfect storm when visibility sucks due to down poor & a front is pushing through at 100% humidity changing baro & switching gusty wind.

 

[hlee]

Let’s imagine that we bore sight our rifle such that it is pointing at a target. In the absence of any external effects (wind, gravity, etc) the bullet will fly straight and true and hit the target.

Gravity, as we know, has the effect of pulling the bullet away from that straight line path. Gravity is a vector- it has magnitude and direction. The magnitude is 9.8 m/s^2. The direction- on earth- is pointed at the center of the earth. On flat and level ground, we will say that gravity acts perpendicular to the ground.

Because gravity is a vector, it can be expressed as a sum of vectors. This vectors that are most appropriate to the discussion are parallel to the bore and perpendicular to the bore.

Where the bore is parallel to the ground, the magnitude of the vector parallel to the ground is 0. The entire magnitude of gravity can be expressed as a single vector- perpendicular to the bore of the rifle. When the rifle is parallel to flat and level ground, gravity has maximum effect of pulling the bullet away from the bore line.

If one were to fire a rifle directly up- with no external effects present. The bullet would fly straight and true until it apexed and came falling back to earth- right back into the shooter. Gravity, while it affects the height the bullet can climb, does not pull it away from that straight path. As such, if you fire a bullet straight up it “should” come straight back down. In this example, the entire magnitude of gravity can be expressed as a single vector acting parallel to the bore line. As such, there is no force to pull the bullet away from that straight line path.

In Every other scenario, the magnitude of the vector acting perpendicular to the bore line (the component actually pulling the bullet away from straight line travel) is some value greater than zero, but less than the full effect. This is why you have “cosine” indicators.

Cosine(angle) = adjacent/hypotenuse.

adjacent = level line distance

hypotenuse = bore line distance

Cosine of 90 deg = 0. Gravity has no effect on bullet path (straight up or down).

Cosine of 0 deg = 1. Gravity has max effect on bullet path (flat and level ground

 

[DocRDS]

+1 on the wind rose on its side.

 

[Carlos Danger]

Nope, but that is a damn fine looking range. We were at Mifflin Sportsman in Reedsville, PA and the hill the long range targets were on was a good bit steeper that it appears is the case for this range.

Where is this, may I ask?

Ridgway Rifle Club. Closest thing to me from Western New York.

 

[Diver160651]

Maybe I just missed it... But going back to basics on what you need when using a PLRF during High angle shooting.

Of course, GD is reduced up or downhill, and for the most part, no need to get totally vacuum-centric in that yes, up is mathematically different than down, but the variance is far inside most people's CEP. Practically stated no different. But you guys got that all charted up.

So you need BOTH the gravity distance and the LOS time for wind,, This really is only a big deal on very high-angle long shots. The biggest bugger in extreme angle shooting is the shooting position itself/

Takeaway on ultra high angle long shots: Remember LOS wind, build a solid position off of well-practiced shots.

Just thought I add those 2 pieces if they were not brought up.

 

[hlee]

A picture representation of what I wrote in post #30.

 

[Nik H]

A picture representation of what I wrote in post #30.

View attachment 7724409

This is the same visual that everyone uses but the bullet's path is parabolic.

This is a more representative way to look at it

The actual equations of motion are 2nd order differential equations.

The derivation of the bullet's motion, unless in a vacuum, is quite complex. No one here has discussed the angles represented above and their importance in understanding the problem. It starts with understanding F=ma...it goes from there to develop the equations of flight in terms of x and y positions of a particle and all the forces acting upon it including gravity and drag.

It is much simpler to remember that when shooting uphill or downhill, the POI is slightly higher. I try not to perform these calculations when shooting at a target

YMMV

 

[DocRDS]

This is the same visual that everyone uses but the bullet's path is parabolic.

This is a more representative way to look at it

View attachment 7724607

The actual equations of motion are 2nd order differential equations.

The derivation of the bullet's motion, unless in a vacuum, are quite complex. No one here has discussed the angles represented above and their importance in understanding the problem. It starts with understanding F=ma...it goes from there to develop the equations of flight in terms of x and y positions of a particle and all the forces acting upon it including gravity and drag.

It is much simpler to remember that when shooting uphill or downhill, the POI is slightly higher. I try not to perform these calculations when shooting at a target

YMMV

While you are correct, if you treat x and y motion as parametric with t being the parameter, the angles become inherent to the problem. When I derived the time of flight, in essence, I was doing the same thing--we often say x and y are independant, but they are in fact 'coupled' and the angles represent that coupling, but so does time.

six of one, half dozen of the other.

