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I'm happy to answer any questions or give advice. If you have some ideas or experiences you would like to share, consider submitting and article for publication on the site. You can contact me at info@slinging.org.

I began with slinging just before some months but had a lot of time to practice since then. So I am a run-away beginner (when writing this article). I first began with slinging for range and now am also aiming at a target. What I missed in the slinging internet site was some physical background. So I have thought myself at how slinging works physically. I will present some formulas which could be interesting for slingers. The reader who has some basic knowledge of physics will probably understand much of it. I also added some words for beginners which would have been important for me.

Just a word about (my) style

From the three basic styles (overhand, underhand, overhead) I use overhand slinging as for me it seems to be the most natural style. The „helicopter“ or overhead style looks „high tech“ but seems difficult to learn and for practicing it one needs a lot of space in all directions (at least at the beginning) which isn’t the case for the two other styles. This two styles require some space behind you and a lot in front of you. I confess that overhead style has at least one big advantage: You can’t send a rock high in the sky which afterwards falls on you own head. That is one reason why I don’t like underhand style. It is very easy to shoot on you own head. I made my first few shots during a winter night over a sea (of course don’t do it in the summer if there could be a swimmer). First I tried some overhead shots but the stones always landed only a few meters in front of me. One also landed just 50cm from my feet as a result of a bad pouch but anyway it wasn’t too big. Then I tried the underhand style and immediatly got two good shots. I turned enthusiastic and made a 3rd shot whirling my arm much more than before. I waited for the splash on the sea but I waited and waited. I stood there with opened mouth and was wondering. „Strange ! I don’t understand.“, I thought. And suddenly there was a big plump into the bushes behind me! And this stone was of the size of a hen egg ! Another reason why I don’t like underhand is that not only the vertical but also the horizontal direction seems hard to manage. In fact, later, at the time I was already able to sling overhand I tried again some underhand shots. Not only some of them went again high into the sky but they fell far to the right or left of my signal strap (which marked a given distance).

Range and velocity

The range grows as the square of the velocity at which you release you stone.
(On the other hand the range and the release angle give you the lower limit for the release velocity.)

In the vacuum the release angle of 45deg gives the largest range. The more air friction the flatter this angle will of course be. This is due to the fact that a flatter orbit has a shorter path length. The higher the density of the bullet or generally the surface/mass ratio, the closer we come to the vacuum conditions. That means that a lead bullet as having a much higher density than stone will come much closer to the vacuum conditions. A big stone also comes closer to vacuum conditions than a little one.

To better understand it: Imagine a little cube that has the same mass as a mouse. Take so many of these little cubes you need to get the mass of an elephant. When you put all these little cubes together to get one big cube, most of the surfaces will disappear. Only a little part of them will be seen from outside. So for a big mass the air friction in respect to the volume (or mass) will also be lower than for a little mass. In more physical words: The surface of a sphere growths proportionally to the square of its radius (radius*radius) whereas the volume and mass growth proportionally to the cube of the radius (radius*radius*radius). So, as the kinetic energy is also proportional to the mass and the friction only proportional to the surface, if you double the radius you multiply the kinetic energy by 8 (for a given velocity) whereas the energy which is eaten by the friction during a given time is only multiplied by 4. That means that the optimal bullet is the one which isn’t to light and at the same time does not yet slow down you throwing movement. (The impression I got when aiming at my target at a distance of 25m is that in the range of mass of 80-150g heavier bullets don’t slow down my arm significantly.)

Details

(If you have no physical and mathematical background don’t try to understand. In this case the only interesting thing is the velocity versus range formula)

The bullet when released has the velocity value V and the release angle theta:

horizontal component: Vx0, vertical component: Vy0 (in other words: Vx0=V*cos(theta), Vy0=V*sin(theta))

In the vacuuum we have:
Horizontal velocity is constant: Vx=Vx0
Vertical velocity: Vy=Vy0-Gt (G: gravitational constant, t: time)

This gives for the position:
x= Vx0*t
y= Vy0*t-(1/2)*G*square(t)

The stone falls back to earth when we have again y=0, giving t=2*Vy0/G

This gives the range:
x=Vx0*t=Vx0*2*Vy0/G=V*cos(theta)*2*V*sin(theta)/G=
2*square(V)*cos(theta)* sin(theta)/G

To find out for which release angle theta the range is the highest we must derive x in respect to theta. dx/d(theta) is zero when theta is Pi/4 (in rad) or 45degrees. In this case Vx0=Vy0. As sin(45deg)=cos(45deg)=1/SquareRoot(2)

This gives: x=square(V)/G

or x being the range:
V=SquareRoot(range*G), (SI units assumed: V in m, range in m and g=9.81m/(s*s))

That means: As you double you velocity, you quadruple you range (and your energy)! (That is the same principle as with the stopping distance for a car: As you double you velocity, you quadruple the distance you need to stop.)

