The Technical Archer
Austin Wargo
May 14, 2010
Abstract
A mathematical model of the interactions between a long bow and an arrow.
The model uses the Euler-Lagrange formula, and is based off conservation of
energy. It is a valid model from the time the arrow is shot to the time the
arrow reaches brace height. Matlab will be used to solve the equation for initial
velocities and positions as the arrow is shot.
1
Introduction
Archery has exploded in popularity in the last ten to fifteen years. This is
mostly due to the introductions of easier shooting compound bows. Yet, little
is known by the general population of what actually happens when the string
of the bow is released. This paper will deal with the simple case of a long bow
being shot, and will mathematically model it. It will be a valid model from the
point where the arrow is released to the point where the arrow reaches the brace
height and leaves the string. The model will be based on conservation of energy,
and use the Euler-Lagrange formula to find a governing differential equation.
This equation will then be numerically solved to find the initial velocity of the
arrow, and the position of the arrow at certain times as the arrow is being
shot. The paper will be broke down into six sections. The first section is The
Model and deals with the derivation of the model, and assumptions made in the
process. The second section is the Set Up of the Equation, explaining how the
energy equations became a differential equation. The third section is Solving of
The Equation. Stating how the equation was solved, and the results that where
found. The fourth section is Checking the Model, and does an analysis of how
close the model is to real life. The fifth section is Other Topics of Research, and
will give brief account of some other topics that pertain to archery, but will not
go into the solving of these topics. The sixth section is the Conclusion which
will analyze what was learned from doing this project. Before the modeling
process is started, some needed definitions of archery terms will be given.
1.1
Archery Terms
These archery terms are terms any archer will know, but for those who are not
familiar with archery will need to know before reading this paper.
Full Draw
When the bow is pulled all the way back.
Knocking Point Where the Arrow attaches to the string.
Knock
The clip on the end of the arrow that attaches to the string.
Draw Length
Is the length between the knocking point and the handle of the bow, at full draw.
Brace height
The length between the handle and the string when the bow is not drawn.
Riser
The stiff inflexible part of the bow between the bow’s limbs.
Limbs
The flexible part of the bow.
Draw Weight
The amount of force required to pull the bow back to full draw.
These terms will be referred to throughout the paper, and will need to be known
to fully understand the model.
2
The Model
In this section of the paper assumptions will be addressed, and the energy
equations will be set up. The first part of this section will deal with assumptions.
The assumptions will be the major ones made in the paper, but some smaller
ones will be made as the energy equations are derived. The following derivation
1
of the energy equations and assumptions are not original work, but were taken
from reference [1].
2.1
Assumptions
The model starts out with numerous assumptions. The first major one is the
archers paradox is ignored. The archers paradox is the bending of the arrow
as it is compressed by the forward action of the bow. This is an enormous
assumption, but in the short time frame this model will be valid it is acceptable.
Also, the math involved with the archers paradox is so complicated, for the little
effect it will make, it will be easier to leave it out of the model. The next major
assumption is the limbs of the bow will be treated as two inflexible beams, with
a hinge point at the ends of the riser. This does not follow the exact motion a
bow’s limb will take when being shot, but due to the complexity of the math
involved, different parts of the limbs changing at different rates and angles,
this assumption will be made. The potential energy of the bow’s limbs will be
assumed to be proportional to the square of the deflection angle, this will be
explained later. The model assumes the string of the bow stays straight as the
bow is being shot, and does not bend. The model will not take into account
any dampening effects due to outside forces or energy losses. In other words
air resistance and sound, the two main outside effects on the bow arrow system
will not be taken into account. Air resistance can be neglected for the very
fact of the shortness in duration the model is valid. So it’s effects should be
minimal, and if taken into effect it would most likely slow the system down only
slightly. Sound will effect the system by energy leaving the system. Taking it
into account however, would enter too many other variables not wanted to be
dealt with. The model will also, ignore stretch of the string. String stretch can
be figured out, but due to time constraints this will be ignored. Also, do to
modern materials string stretch is not nearly as big of a factor as what it was
in years past. Finally, the arrow is assumed to be a point mass concentrated at
the center of the string. This takes care of most of the major assumptions, and
the model can now be set up.