I also became interested in the 'true' equations of motion, including the proper drag equation (not drag coefcients) which is suprisingly tough to find, simply because I do enjoy that kind of mathematics and also finding quick approximations (aka for small x, sin x ~ x, or (1-x)^n = 1-nx, etc). That's actually why the rifleman's rule works is because the 'theta' angle is very very small, and thus the 1- tan (junk)) is basically 1 until extremely large 'alpha' and large 'theta' in the 'true derivation'

Rifleman's rule - Wikipedia

en.wikipedia.org

I came across this:

INCLINED FIRE - Sierra Bullets

www.sierrabullets.com

Of course their 'third way' is very confusing with both "drop" <drop from inclined line of sight> and "bullet path <aka Bullet drop>" No way that is usable in the field.

 

[Nik H]

While you are correct, if you treat x and y motion as parametric with t being the parameter, the angles become inherent to the problem. When I derived the time of flight, in essence, I was doing the same thing--we often say x and y are independant, but they are in fact 'coupled' and the angles represent that coupling, but so does time.

six of one, half dozen of the other.

I also became interested in the 'true' equations of motion, including the proper drag equation (not drag coefcients) which is suprisingly tough to find, simply because I do enjoy that kind of mathematics and also finding quick approximations (aka for small x, sin x ~ x, or (1-x)^n = 1-nx, etc). That's actually why the rifleman's rule works is because the 'theta' angle is very very small, and thus the 1- tan (junk)) is basically 1 until extremely large 'alpha' and large 'theta' in the 'true derivation'

Rifleman's rule - Wikipedia

en.wikipedia.org

I came across this:

INCLINED FIRE - Sierra Bullets

www.sierrabullets.com

Of course their 'third way' is very confusing with both "drop" <drop from inclined line of sight> and "bullet path <aka Bullet drop>" No way that is usable in the field.

I prefer to solve the PDQs myself. Until you can actually set up the problems and derive the equations of motion from 1st principals, you really don't get a full appreciation of how complicated it is to solve but then again, that is why I have grad degrees in Physics, EE and Applied Math

 

[Baron23]

Hi guys - wow, sorry I didn't participate yesterday but was out and about and in fact I also don't have a lot of time today before I have to run.

I did read @Nik H suggestion of reading the Rifleman's Rule on Wikipedia and yes, this seems very applicable when try to determine the equivalent horizontal distance (in order to calculate drop) from a an angled shot.

Yes, range of

but what I tried to address and understand is a hypothetical situation where R subH = R subS. This the case that Frank used in his hand drawn model during the clinic, if I understood correctly.

This is from the Wikipedia page: Apply the "rifleman's rule" to determine the equivalent horizontal range

and then go to your dope table. I understand that this is how to calculate the reduced drop but not why the drop is reduced....and its not because of the horizontal component of the angled flight path is less than the horizontal flight path which is what we discussed at Mifflin. There is no distance feature when calculating the effect of gravity....except to the extent that distance and volocity are used to calculate time of flight as its time and not distance that determines the magnitude of gravity's effect.

I want to say once again...I'm not saying at all that Frank/Marc are incorrect or don't know this subject inside and out. I adored the clinic, really enjoyed it, and have the utmost respect for their expertise. I personally believe that what Frank showed was just a simplified model useful in conveying the general concept that drop is less when shooting the same distance uphill vs horizontal because....well, nobody wants to go thru this level of discussion during a clinic! I'm not even sure I want to go thru this online! haha

But I still don't think its a truly accurate of the actual forces at work.

Please let me say again that for my purposes I have dramatically simplified the situation as follows:
1. in a vacuum....no loss of V due to aerodynamic drag. V is constant.
2. Bullet is a laser and LOS and projectile flight path are the same so I dispense with that consideration.
3. I'm simply comparing drop for horizontal range compared to the exact equivalent slant range distance
4. I used 3,000 fps V, 1,000 yd range (for both slant and horizontal so one second time of flight for both), and showed 45 degrees as the up angle just because it makes it easy to see vector components....one up for every foot forward.

My question was exactly what force acts to reduce drop in a slant angled shot. Although the change in dope can be calculated via trig...ala' the Rifleman's Rule....that's not what reduces drop as distance is identical for both and velocity stays constant and identical for both.