The air friction will of course diminish the vacuum range.

Speculation about the slinging stone velocity

I have read a lot of speculation about the velocity of slinging stones in the forum. In my opinion the best approach is to take the range. If you now you range for a given shot you can compute with the formula V=SquareRoot(range*G) the absolute minimum starting velocity to get this range. If you orbit was very flat, that means that you starting velocity was in fact much higher.

Of course the problem of the velocity at the landing point isn’t solved. But if you do target aiming and the target is placed at a distance which is only a part of you range you still have when hitting the target an order of magnitude for the velocity corresponding to the starting velocity.

Some examples:
Range: minimum needed release Velocity (vacuum conditions):
26m: 16m/s
50m: 22m/s
80m: 28m/s
100: 31m/s
200: 44m/s
400: 63m/s

As slinging is always compared to bow shooting: I read that a typical velocity for target aiming at 100m for bow arrows is about 70m/s (In www.archery.de where there speak about traditional barebow - bows without sighting device - I have found the velocity number 160km/h=44m/s but also the range number 250-300m which corresponds to a start velocity of at least around 50m/s. (By the way: They say that to kill a deer one should come as close as 15-20m). That means that if one slings 100m and more and taking also in account an air friction factor, we are coming close to arrow velocities. If anybody is able to sling at 400m, the starting velocity of his bullet will be in fact much higher than 63m/s or even 70m/s, as at this latitudes the air friction is very important.

Principle of the sling relative to the range

The principle of the sling (at the difference of the staff-sling) is the accumulation of velocities. Of course that is clear for everybody. But how does this accumulation occur ? I will just present three explanation models. The first one is a bit too simplistic but is probably understandable by everybody and contains what is the most important. In all models I neglect the earth gravitation because at the velocity the pouch of our sling moves it is really neglectable. After the release the gravitation is of course the most important factor. The higher the velocity the higher of course the air friction is, but I’ll neglect it. Perhaps we can really do that with very good bullets (well shaped lead bullets ?) as long as we speek of ranges not over 100m.

Just think that you are the observer who looks at a right-handed slinger from his right side. To describe the directions I will take the numbers of an analog clock. So 3o’clock is the direction where the slinger looks (if he looks in horizontal direction), 6o’clock is downward, 9o’clock is backward, 12o’clock is over the head of the slinger upward, etc. The slinger is slinging in overhand-style.

1st model: Infinite acceleration of the hand which afterwards moves with a constant velocity in a straight line.

It is assumed that the sling already has the preliminary velocity Vprelim when it passes at the 9 o’clock point. Just at this moment one makes the infinite acceleration forward.

Admitting the infinite acceleration is produced by the fast snap forward which most of the slingers use just before releasing and admitting that the sling is long enough to allow that the hand reaches its fastest possible velocity (which I’ll just call VhandMax which then corresponds to 16m/s if you reach a range of 26m), now the pouch moves at a velocity which is the vectoriel addition of Vprelim and VhandMax. Immediatly after our infinite acceleration both vectors are orthogonal to each other. Assuming that the value of Vprelim is just the same as the value of VhandMax (that is no problem because with a sling of 1m it just corresponds to 2.5 rotations/s - I confess that with a short sling with just 50cm to require the necessary 5 rotations/s is less simple) and the release occurs immediatly, we will have for the sum velocity:

square(ValueOfVsum)=square(ValueOfVprelim)+square(ValueOfVhandMax)
=2*square(ValueOfVhandMax)

This gives: ValueOfVsum=squareroot(2)*ValueOfVhandMax

As we have already seen the range increases as the square of the velocity. And if our fast snap forward was horizontal, Vsum just has an angle of 45deg with the horizontal (because ValueofVhandMax=ValueOfVprelim) which corresponds to the furthest range (at least in the vacuum). So we will get a range of 52m.

