I
began with slinging just before some months but had
a lot of time to practice since then. So I am a runaway
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 80150g 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=Vy0Gt (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 250300m 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 1520m). 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
staffsling) 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
righthanded 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 overhandstyle.
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:
Sling Behaves Like a Pendulum
(in respect to the hand)
(9) AngulAcc+(aHand/L)*sin(theta)=0
This is just the equation for the pendulum which is covered
in any book of basic physics!
(Of course in this case theta is the angular deviation
from the vertical and not from the horizontal.)
So the sling dragged by the accelerating hand in respect
to this hand behaves mathematically as a pendulum submitted
to the gravitation. If we take the hand as the origin
of our coordinate system, in this system we get a fictive
acceleration towards the opposite direction of the real
acceleration
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 staffsling
The staffsling 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 staffsling 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 staffsling. 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 6065m.
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