2.2
Derivation of the Energy Equations
The differential equation will be set up using the Euler-Lagrange formula. Which
is first set up by getting kinetic energy minus potential energy. The kinetic energy of the string, arrow, and limbs need to be found, and the only potential
energy is the energy stored in the limbs.
The kinetic energy of the arrow is simple enough to figure out. Assuming
the mass of the arrow is a point mass concentrated a the center of the string.
𝐾𝐸𝐴 =
1
𝑀 𝑥˙ 2
2
(1)
Where 𝑀 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑟𝑟𝑜𝑤 and 𝑥˙ is the rate of change with respect to time
in the 𝑥 direction. There is no 𝑦-component because the arrow is on a rest and
2
Figure 1: Bow Diagram 1
resting on the string, perfectly horizontal. In the time frame the model is valid
there is no change in the 𝑦 direction assuming the arrow is shot directly in the
𝑥 direction and not at an angle.
Finding the kinetic energy of the string is a little more difficult. The mass
of the string cannot be considered a point mass at the center of the string. The
velocity is not constant along the whole length of the string at a specific time,
so an integration is needed to find the kinetic energy. To do this integration a
uniform linearly mass density is needed to be assumed as 𝜆. 𝑠 will be defined
as the half length of string. The mass of the string can then be defined as the
following 𝑚𝑠 = 2𝜆𝑠. Another variable is needed to be introduced for the point
(𝑥𝑠 , 𝑦𝑠 ) on the string which is 𝜉, and 𝜉 = 𝜖𝑠 where 𝜖 is the fractional length of the
string; remembering, the string is assumed to stay linear no matter what force
is applied. The kinetic energy can now be found from the given information.
∫ 1
𝜆𝑠(𝑥˙𝑠 2 + 𝑦˙𝑠 2 ) 𝑑𝜖
(2)
𝐾𝐸𝑠 =
0
3
Now 𝑥𝑠 and 𝑦𝑠 can be defined by Figure 1.
𝑥𝑠 = 𝑥 + 𝜉 sin(𝜙)
𝑦𝑠 = 𝜉 cos(𝜙)
From this information 𝑥˙𝑠 and 𝑦˙𝑠 can be determined by taking the derivative
with respect to time.
𝑥˙𝑠 = 𝑥˙ + 𝜖𝑠𝜙˙ cos(𝜙)
𝑦˙𝑠 = −𝜖𝑠𝜙˙ sin(𝜙)
Now there is an unwanted 𝜙˙ that can be be replaced by 𝜙˙ = 𝜙,𝑥 𝑥˙ were the
comma subscript means partial derivative. Then by making some substitutions
the kinetic energy of the string becomes.
∫ 1
𝐾𝐸𝑠 =
𝜆𝑠((𝑥˙ + 𝜖𝑠𝜙,𝑥 𝑥˙ cos 𝜙)2 + (𝜖𝑠𝜙,𝑥 𝑥˙ sin(𝜙))2 ) 𝑑𝜖
(3)
0
By carrying out this relatively simple integration, and substituting the mass of
the string in the results are shown in equation (4).
𝐾𝐸𝑠 =
1
1
𝑚𝑠 (1 + 𝑠𝜙,𝑥 cos(𝜙) + 𝑠2 𝜙,𝑥 2 )𝑥˙ 2
2
3
(4)
This sums up the kinetic energy of the bow string, and now it’s time to move
onto the kinetic energy of the limbs of the bow.
Working off of the diagram in Figure 2 the kinetic energy can be determined.
𝑥 will be the axis of symmetry cutting the bow in half. 𝑙 is the length of the
moveable part of limb. The riser or handle of the bow will be considered to be
2𝐿. The change in the position of the limb will be characterized by the angle
in between limb and a vertical line that goes down from where the string meets
the limbs. The limbs will be assumed to have a constant density called 𝜌. A
constant thickness will also be assumed and will be called 𝜏 . Finally the width
can vary uniformly as it travels up the limb. 𝜔𝑜 will the width at the top of the
limb, and 𝜔𝑜 + 𝜔1 is the width at the hinge point of the limb. Also, 𝑞 is the
length along the limb at a given point as seen in Figure 1. The width of one
∫𝑙
limb can now be described as the following 0 (𝜔𝑜 + 𝜔1 − 𝜔1 𝑞𝑙 ) 𝑑𝑞. The Mass of
the limb can now very easily be taken from here.