If we don’t release immediatly after the fast snap but continue to move our hand at the velocity VhandMax until the pouch is just at the vertical over the hand, where we release, we will just change the direction of Vprelim. At the release Vprelim and VhandMax will be parallel. So the values are completely added: ValueOfVsum= 2*(ValueOfVhandMax). This value would give a range of 100m if the Vsum had an angle of 45degree with the horizontal instead of 0. So one should do the fast snap earlier (when the poach is at 7.30 o’clock) and directed at 45degree.

Of course this model is a bit too simple. You can’t really accelerate you hand fast enough to get VhandMax during a time where the pouch just moves at a maximum of a few degrees. This would require a sling of several meters. To continue to move the hand at the maximum velocity VhandMax for a while is not possible as the hand is fixed to the body. Instead it will be slowed down very fast after having reached its maximum velocity.

But even if this model is very simplistic the 2nd model of explanation will show that our 1st model is a very good approximation. Only at very low - and uninteresting - velocities the 1st model is unsufficient.

2nd model: Constant acceleration of the linearly moving hand

As the first model seemed too unrealistic to me I wanted to know if it works in the same manner if we have a more realistic acceleration. I confess that it took me a lot of time until I got a real result. First I didn’t find any solution myself. So I searched in a lot of physics books at the university. But such a thing wasn’t treated anywhere. At the end I found myself an interesting equation which says a lot and which in fact is very easy to obtain (once one knows how to find it).

One starts from the following two equations which give the position of the sling in respect to the time:

x=0.5*aHand*square(time)-L*cos(theta)
y=L*sin(theta)

where:
- x : horizontal coordinate
- y : vertical coordinate
- L : length of sling
- aHand: acceleration of the hand
- theta : angle of the sling (0 when horizontal and in backward direction, pi/2 when vertical and over the head, etc)

After deriving this two equations twice in respect to the time we get the components of the acceleration of the pouch:
(1) ax=aHand+L*cos(theta)*square(AngulVel)+L*sin(theta)*AngulAcc
(2) ay=-L*sin(theta)*square(AngulVel)+L*cos(theta)*AngulAcc

where:
- AngulVel : angular velocity of the sling (or derived of theta in respect to time)
- AngulAcc : angular acceleration of the sling (or twice derived of theta in respect to time)

On the other hand we have following equations (which are as evident as the two for x and y, at least if you know a little bit of physics and maths):

(3) ax=aHand*square(cos(theta))
(4) ay=-aHand*cos(theta)*sin(theta)

So, as the left side of equation (1) is equal to the left side of equation (3) and as the left side of equation (2) is equal to the left side of equation (4), the right sides must also be equal:

(5) aHand+L*cos(theta)*square(AngulVel)+L*sin(theta)*AngulAcc=aHand*square(cos(theta))
(6) -L*sin(theta)* square(AngulVel)+L*cos(theta)*AngulAcc= -aHand*cos(theta)*sin(theta)

After multiplying (5) with sin(theta) and (6) with cos(theta) we get two similar terms in both equations:

(7) sin(theta)*aHand+sin(theta)*L*cos(theta)*square(AngulVel)+ sin(theta)*L*sin(theta)*AngulAcc= sin(theta)*aHand*square(cos(theta))
(8) -cos(theta)*L*sin(theta)*square(AngulVel)+cos(theta)*L*cos(theta)*AngulAcc= cos(theta)*aHand*cos(theta)*sin(theta)

When adding the left side of (8) to the left side of (7) and the right side of (8) to the right side of (7) we get (we also need to know that square(cos)+ square(sin)=1):

aHand*sin(theta)+L*AngulAcc=0

which is equivalent to (9) below:

Remark

For the pendulum they always make the assumption that it does only do little deviations from the vertical so that sin(theta) can be assumed equal to theta. In this case the equation becomes:

AngulAcc+(aHand/L)*theta=0

which has the general solution: theta=sin(omega*time+phi)=sin((2*pi/T)*time+phi)

(where phi depend on the position of the pendulum at time=0, T is the period of an oscillation)

omega=SquareRoot(aHand/L)

In the case of a pendulum instead of aHand we have the gravitation constant and L is of course the length of the pendulum)

But for us this solution for little theta is uninteresting as we have big angles. But nevertheless the equation (9) tells us a lot.

First, as I already said, our system behaves like a pendulum, but in our case as a pendulum which makes whole rotations around its fixed end. Such a pendulum, when moving down, gains velocity and when moving up looses velocity. As anybody sees from the equation - as sin(theta)=-sin(-theta) - the increase of the angular velocity on one side is compensated on the other side. That means that the angular velocity is the same for a given height (the height in my special case of the sling is the horizontal distance from the pouch to the hand).

