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External Ballistics: Elevation/ Declination

Posted by ZDP-189 , in Techniques, Technical 03 December 2011 · 1,356 views

external ballistics elevation declination drop gravity angle tables accuracy
I have previously touched on external ballistics and drop, but as my focus is target shooting and not hunting, I had not made a study of elevation. In this blog entry, I discuss:
  • Compensation for drop by aiming above the target
  • How a target that is higher or lower than the shooter affects the flight of the projectile
Please note! When it comes to compensating for target range and elevation, I would say it's better to judge from experience than to use a table of numbers. Tables and charts multiply measurement and calculation inaccuracies, whereas the human eyes and brain are naturally very good at integrating all this information and predicting an aiming solution. However, by understanding the way that external ballistics works and is computed, we can better equip ourselves to visualise the projectile's flight, particularly in unusual situations of extreme elevation.

A Simplified Approach

External ballistics is an incredibly sophisticated and detailed science. Firearms and artillery shooters can compensate for so many factors that the wind direction may have changed or the target moved on by the time the final variable is measured and entered into the ballistics computer. Fortunately many of these factors become significant only at high velocities and for a typical slingshot with a dense projectile, just about the only significant force acting on the projectile is gravity as drag is immaterial compared to the momentum of the projectile. The only variables that we need to consider are:
  • velocity of the projectile leaving the pouch;
  • angle of elevation;
  • distance to the target; and
  • height of the target relative to the slingshot
Gravity

After you shoot a slingshot, gravity acts on the projectile. Gravity acts as if a force constantly accelerates the projectile faster and faster towards the earth at 9.81 metres per second for each second of flight time. The longer the ball remains in flight, the faster the projectile will fall towards the ground. Seeing as a round ball that is not spinning does not generate any lift, then within reasonable limits, it doesn't matter how big or heavy the projectile is, it will always have the same acceleration due to gravity of 9.81m/s/s. The flight time and hence the distance dropped is therefore down to the distance to target (range) and the projectile's speed (velocity).

The following chart demonstrates this:

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Projectiles shot at targets that are further away see exponentially more drop. You can compensate by increasing velocity. If you can shoot twice as fast at a target twice as far away, the drop will be the same, as the flight time will be the same. For example, a projectile shot at 80m/s at a range of 80m drops the same distance as one shot at a velocity of 40m/s and at a range of 40m because they both travel for a total flight time of 1 second. However, it's not all that easy to vary the velocity exactly as we want it, so we instead must aim upwards and use some of the projectile's forward momentum to compensate for drop.


Elevation

At the short distances and reasonably high speeds typical of target shooting, not much drop is experienced. Also, the distance does not vary. If you're the sort of shooter that likes to aim the fork at the target, then the easiest way to compensate for drop is to measure how many fork widths above or below the target you must place the top edge of the slingshot and always use the same offset. If you stick to the same bandset at the same draw length firing the same weight of projectile and you hold the fork the same distance from your eye, the velocity will be the same so the elevation compensation will not change from shot to shot.
Posted Image

It gets a lot trickier when the target range is very far or you are flinging heavy rocks. This is often the situation of hunters whose quarry will not let them approach and who must shoot heavy rocks or ball. In this case they will find themselves arcing up into the air and down onto their target.

Objects thrown up in this way to fall under gravity follow a parabolic arc.

A Bouncing Basketball (Wikipedia)

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The horizontal distance travelled by a projectile, the maximum height attained and the flight time depend on the angle of elevation and are easy to understand.

Effect of Angle of Elevation on Range, Maximum Height and Flight Time (Wikipedia)

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High arcs give high heights and long flight times, but do not travel far. Low angles result in a flatter trajectory, but they soon hit the ground. If you wish to shoot as far as possible, you would shoot upwards at a 45 degree angle. This is not the longest lasting flight, but it offers the longest range.

These are described as:

R=(vi2sin2Θi)/g
h=(vi2sin2Θi)/2g


and

Θ=0.5sin-1(gR/v2)

where:
  • R= max range
  • h= max height
  • vi= initial velocity
  • g= acceleration due to gravity
  • Θi= initial angle of elevation from the horizontal
Note that there are two possible solutions for the angle of elevation. One is a shallow arc and one is a high arc. For example, in the chart above, you can see that both the 75 degree and 15 degree arcs land in the same place, as do the 30 degree and 60 degree arcs. A slingshot shooter would generally select the most shallow angle.

Don't fret about the math. I've made a chart and if you check the references section HyperPhysics has a very useful calculator for calculating the initial angle of arc.