∫ 𝑙
𝜔1
𝜔1 𝑞
𝑚𝐿 = 𝜌𝜏 𝜔𝑜
(1 +
−
) 𝑑𝑞
(5)
𝜔𝑜
𝜔𝑜 𝑙
0
From here, the kinetic energy of the limbs can be determined, but one more
factor needs to be taken into account. The limb travels in an arc as seen in
Figure 1. This distance of the arc can be defined as 𝑟 = 𝑞𝜃. 𝑟˙ is needed for the
kinetic energy of the limbs, 𝑞 is an independent variable and 𝜃 is the dependent
˙ This can now be substituted into the kinetic energy equation,
variable so 𝑟˙ = 𝑞 𝜃.
and kinetic energy can be solved for.
∫ 𝑙
𝜔1
𝜔1 𝑞 2 ˙2
)𝑞 𝜃 𝑑𝑞
(6)
𝐾𝐸𝐿 = 𝜌𝜏 𝜔𝑜
(1 +
−
𝜔
𝜔𝑜 𝑙
𝑜
0
4
Figure 2: Bow Diagram 2
By integrating this equation the following kinetic energy equation is obtained.
(
)
1 𝜔1 ˙2
31
𝐾𝐸𝐿 = 𝜌𝜏 𝜔𝑜 𝑙
1+
𝜃
(7)
3
4 𝜔𝑜
This equation needs to be in terms of 𝑥 so 𝜃˙ = 𝜃,𝑥 𝑥˙ by making this substitution,
and substituting in for 𝑚𝐿 the equation can be simplified down into a more
useable form.
𝜔
1 1 + 4𝜔1𝑜
2 2
𝐾𝐸𝐿 = 𝑚𝐿 𝑙2 (
˙
(8)
𝜔1 )𝜃,𝑥 𝑥
3 1 + 2𝜔
𝑜
To simplify this equation define 𝛽 as,
(
1 1+
𝛽=
3 1+
𝜔1
4𝜔𝑜
𝜔1
2𝜔𝑜
)
then substituted into the kinetic energy equation.
𝐾𝐸𝐿 = 𝑚𝐿 𝑙2 𝛽𝜃,𝑥 2 𝑥˙ 2
(9)
The kinetic energy equations of the system are now all derived, and before the
Euler-Lagrange formula can be used the potential energy of the system needs
to be found.
5
Before the potential energy equation can be found, a geometric relationship
will need to be made to get 𝜃, 𝜙 in terms of 𝑥 and constants. Referring to Figure
2 the following can easily be equated.
𝐷 − 𝑥 = 𝑙 sin(𝜃) + 𝑠 sin(𝜙)
(10)
𝐿 + 𝑙 cos(𝜃) = 𝑠 cos(𝜙)
(11)
At this point it is time to find 𝜙,𝑥 , 𝜃,𝑥 , 𝜙,𝑥𝑥 ,and 𝜃,𝑥𝑥 , which will be used later on in
the differential equation and solving for potential energy. These can be obtained
by equations (10) and (11). In taking the partial derivatives of equations (10)
and (11), it is important to make sure to take the partial of 𝜙 and 𝜃 because
they are both in terms of 𝑥. With the two partial derivatives expressions, and
two unknowns, 𝜙,𝑥 , 𝜃,𝑥 , the partial derivatives equal.
𝜙,𝑥 =
− sin(𝜃)
𝑠 sin(𝜃 + 𝜙)
(12)
𝜃,𝑥 =
− sin(𝜙)
𝑙 sin(𝜃 + 𝜙)
(13)
From the first derivatives the second derivatives can be found.