So, if you take the hand as the origin of coordinates the sling behaves like a pendulum submitted to a high gravitation force - much higher than the earth gravitation force. What is the value of this fictitious gravitation force ? It can be coarsely estimated. If you throw a stone with you nacked hand at 26m (which is my limit) you need a release velocity of 16m/s. This velocity is reached in a fraction of a second. If we suppose that this fraction is 0.1s then our acceleration was 160m/(s*s) (whereas the acceleration by the earth gravity is G=9.81 m/(s*s)). As with the real gravity our fictitious super gravity is bound to (fictitious) potential energy. A difference of height H for a given mass M corresponds to a potential energy M*G*H. If the pendulum looses potential energy it gains kinetic energy. The same reasoning applies to the sling. When the pouch is at the 9 o’clock position - which corresponds to the lowest height of the pendulum - we have M*aHand*L more kinetic energy than when the pouch is just over or under the hand. To what amount of velocity does this correspond ? It depends on the total kinetic energy. It makes only an appreciable difference when the sling is rather moving slowly or not at all. So, we will see in the following in a presentation of different cases that with only a fast snap (and no preliminary velocity of the sling) it is possible to throw further than just by nacked hand: With a sling of 1m, doing the snap downwards (instead of forwards) in a line having 45degrees, and releasing at the right moment it may be possible to get a range of 40m (assuming a hand accelerartion of 160m/(s*s)). The higher the amount of preliminary velocity when you start you fast snap the more neglectable becomes this pendulum effect and our model of constant acceleration gets similar to the model with an infinite acceleration.

I always assume:
- the fast snap goes toward the direction given by the 3 o’clock position.
- the fast snap begins when the pouch is at the 6 o’clock position
- the length of the sling permits to reach just vHandMax when the sling is or would be at the 12 o’clock position.
- vHandMax is 16m/s.

What is also important to know is that the time the pouch takes to get from the 6 o’clock position to the 9 o’clock position is equal to the time it takes from the 9 o’clock to the 12 o’ clock position. (This should be intuitively clear: When you let you pendulum fall down it will need the same time to reach the lowest position than it needs to move up on the other side until it reaches the initial height.) That means that when the pouch is at position 9 o’clock, half of the time for the acceleration has passed and that this means that at this instant the hand has reached the velocity vHandMax/2 (as for a constant acceleration we have: velocity=acceleration*time).

A formula which will be used is how to compute the velocity at the 9 o’clock position as a function of the velocity at the 6 o’clock position Vprelim and of the sling length. To get this formula I use again the mathematical identity with the pendulum physics. So, if we take the accelerating hand as our origin, the pouch sees a (fictitious) force trying to pull it towards the direction opposed to the hand. There is also a (fictitious) potential. The farer the pouch is from the line 6 o’clock to 12 o’clock the lower its potential energy and higher its kinetic energy. Especially for the 9 o’clock position we have (V9 being the velocity at this position):

M*aHand*L=0.5*M*square(V9)-0.5*M*square(Vprelim)

This gives:
V9=SquareRoot(square(Vprelim)+2*aHand*L)

V9 grows with the length of the sling L. So let’s just assume L=1, which is the biggest length still more or less well manageable.

V9=SquareRoot(square(Vprelim)+2*aHand)

Case 1: Vprelim=0, release at 12 o’clock position:

The angular velocity gained from the 6 to the 9 o’clock position disappears again completely when the pouch comes to 12 o’clock. So the total velocity of the pouch is equal to Vprelim + vHandMax = vHandMax. So it is just as throwing without sling. By making our snap 45degrees upwards we get the range of 26m/s.

The 1st model would give the same result.

Case 2: Vprelim=0, release at 9 o’clock position:

The angular velocity gained from the 6 to the 9 o’clock position is fully preserved. So the total velocity vector is the sum of the velocity due to angular motion and the reached hand velocity.

As Vprelim=0 we get:
V9=SquareRoot(square(Vprelim)+2*aHand*L)= SquareRoot(2*aHand)=SquareRoot(2*160)=17.9m/s

The total velocity is the sum of V9 and vHandMax/2. As the vectors are orthogonal we can’t just add the values but have to use the formula:

vTotal=SquareRoot(square(V9)+square(vHandMax/2))=19.7m/s

That is a velocity which allows us to get a range over 40m. So theoretically it is possible, even when having Vprelim=0, by doing a snap which goes mostly downward (if it goes only forward we get a velocity vector which is nearly vertical) and releasing at the right instant we get a range 15m higher than by throwing without sling.