Posted Image

I don't expect many people to actually use these charts and formulae to rain bullets down onto their targets, although they can be used to come up with a rough idea of how high to shoot. They are more important to illustrate general concepts, such as:
  • For small elevations of under say 15 to 20 degrees, the ratio of elevation angle to range is almost linear.
  • For angles above 30 degrees, much bigger increments of arc are needed to get a little more range.
  • It gets tricky to hit the target using high elevations. At near 45 degrees, consider yourself lucky to reach.
  • You can't reach a target if you can't get that far with a 45 degree elevation.
  • Arcs above 45 degrees can be used to clear obstacles, but there'll be more vertical travel component than horizontal.
  • Fast projectile-bandset combinations shot within ranges that one'd be accurate enough to hit a small target (say up to 100m) can do so with little drop and small angle changes
  • Slow projectile-bandset combinations have significant limitations at long ranges. You cannot reach a target 100m away with a projectile travelling 30m/s.
  • Above about 70% of the maximum range, estimating the required elevation becomes tricky.
  • Any velocity produces a corresponding maximum range at a given angle.
  • If you shoot a slingshot straight up, it will come straight back down. The speed it hits you on the noggin is the speed you shot it, less air resistance and never more than its terminal velocity.
  • If you shoot your slingshot at say, 45 degrees and the slingshot is close to the ground on flat terrain, you can estimate its velocity based for far the bullet flies before it first lands.

That last comment sounds useful; we can estimate initial velocity without a chrony. Let's take a closer look.

Rmax = vi2/g

so

vi = (gRmax)-2

e.g. if the maximum range = 90m, then the initial velocity is the square root of 9.81m/s/s times 90m or ..... 29.7m/s

These are the velocities implied by a given range at a 45 degree arc.

41m=>20m/s
92m=>30m/s
163m=>40m/s
255m=>50m/s
367m=>60m/s
499m=>70m/s
652m=>80m/s
826m=>90m/s
1,019m=>100m/s

And here's a chart.

Posted Image

Yes, that implies that a projectile travelling a not-unheard-of speed like 100m/s has a potential range of over one kilometre. Of course, at these velocities and with light projectiles, air resistance is a factor, so you have to dial down your expectations a lot, but at the lower end of the scale, it should be indicative.

In my comment below, I will address some more advanced topics.

Blog References:
http://slingshotforu...-of-slingshots/
http://slingshotforu...timal-velocity/

External References:
http://hyperphysics....raj.html#tracon
http://www.pha.jhu.e...m/l5/node3.html
http://en.wikipedia....wiki/Trajectory
http://en.wikipedia....fleman%27s_rule
http://web.archive.o...ory_part_1.html



Somewhat More Advanced Concepts

A common question is: If I have an elevated target, say something sitting in a tree, how high do I have to aim to hit it? Unfortunately, even without considering aerodynamic drag and spin, it's not an small calculation to do.

Θ=tan-1(v2±(v4-g(gr2cos2ϕ+2v2rsinϕ))-2)/grcosϕ)

or so says Wikipedia.

In practice, there is a simplification that rifle shooters use, called the rifleman's rule and it applies when the inclination angle to the target is small, but it kind of presumes that you are packing a laser rangefinder and not many of us do.

RH = RScos(α)

Where:

RH is horizontal range
RS is slant range
α is the inclination angle

as illustrated below (source: Wikipedia)
Posted Image

To save calculation, I have drawn a chart:

Posted Image

So you take either a measurement, or more likely, a rough guess of the angle of inclination to the target and adjust down the range you put into your drop table/chart or angle or elevation table/chart. You still need a good estimation of the range to the base of the tree, or it's all for naught.

The fact that you have to reduce the distance estimation explains why a lot of us have trouble hitting elevated targets. We see something high above us, say 10m up in a tree and incorrectly guess that we need to shoot higher to compensate, when really we need to reduce the amount we have allowed for drop. Another interesting conclusion is that moderate elevation should work to the advantage of slower hunting slingshots (which have more drop), as long as the shooter remembers not to over compensate and shoot high.

Before slower slingshot shooters celebrate, it is still worth noting that the maximum height (and therefore the remaining kinetic energy at a given height) will be limited by the initial velocity.

Hmax = vi2/(2g)

Posted Image

Remember the same caveat about drag and high velocities applies as before.

External References:
http://en.wikipedia....Rifleman's_rule
http://www.pha.jhu.e...m/l5/node3.html
http://microgravity..../ballflght.html
Another great post! Thanks!

whipcrackdeadbunny
Dec 04 2011 01:00 PM
Very nice; am I correct in saying, it will mean, if shooting at an animal in a tree, it would be best to attack from a position where you can hold the slingshot at 45%?

whipcrackdeadbunny, on 04 December 2011 - 01:00 PM, said:

Very nice; am I correct in saying, it will mean, if shooting at an animal in a tree, it would be best to attack from a position where you can hold the slingshot at 45%?

If you were directly under the target, shooting up at 90 degrees (vertical) you would not have to compensate for drop at all.

Hrawk, on 04 December 2011 - 05:44 PM, said:

whipcrackdeadbunny, on 04 December 2011 - 01:00 PM, said:

Very nice; am I correct in saying, it will mean, if shooting at an animal in a tree, it would be best to attack from a position where you can hold the slingshot at 45%?
If you were directly under the target, shooting up at 90 degrees (vertical) you would not have to compensate for drop at all.