I do understand the suggestion of rolling a wind rose on its side....but I don't see that as all that applicable as gravity (instead of wind) will always come the 12 o'clock position...that is, there is only one vector with any magnitude and that's perpendicular to a plane tangential to the earth's surface (that is, absolutely downward). Again, showing some component of gravity in other than the absolute vertical may well be helpful in visualizing and understanding this situation, it is not a accurate depiction of gravity.

I'm perhaps on the edge (or past it haha) of becoming argumentative after asking a question because I don't know....but this I do feel certain of: gravity is only a downward force purely perpendicular to the surface of the earth (more or less). If you try to break it down into vector components to be added, all vector components other than straight down will be a value of zero. I just don't think that can be disputed (but I'm probably wrong and am prepared to be schooled by a real physicist which I am NOT LOL).

And gravity has no care at all about projectile shape, attitude in space, or anything like that. Its isolated effects are identical if its a bullet with a bit of nose up or a beach ball. Same acceleration in the same direction.

In order to reduce the effect of gravity on our projectile there must be some force applies in the upward vertical direction and all I can see is as I showed above. The vertical vector component of the projectile's flight is in direct opposition to the force of gravity and when summed the net is a reduction of the impact of gravity...that is, reduced drop.

My head hurts from this (and yes, I'm an old idiot who used to know a lot more about this kind of stuff than I do now...sigh haha) and there is really no practical value....unlike the concepts of the Rifleman's Rule which is practical and applicable and has no need for understanding of the forces at work that make it valid.

Please be kind in identifying any utter ignorance in this post (or don't...after all, it is the Hide! haha) but this is still how I see it. I just don't see any other forces at work in my simplified scenario.

Cheers and have a brilliant day to all.

 

[DocRDS]

You are right its not turly accurate. But its 'close enough'. That's one of the keys of physics that a lot of people get lost in. After all, we don't calculate a special relatavistic correction to bullet travel, even though to be precise, bullets experience relativity, the correction is so small--its not worth the effort.

We are very very accurate and precise with how all these things contribute, but what we forget it many times the 100% solution if you will with all the details, is actually very close to a 'rough approximation' in which case the approximation is very rough--its actually quite good.

My question was exactly what force acts to reduce drop in a slant angled shot. Although the change in dope can be calculated via trig...ala' the Rifleman's Rule....that's not what reduces drop as distance is identical for both and velocity stays constant and identical for both.
[\quote]

Its not a force, its what is known as an initial condition (initial velocity). Gravity is always in the vertical at 9.8 units.

Normally the vertical velocity is very small. If you drop a bullet from 19.6 meters it takes 1 second to hit the ground.
Now toss the bullet up--it takes longer to hit the ground. Thrown the bullet down, it takes less time to hit the ground.

Why does this matter? because what stops the horizontal motion is when it impacts something. The horizontal motion in the 'honey badger' here--0 fucks given. I'm just going along until i smack into something. Until gravity pulls it into the ground, horizontal or x motions is just cruising along (ok drag is affecting it, but that i s it).

Now here are the free body diagrams of the forces on a bullet for level and inclined shooting

Level shooting drag, and gravity are independent. Inclined motion, drag is actually 'assisting' gravity is accelerating the bullet fall. Shooting down, drag is actually 'opposing' gravity and reducing the vertical acceleration.

Now since forces cause changes in velocity, lets look at the velocities

The velocity (red) is being changed by the forces (blue). The changes (green) will be different depending on how much drag is in the vertical versus horizontal but always down and left (assuming right bullet travel). The velocity also depends on its initial direction too.
So shooting uphill you have increased 'downward force' of drag+gravity. shooting downhill you have 'reduced force' of 'gravity-drag'--green always looks the same, but its size and specific direction are not.

All this time, horizontal velocity is only experiencing drag proportional to its horizontal velocity.

 

[hlee]

If you are solving partial differential equations on the firing line, you have already lost. Close enough is close enough, and high school trig is all that is required to explain why we use the flat distance and not line of sight distance for elevation adjustments in high angle shooting.

 

[Baron23]

If you are solving partial differential equations on the firing line, you have already lost. Close enough is close enough, and high school trig is all that is required to explain why we use the flat distance and not line of sight distance for elevation adjustments in high angle shooting.

hahaha...I agree utterly.