Case 3: Vprelim=16m/s, release at 12 o’clock position:

The angular velocity gained from the 6 to the 9 o’clock position disappears again completely when the pouch comes to 12 o’clock. So the total velocity of the pouch is equal to Vprelim + vHandMax=16m/s+16m/s=32m/s. By making our snap 45degrees upwards we get a range of 100m/s.

The 1st model would give the same result.

Case 4: Vprelim=16m/s, release at 9 o’clock position:

The angular velocity gained from the 6 to the 9 o’clock position is fully preserved. So the total velocity vector is the sum of the velocity due to angular motion and the reached hand velocity.

V9=SquareRoot(square(Vprelim)+2*aHand*L)=SquareRoot(16*16+2*160)=24m/s

The total velocity is the sum of V9 and vHandMax/2. As the vectors are orthogonal we can’t just add the values but have to use the formula:

vTotal=SquareRoot(square(V9)+square(vHandMax/2))=25.3m/s

Conclusion

Case 3 and case 4 differ only in the fact that we have a very early release in case 4. As the release velocity in case 3 (32m/s) is much higher than in case 4 we see that here such an early release doesn’t make sense. Only with Vprelim<4m/s do we get for a 9 o’clock release a velocity that is higher than for a 12 o’clock release. The effect is the highest with Vprelim=0 but to reach the maximum range of about 40m (or possibly a few meters more if aHand were in fact higher than my coarse assumption of 160m/(s*s)) in this case doesn’t present any advantage. So at the end this angular acceleration due to the pendulum effect is rather uninteresting for a slinger. The slinger can always think of his sling as functioning as in the simple 1st model if he makes a snap with his hand which is a straight line and if his sling had some preliminary velocity. If he just releases when the force on his finger disappears the bullet will just leave parallel to his snap direction and the value of its velocity will be the addition of the value of Vprelim and vHandMax.

3rd model: Hand exercices a central force with diminution of the distance from the center of the force to the pouch

If an object is submitted to a force which always points to the same point (central force) the product of the square of the distance from the object to the center of the force (DistanceToCenter) with the angular velocity (omega) is constant.

Square(DistanceToCenter)*omega=constant

This implies that if DistanceToCenter is for example reduced to the half then omega is multiplied by 4.

Example: Shoulder as center of the force. Assuming that at a certain moment you arm is extended backward and the hand is a little higher than the shoulder and the sling is in one line with the arm, if from that point on you can move you hand/arm in a manner that the cords always show to the shoulder and at the release instant the hand is just next to the shoulder, that would mean that the DistanceToCenter has diminished by the length of you arm. For me this is around 0.6m. If you sling has the same length (to simplify), at the release you have reduced DistanceToCenter by half and at the same time quadrupled the velocity of the pouch.

(Sometimes I have the impression that I use this effect to some extent.)

Principle of the staff-sling

The staff-sling is more a staff than a sling and its physical principle is different. The staff gives you a „long arm“. If you arm has a length of 60cm and the stick is only twice as long (120cm) and you manage to move this long arm at the same angular velocity as you real arm alone, the tip of the stick will have the double velocity of the hand, meaning you will theoretically reach four times the range. A projectile you throw at 25m with nacked hand could theoretically reach 100m with this staff.

The sling of the staff-sling is in fact a device which makes it possible to charge the staff with the projectile and which provides and automatic release (other kind of release mechanism are possible and in fact they exist in different kind of sports and I read about something like this somewhere in the forum). A short sling should be sufficient as a longer sling doesn’t give more velocity.

What would be very interesting would be the combination of both principles, the principle of the sling and the principle of the staff-sling. I personly made a few shots with a rather long sling. It had a length of 1.30m. I was tedious to load and I had to sling a the height of my shoulder. But I got a range of over 100m very easily. I could imagine an even longer sling. But to sling I would „elongate“ my arm with a stick bound to my hand, bind one cord at the end of the stick and let the release cord pass through the end of the stick and then come to my hand. (I wonder if anyone ever tried this.)

Centrifugal force

That is the force which pulls the knot of the release cord out of you hand before you decided to release yourself, and lets the stone fly to the vertical and fall back on you head. And it is the same force which makes it necessary to put tape on you retain finger. (To avoid this just make a longer sling.)