𝜙,𝑥𝑥 =
−𝑠 sin(𝜃 + 𝜙) cos(𝜃)𝜃,𝑥 + 𝑠 sin(𝜃) cos(𝜃 + 𝜙)(𝜃,𝑥 + 𝜙,𝑥 )
𝑠2 sin2 (𝜃 + 𝜙)
(14)
𝜃,𝑥𝑥 =
−𝑙 sin(𝜃 + 𝜙) cos(𝜙)𝜙,𝑥 + 𝑙 sin(𝜙) cos(𝜃 + 𝜙)(𝜃,𝑥 + 𝜙,𝑥 )
𝑙2 sin2 (𝜃 + 𝜙)
(15)
Equations (14) and (15) have 𝜃,𝑥 and 𝜙,𝑥 in the equations, but these can be
substituted in from equations (12) and (13). Now the potential energy of the
bow can be found.
There is no exact potential energy equation for a bow. So the potential
energy will be assumed to be proportional to the square of the deflection angle,
𝜃. The potential energy will also be assumed to vanish at the brace height;
which is not exactly accurate, because the limbs are still flexed at that point.
In the context of this model though it is an accurate assumptions that should
not effect the outcome, because the energy cannot be used in the system. from
this information the following equation can be derived.
2
𝑃 𝐸 = 𝐶(𝜃2 + 𝜃𝐵
)
(16)
In this equation 𝐶 is a constant. To find 𝐶 the second derivative with respect
to 𝑥 of the potential energy will be taken. By taking the second derivative the
spring constant 𝑘 can be obtained by evaluating the equation at a point. The
point will be at 𝑥𝑜 which is the point where the bow is at brace height. In other
2
(𝑃 𝐸)
at 𝑥 = 𝑥𝑜 . By taking the second derivative the following
words 𝑘 = 𝑑 𝑑𝑥
2
equation is obtained.
𝑑2 (𝑃 𝐸)
2
= 2𝐶(𝜃𝜃,𝑥𝑥 + 𝜃,𝑥
)
𝑑𝑥2
6
At the point 𝑥 = 𝑥𝑜 , 𝜃 = 𝜃𝑜 , and 𝜙 = 0. By plugging in and solving for what is
known 𝐶 = 21 𝑘𝑠𝑙 sin(𝜃𝑜 )/𝜃𝑜 . 𝐶 can then be substituted into equation (16) the
potential energy is following equation.
𝑃𝐸 =
sin(𝜃𝑜 ) 2
1
𝑘𝑠𝑙
(𝜃 − 𝜃𝐵 2 )
2
𝜃𝑜
(17)
The energy equations have all been solved for, and the Lagrangian can now be
set up for the system.
3
Set Up of Equation
With the kinetic energy and the potential energy found the differential equation
can now be set up using the Euler-Lagrange formula. The Euler-Lagrange is
a way to take the energy equations of a system and turn them into a system
of differential equations. In our case however, there is just one differential
equation, but the process is the same. By taking some partial derivatives of
the lagrangian, which is shown in equation (19), The differential equation can
be found. First, a simplification will be made to make the equation more user
friendly. 𝐾 will be defined as follows.
1
𝐾 = 𝑚𝑠 (1 + 𝑠𝜙,𝑥 cos(𝜙) + 𝑠2 𝜙,𝑥 2 ) + 2𝑚𝐿 𝛽𝑙2 𝜃,𝑥 2
3
(18)
The Lagrangian can now be written as,
𝐿=
1
(𝑀 + 𝐾)𝑥˙ 2 − 𝑃 𝐸
2
(19)
The Euler-Lagrange formula is now used. First, the partial derivative with
respect to 𝑥 is taken so the equation becomes.
∂𝐿
1
= 𝐾,𝑥 𝑥˙ 2 − 𝑃 𝐸,𝑥
∂𝑥
2
Where 𝑀 is a constant and does not depend on 𝑥, but 𝐾 depends on 𝑥, and so
does 𝑃 𝐸.
The next step is to take the partial derivative with respect to 𝑥˙ and differentiate with respect to time, which gives the following.
𝑑 ∂𝐿
= (𝑀 + 𝐾)¨
𝑥
𝑑𝑡 ∂ 𝑥˙
˙
Where only 𝑥˙ 2 depends on 𝑥.