But I do enjoy the aspect of discussing and understanding ballistics and other subjects related to longer range rifle shooting as well as the shooting itself.

Hey...I was the kid who was always asking "why is the sky blue" haha....yeah, I was and prob still am pretty annoying LOL

Have a great day.

 

[hlee]

Because light diffraction is proportional to the inverse of the 4th power of the wavelength, and blue light has a short wavelength, thus it is more easily diffracted. Also why sunsets are red- light passing through more air, diffracting longer wavelengths. And, why the sky can “go green” before a tornado- increased particulates in the air diffracting the longer (than blue) green light.

 

[dutchboy53]

The problem is stated in your premises
you assumed no parabola and a laser
that is not what we are working with
we have a projectile and gravity and a parabola
no projectile fired from a weapon will ever be in a straight line for any significant distance
see Nik H post # 35
I believe for a beam of light over one thousand yard range your confusion would be substantiated but we have a bullet and a parabola
just my 2 cents after staying at a Holiday Inn Express

 

[Nik H]

If you are solving partial differential equations on the firing line, you have already lost. Close enough is close enough, and high school trig is all that is required to explain why we use the flat distance and not line of sight distance for elevation adjustments in high angle shooting.

The rest is all mental masturabation

 

[DocRDS]

If you are solving partial differential equations on the firing line, you have already lost. Close enough is close enough, and high school trig is all that is required to explain why we use the flat distance and not line of sight distance for elevation adjustments in high angle shooting.

This. 1000X this. That is why the '3rd' way of Sierra is so frustrating, there is no easy way to calculate it in your head. Kestrels do fail, so its good to have a backup.

(Also its a ODE)

hahaha...I agree utterly.

But I do enjoy the aspect of discussing and understanding ballistics and other subjects related to longer range rifle shooting as well as the shooting itself.

Hey...I was the kid who was always asking "why is the sky blue" haha....yeah, I was and prob still am pretty annoying LOL

Have a great day.

You are not alone--part my enjoyment is understanding the mathematics--most people find it off putting, but I enjoy it. No joking, I went off to find the actual drag equations simply to do some of my own work 'for fun'. I know they do numeric integration to find solutions, but that's the same thing I do, so its right up my alley.

I am a very sick individual.

 

[DocRDS]

The rest is all mental masturabation

Me after getting to do math:

 

[Nik H]

Me after getting to do math:
View attachment 7724757

You're a strange person but no judgement....LOL

I always say that you should hope to learn something new everyday. In my professions, that is key

 

[Diver160651]

You're a strange person but no judgement....LOL

I always say that you should hope to learn something new everyday. In my professions, that is key

Good advice, but sometimes as I mentioned earlier the doing is more important. Especially because the distances that actually produce high angels, tend to also produce shorter distances than theoretically probable. Couple this with a very fast-moving (low TOF) low drag projectile and the shooter's error is going to be what everyone should worry about, not the slight difference in GD up vs down. Better money/time spent getting a quick process to easily identify rough GD and wind distance, on a practiced possition.

If you guys we're discussing high drag arrows often flying less than a 1/10 of the speed (160-290fps) of what most of us are shooting, ya, pic away.. at least where it makes sense to know the end of the curve at a high-angle is noticeably different up, vs down. But even here, a lot of the shot is made or broke on how you hold the bow and they are drastically different up and down.

 

[Nik H]

Good advice, but sometimes as I mentioned earlier the doing is more important. Especially because the distances that actually produce high angels, tend to also produce shorter distances than theoretically probable. Couple this with a very fast-moving (low TOF) low drag projectile and the shooter's error is going to be what everyone should worry about, not the slight difference in GD up vs down. Better money/time spent getting a quick process to easily identify rough GD and wind distance, on a practiced possition.

If you guys we're discussing high drag arrows often flying less than a 1/10 of the speed (160-290fps) of what most of us are shooting, ya, pic away.. at least where it makes sense to know the end of the curve at a high-angle is noticeably different up, vs down. But even here, a lot of the shot is made or broke on how you hold the bow and they are drastically different up and down.

I am in total agreement. I view this entire discussion as FYI only. The last sentence in my post (#35) kind of sums it up for me.

@Baron23 is a troublemaker...LMAO

 

[DocRDS]

Nah, its good to try and understand things in depth--it helps with what matters and what is trivial. But as an experimentalist, I whole heartadly agree with @Diver160651.

In ballistics, putting lead (or copper solids!) on target has no substitute.