F=m*Square(V)/R, where m=mass, V=velocity, R=radius of rotation

In other words the stress on you hand (or finger) is proportional to the mass, decreases as the radius of rotation increases and is proportional to the square of the velocity. That means for example - as the range is proportional to Square(V) - that throwing a bullet at 100m with a sling of 1m puts the same stress on your fingers than throwing the same bullet at 50m with a sling of 50cm.

Accuracy: necessity for a technique which gives a standard velocity

Accuracy

After having reached my range goal of 100m, I decided to do some target aiming. I put a stick at over 70m and now used some round cement bullets instead of potatoe shaped zigzaging stones. I was very astonished about the horizontal accuracy at this distance. A great part of the bullets passed closer than one meter from the stick. I even destroyed my stick already at the 5th or 6th shot and find the upper part some meter further (but I confess that it was already damaged)!

Somehow my body and subconscious had already learned how to get a good precision in the horizontal dircetion during my practice for long range shots, as I was always shooting at a given direction. And anyway, horizontal accuracy by using overhand style is reached in the same manner as throwing stones or snowballs by hand. So it is not really a surprise that a good precision is possible from the start. You eyes need only to find the right gap between the target and the pouch (or hand). And this gap can be taken as the same at all distances that are really interesting. At very low distances (when you are throwing a distances under 20m) the gap diminishes until reaching 0 at the distance corresponding to the length of arm+sling.

Vertical Accuracy

Of course, at the beginning I had no vertical accuracy at all. This was what I expected. For the horizontal direction, once you have got the right direction - neglecting details as the wind, air friction which can change the direction of bullets which aren’t ideally shaped - the bullet flies in this direction independently of its velocity. For the vertical direction comes the gravitation which lets you bullet fall down a given distance in a given time. So, to hit a target you need to develop a throwing pattern which gives you a defined velocity.

To illustrate this problem in more physical terms and some numbers:

Horizontal position = Vrel*t
Vertical position = 0.5*G*square(t)
Vrel: velocity of your bullet when released
G: gravitational constant (=9.81 or about 10)
t: time

1. case: value of Vrel=22m/s is the velocity you need for a (vacuum) range of 50m
2. case: value of Vrel=28m/s is the velocity you need for a (vacuum) range of 80m
3. case: value of Vrel=31m/s is the velocity you need for a (vacuum) range of 100m

Assuming there is no gravitation, if you aim at a target at 30m from where you are standing you’ll need approximately 1.5seconds in the 1st case, a little bit more than one second in the 2nd case and a little bit less than one second in the 3rd case. In these times, if switching again the gravitation on, we would have a falling down of:

1. case: 0.5*g*square(1.5)=11m
2. case: 0.5*g*square(1.07)=5.6m
3. case: 0.5*g*square(0.97)=4,6m

That means, in order to hit you target you can’t release the stone when it is just moving straight to the target, but have to release it when its velocity has an upward component. The slower you release, the higher you’ll have to aim. (At the same time the orbits will now be curved, so that the falling down until reaching the target will even increase especially at lower velocity.)

As the 3 cases show and what is of course trivial, with higher release velocities the orbit becomes more straight. So, theoretically it would always be the best to used the highest possible velocity. But in my opinion this is not the best choice because to get the highest possible velocity one also make a lot of body distortion which will instead reduce the accuracy (and it is of course too tiring.) But in fact it is necessary to develop some kind of throwing technique which gives you always a defined (possibly high) release velocity. (This problem doesn’t exist with guns because the explosion energy of a bullet is always the same amount; so the falling down at a given distance is always the same. But with bows the problem also exists: if you pull too much on the bow the arrow will pass above the target and if you pull too less it will pass below the target. So, you have to find the right pulling force which suits you most.) Once you have established such a technique (or at the same time), you’ll have to practice to find the release point which allows to hit at a given height. A lot of practice will program you brain to use automatically the right pattern.

I reduced my own target aiming distance to 25m and after around 100h of training was able to hit my target (towel hanging on a stick, width=45cm, height=60cm) in more than 10% of the shots. With around 80h more I’m at 15%. That seems to be a lot of time for a thin result. But if you look at how much time you need to be a master of a martial arts like karate or aikido it is not very much. There you are still nearly a bloody beginner after 200h of practice. As with slinging you need a lot of time to learn the right movement patterns and to automate them.

The weight of my bullets ranges from 80 to 150g, but I haven’t the impression that different weights in this range do really have an impact on my throwing pattern.