By setting these two equations equal to each other the governing differential
equation is as follows, according to the Euler-Lagrange formula.
(𝑀 + 𝐾)¨
𝑥=
1
𝐾,𝑥 𝑥˙ 2 − 𝑃 𝐸,𝑥
2
7
(20)
Substitutions can be made to make the differential equation into useable form,
but due to the length of the equation, I will leave it separate. However, equation
(18) can be substituted in for 𝐾, and 𝐾,𝑥 can be substituted in from equation
(21). Which is a matter of taking the partial derivative of 𝐾 with respect to 𝑥.
In taking the partial derivative with respect to 𝑥 one has to be careful because
𝜙 and 𝜃 are defined in terms of 𝑥 so the partial of 𝜙 and 𝜃 needed to be taken,
again.
2
𝐾,𝑥 = 𝑚𝑠 (𝑠𝜙,𝑥𝑥 cos(𝜙) − 𝑠𝜙2,𝑥 sin(𝜙) + 𝑠2 𝜙,𝑥 𝜙,𝑥𝑥 ) + 4𝑚𝐿 𝛽𝑙2 𝜃,𝑥 𝜃,𝑥𝑥
3
(21)
Next the partial derivative of potential energy can be taken. Again, taking into
account 𝜙 and 𝜃 are defined in terms of 𝑥.
𝑃 𝐸,𝑥 = 𝑘𝑠𝑙
sin(𝜃𝑜 )
𝜃𝜃,𝑥
𝜃𝑜
(22)
The differential equation is now set up, but it can be seen solving the equation
will be rather difficult with the equation in terms of three variables. This problem will be resolved in the next section with some geometry already displayed.
4
Solving of Equation
The differential equation being in terms of three variables creates problems for
any way of trying to solve this equation. Also, there are numerous partial
derivatives in the equation, but they have already been found from earlier. By
referring to equations (12), (13), (14), and (15) they can be substituted into the
differential equation.
This takes care of all partial derivatives, but the problem of having the equation in three different variables is still there. I chose to solve for sin(𝜃), sin(𝜙), cos(𝜃),
and cos(𝜙). To solve for these four terms from equations (11) and (12) which are
𝐷−𝑥 = 𝑙 sin(𝜃)+𝑠 sin(𝜙), 𝐿+𝑙 cos(𝜃) = 𝑠 cos(𝜙), and knowing sin2 (𝜃)+cos2 (𝜃) =
1 and sin2 (𝜙) + cos2 (𝜙) = 1, it takes a not quite so simple algebra step. Results
however, of solving for these terms are as follows. To fit it on the paper, first,
a term has to be defined.
𝑠2 − 𝑥2 + 2𝐷𝑥 − 𝐷2 − 𝐿2 − 𝑙2
2𝐿
√
2𝐿𝐴 + 4𝐿2 𝐴2 − 4(𝐴2 − (𝑥 − 𝐷)2 )(𝐿2 + (𝑥 − 𝐷)2 )
cos(𝜃) =
2(𝐿2 + (𝑥 − 𝐷)2 )
√
2𝐴(𝑥 − 𝐷) + 4𝐴2 (𝑥 − 𝐷)2 − 4(𝐴2 − 𝐿2 )(𝐿2 + (𝑥 − 𝐷)2 )
sin(𝜃) =
2(𝐿2 + (𝑥 − 𝐷)2 )
√
2𝐴(𝑥−𝐷)+ 4𝐴2 (𝑥−𝐷)2 −4(𝐴2 −𝐿2 )(𝐿2 +(𝑥−𝐷)2 )
𝐷 − 𝑥 − 𝑙(
)
2(𝐿2 +(𝑥−𝐷)2 )
sin(𝜙) =
𝑠
𝐴=
8
(23)
(24)
(25)
(26)
√
𝐿 + 𝑙(
cos(𝜙) =
2𝐿𝐴+
4𝐿2 𝐴2 −4(𝐴2 −(𝑥−𝐷)2 )(𝐿2 +(𝑥−𝐷)2 )
)
2(𝐿2 +(𝑥−𝐷)2 )
(27)
𝑠
These equations are not simplified, but how I chose to solve for the differential
equation, simplifying was not needed. At this point it can be seen why I did not
write out the whole equation in complete form, because it would be too long to
comprehend.