Not necessarily, Hrawk; practical experience and physics both agree that this is a hard shot.

Let's look at three extremes of elevation and we can see the effects of gravity and other practical considerations. I will presume the target is high up and being shot at from very close to ground level.
  • Straight under the target
    • Maximum energy at impact (but very little difference as drag has little effect)
    • No drop adjustment (but your trained sight picture will include drop e.g. 10m) so you'll effectively have to aim down
    • Flight path to animal obstructed by branch
    • Headshot obstructed by body
  • 45 degrees
    • Medium drop compensation
    • Likely obstruction by other branches
  • Far away
    • Maximum obstruction by other branches
    • Maximum range to target; hard to hit
    • Maximum drop deviation from straight line of sight
    • Maximum chance of a target seeing the incoming projectile
To be practical, I suggest you just manoeuvre into a position where you can get a clean shot, make a quick mental adjustment of your drop compensation based on the angle of incline to target. Nobody's going to plot a complete firing solution on a squirrel in a tree, even on a smartphone, but if you do save picture of the elevation fan below onto your phone, it'll give you a quick estimation of the range compensation to build in.

Posted Image
Remember this is how you'd adjust range not drop. It presumes you have experience training at different ranges. Math is no substitute for experience, so you may as well also practice shooting into trees. Where this blog post is intended to help you is to give you a better idea of the relationships of velocity, elevation, range and height, so that when you practice, you will have a framework to help differentiate between errors of shooting accuracy and errors of estimation, because as you can see, it may appear a little counter-intuitive at times. For example, if I hadn't researched and done the math, I'd be thinking that I had a tendency to accidentally shoot high at high elevations, when in fact I should have been aiming down.

whipcrackdeadbunny
Dec 05 2011 08:12 AM
Ok; I think I get it, thank you. And personally, I think there's a chance here, to cut out the extremities of the equation, as I wouldn't shoot an animal from underneath (unless a headshot was offered, in which case the shooting is relatively straight) and I don't think I'll be shooting lower than #sic# 2-3 metres (at an animal in a tree) where the deviation is very small. Soooo 50-94% probably. And with experience (and a uniform shooting set), I can learn to judge maximum distance before droppage, therefore compensating quite easily in the field. Yes, I think I have something useful here. Thanks again.

Deltaboy1984
Dec 05 2011 12:35 PM
Very interesting.
Good post! This is very interesting!
Hi mate, thank you very much for this very detailed information, my profession is, Master's and professional golf player, the calculation of flight distance and destination is my life.
A hug

alfshooter, on 07 December 2011 - 05:59 PM, said:

Hi mate, thank you very much for this very detailed information, my profession is, Master's and professional golf player, the calculation of flight distance and destination is my life.

Unfortunately, while golf and slingshots are subject to the same laws of physics, many different variables apply. A long time ago (the 1980's) I made the kinetics model for a golf video game. The flight model was very different. The parabolic trajectory would seem to imply that a 45 degree pitching wedge would out-carry all other irons. We all know that isn't so. What's going on?

One factor is the length of the iron. Longer irons make for more energy at impact. Another is the angle of attack. If you place the ball farther forward relative to bottom dead centre of the swing arc, then the angle of attack is increased, and combines with the loft. The combination of angle of attack and loft and friction of the club head results in spin. Spin affects projectiles of larger radius and lower density such as golf balls, more. The imparted projectile velocity may be similar to slingshots, but the time of flight is much longer, so form and surface drag become material. A slingshot projectile hardly loses any velocity during its flight, while a golf ball may lose two thirds of it's velocity since leaving the club face.

The terminal ballistics is also very different. Slingshot shooters are concerned with the impact coordinates and kinetic energy transfer, but golfers are concerned with the ultimate position of the projectile. A golf ball is much larger than a slingshot bullet, which is great news, or you'd plug every single shot deep in the turf. Possessing a generally low angle of incidence, low density, large wetted area, resilience and topspin, golf balls tend to carry. To calculate carry, you need inputs for the softness (attenuation) and coarseness (friction) of the impact surface. Carry is one big reason why long irons with low loft travel further than short irons, because they affect angle of incidence, kinetic energy at impact, the horizontal component of the kinetic energy and spin. Incidentally, a golf ball cannot have topspin in flight, just less backspin; topspin is created only by momentum vector of the ball and the interaction of the impact surface.

Another interesting difference is the clubhead imparts only a fraction of its kinetic energy to the ball at impact, but as the ball is lighter than the clubhead, the ball may leave the clubface at 50% faster than the clubhead swing speed. That's why it's very rare to get a double hit and why it never happens on a properly struck ball.
Hey Dan, I've written a program to model the trajectories of subsonic projectiles with air resistance. I've made a post about it here: http://slingshotforu...elling-program/.

I've open-sourced the code and created a GitHub repo for it, you can view the page here:
http://samchristy.gi.../trajectory.js/

Please tell me what you think, I'm eager to hear your input on it!

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