The next step would be - once one gets a high hit rate for a target at a given distance and height - to variate the distance to the target or its height.

Ammunition

At the beginning my aim was to get a high range, at least 100m. My stones had any potatoe like shapes and where also of very different size. I always tried to throw towards and beyond a signal strap which gave me the distance I wanted to exceed. I realized that even at low ranges like 60m or even less some stones were zigzaging around like fool. Sometimes I wondered about the accuracy of an orbit going straight to the signal and just in this moment the stone changed direction and landed far away to the left or right.

So, as I had reached my goal of 100m, I decided to do some target shooting. But for this purpose I wanted to used some round projectiles. I bought one golf ball for trying but it had a too low density (only slightly more than water). So I made cement bullets. I have been slinging during very windy weather and such a 100g cement bullet is rather unsensible to wind whereas the golfball is just blown away.

Some pieces of advice for beginner

Slinging is dangerous and not only for other persons but also for the slinger himself. But some simple precautions can nearly bring this danger to vanish.

Pouch
VERY IMPORTANT: The material must be unstretchable!

As you can see on the site some people make cupped pouches, other just use some strip of material. Better begin with a cupped pouch. That gives you some work (sewing) but also gives you more security. And you can use stones of any shape or size. I just used any material at my disposition. I had also some very stiff material. As long as you make a cup this doesn’t really matter at the beginning. Later on if you want to do some aiming, I can’t say if this stiffness would blur the precision. For my newer pouches I used some pieces of an old judo trouser (that is really unstretchable and resistant!).

If you just want to use a simple strip of material, it must be extremly soft, so that it embraces you stone as the sea envelops a swimmer. (And/or use elongated stones.) If you put you first stone in and it immediatly rolls down, that is exatly the opposit. Perhaps you can still make a nicely cupped pouched with the same type of material. But don’t make it too slightly cupped. In this case the stone will perhaps stay in the pouch when you load it but leave it during the rotation phase (especially if the material is still a bit stretchable) and than you don’t now where it has gone. Perhaps it will fall on you own head, as it nearly happened to me more than one time. So to assure you pouch is ok, do some shaking test. Put a stone in you cup, let the sling hang down and now shake the cords and make the stone hop in the cup. If it does’t fall down, the cup is good.

And don’t forget: use unstretchable material! I once took a piece of an old jeans and made a very nice cup and first didn’t understand why my stones were flying away too early. The reason was that the jeans was made of strechable material!

Let’s just come back to the very simple pouch consisting only of a strip of material hanging on the two cords. As I said, it should be very soft in order to embrace the stone. So, on one hand such a pouch requires only a minimum of work. On the other hand the material must be chosen very carefully. If the material presents some little stiffness and does’t neatly embrace you stones, forget it. Think at the danger for other persons, animals, properties and if none of these are present think at you own head. In my opinion such a pouch is more something for specialists, in other words advanced slingers.

Length of the sling

In my opinion a beginner should start with a sling which is as long as possible but is still easily manageable. And that is one which goes from the elbow nearly to the ground. For me this is around 1m. Much shorter ones are dangerous at least if you want to sling for range. Longer ones are cumbersome to use. They take much time to reload and you must hold you wrist very high in order not to touch ground.

A 1m sling gives an immediate positive appreciation of slinging. Just by rotating at 3 rotations/sec, which isn’t very much, one gets a velocity of 3*2*PI=6*3.14=19m/s. This corresponds to a range of over 36m, which is probably much more than most people are able to throw just with their hand. Of course, to get this range one must release around 45deg (at this velocity the air friction is still neglectable, so this is the angle for the longest range). At a rate of revolutions of 3 I think that it should even be possible to look at the rotating sling and just release by eyesight when the right angle is reached.

Shorter slings need a correspondingly faster rotation. If you sling has just a length of 50cm, you will have to rotate at a rate of 6 revolutions per second to get 36m. This already is a lot and it will be much more difficult to release at the right position.

I personly began with very short slings and it was a big error. I first learnt to sling overhand with a sling which had just around 55cm. When I was able to start my rocks without bumping them just 3m in front of me I past to a higher length: about 63cm (from the bottom of the pouch to the end of the retain loop). With this sling I began to really practice for range. My stones began to land at 60m and a little more. But that was the limit. That wasn’t too bad but was far grom the 100m and much more I had read about. The problem was that I was first misleaded by the little slinger icons on the forum, where the sling is very short, corresponding to my so far used 55 or 63cm lengths. Second, at that time I thought that a sling reaching from the down hanging hand nearly to the ground would be the longest still well manageable sling. So I made a new sling with 75cm (bottom of the pouch to the end of the retain loop). I was very frustrated as I realized that I got no apreciable increase in range. I thought that this was the longest still easy managable sling and wanted to get a much higher range with it. But only a very few stone reached 80m and most were still landing in the range 60-65m. At this point I began to think at the physics and at my own physical condition. It used the launching method „preliminary rotation followed by a fast snap“. I thought that perhaps the problem was that my bad physical condition (for example a lot of overweight) made some of the difference with other people, as that would give a too slow snap. Another assumption was that with practice it should be possible to increase the manageable velocity preliminary to the snap. But nothing helped. I got really angry and threw the stones with so much force that I often lost grip of the release cord before I really wanted to release. The centrifugal force was simply too high. Many stones flew high to the sky and could have hit me. But all this effort didn’t give any progress.

After all that fiasco I rethought about the length parameter. I used a length of 85cm for a few shots and realized first that the sling was still easily manageable and this time noticed a perceptible increase in range. Then I immediatly went to the length elbow/ground and this length also was still manageable. And I easily came to 80m and with little effort to over 100m! I also used a length of 1.30m for a few shots. I easily reached 100m. But with this length the whole thing is getting very cumbersome.

To resume:

Advantages of long slings:
• higher range as the same revolution rate gives a higher velocity than for a shorter sling
• less dangerous: Because the centrifugal force is lower you don’t so easily loose grip of the release cord.
• to reach an appreciable range, you need no tape on you retain finger, because the centrifugal force is lower
• higher precision in the plane of revolution for a given range, as you turn the sling slower to attain a given range

Advantages of short slings:
• you need less place around you (slinging from under a tree...)
• faster reloading (in the case a lion is jumping at you...)
• less tangling of the cords

A simple security technique: Hands on the head!
(and possibly run to the right side if you are right handed or vice versa)

During the few hours of practice with my first sling I developed a security measure which was necessary after the pouch got a bit too narrow because of some reworking on it as it had frayed too much. So now and then a stone went out too early. In such a case I always run some meters to my left and put my hands on the head. At this time this was a very good technique as I still wasn’t throwing with much energy.

Later I was struggling for range. I wanted to reach 100m and I didn’t know that my sling was too short for that. This time it wasn’t a bad pouch but the too high centrifugal force which sent stones into the sky somewhere over me as I lost grip before I wanted to release. That means there were going very high. In such a case I continued to run 10 to 15m to my left side and still put my hands on the head. Most of the time it was a good technique not only for the security. When I had moved to the side and felt at a secure distance I turned towards the position where I launched the stone and often saw where the stone landed. So I was able to find it. Most of them fell very close to the throwing axis. That made me too careless. So one time as again I had run to the side, I was too lazy to put my hands on the head. Then I stood there and tried to see where the stone would land. I waited for some seconds and suddenly the stone plumped on the soil two meters behind me. I felt badly shocked! Theoretically I should have known that my running to the side technique wasn’t enough, as in my struggle for range I was throwing with a distortion of my whole body, so that the plane of revolution was no longer too close to the vertical. I also already new that some stones were zigzaging. So in fact I should have known that it wasn’t impossible that a stone could land so far to the side. So after this shock I decided, after thinking a bit about it, that running to the right was a bit more secure than to the left and that I shouldn’t be too lazy to put the hands on my head. (Instead I continued to run to the left most of the time as it was already an automatism.) And several stones still landed several meters to the left. In fact, as I thought, none went very far to the right (at least the ones I saw landing).

One last advice: If you are not interested in my advices and instead prefer to use a bad pouch, make range with a very short sling and other crazy things, better use at least a minor helmet (they are very cheap) from the beginning on and not only after you already had a lot of accidents.

And if you don’t wear a minor helmet and it happens that you stone is flying somewhere over you, put you hand or arms on you head. I it wasn’t just a stone that left too early because of a bad pouch when you were just beginning too rotate, but a powerfull underhand shot with too late release or a powerfull overhand shot with too early release, remember that the projectile can take a lot of time to come down. For example: In the vacuum it takes 3 seconds to fall from 50m and 4.5 seconds to fall from 100m. Taking air friction in account it will of course take longer. To this we have still to add the time the stone needs to get at this height. So it is probably a good idea to let the hands on you head at least for 10 seconds.

- George Alsatian