Solving for and substituting these equations in, allowed me to put the differential equation into solvable form. I chose to use Matlabs ode45 numerical
solver to obtain a solution. Before this solution can be obtained though, some
parameters need to be defined. These parameters are not from a specific bow,
but rather are values of what an average bow maybe [1].
Parameters of The Bow Arrow System
Mass of Arrow
𝑀 = 25𝑔
Mass of String
𝑀𝑠 = 7𝑔
Half Length of String 𝑠 = 0.85𝑚
Mass of Limbs
𝑀𝑙 = 93𝑔
Length of Limbs
𝑙 = 0.48𝑚
Limb Constant
𝛽 = 0.271
Draw Length
𝐷 = 0.70𝑚
𝑛
Spring Constant
𝑘 = 620 𝑚
𝜃 at Brace Height
𝜃𝑜 = 𝜋/6
Length of Riser
𝐿 = 0.40𝑚
Using these parameters and plugging into my Matlab file,1 Figure 3 shows
results of velocity. Figure 3 places the max velocity as the arrow leaves the bow
around a 48.5𝑚/𝑠 or 159.1𝑓 𝑡/𝑠. Further analysis of the velocity will be carried
out in the next section.
Figure 4 shows the position of the bow stopping at brace height which is
where the arrow leaves the string and the model becomes no longer valid. A
better analysis of this will be shown in the next section.
5
Checking the Model
With the model running and giving reasonable answers it needs to be compared
against an actual bow. To verify how close the model is to real life a person
would need to find a bow and pull the parameters off the specific bow. Then
test the bow by shooting an arrow through a chronograph, and compare the
results with the model. This paper will not go into the testing of this, but will
accept 159.1𝑓 𝑡/𝑠 with roughly a 65𝑙𝑏 draw weight is about were the bow should
be shooting. 2 However, a few other test of the model itself can show it matches
real life. The tests are to change some parameters to see how the model changes.
1 see
appendix for Matlab file
draw weight can be found from the spring constant of the limbs in the parameters.
2 65𝑙𝑏
9
Figure 3: Graph of Solution of Velocity
If the mass of the arrow is decreased most any archer will agree the velocity of
the arrow will increase. If the mass of the arrow is decreased to 𝑀 = 20𝑔, the
velocity of the arrow increases as shown in Figure 5. By just dropping the mass
of the arrow by 5𝑔 the velocity of the arrow increases to approximately 51𝑚/𝑠,
almost 3𝑚/𝑠 difference. This matches a bow in real life shooting a lighter arrow.
It also adds more validity of the model.
Another quick check of the model is to increase the draw weight of the bow.
This will be done by increasing the spring constant from 620𝑁/𝑚 to 700𝑁/𝑚
by making this change the velocity of the arrow should change like it did when
the mass of the arrow was decreased. By changing the spring constant with
all other original parameters, velocity of the bow can be seen in Figure 6. The
velocity was again increased to approximately 51𝑚/𝑠. This goes with the basis
the heavier draw weights on the bow will make the arrow go faster, and the
model follows this basis.
Most archers know the longer the draw length a person has, the faster the
arrow will release from the bow. It may seem strange at first to people who have
never shot a bow before, but after thinking about it, it should make sense. The
arrow flies faster when released because in was on the string a longer distance,
and was accelerated longer. Also, the bow was pulled back further making the
deflection angle greater, and adding more potential energy to the bow. So by
adding a longer draw length in the parameters the bow should shoot an arrow
faster. As seen from Figure 7 the velocity of the arrow was greatly increased
by just adding 5𝑐𝑚 onto the draw length, with all original parameters. The
velocity increased to near 52 𝑚
𝑠 . Which is a drastic increase, but is in the range
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Figure 4: Graph of Solution of Position
Figure 5: Graph of Velocity With Mass of 20𝑔
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𝑁
Figure 6: Graph of Velocity With a spring constant of 700 𝑚
Figure 7: Graph of Velocity With a draw length of 75𝑐𝑚
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Figure 8: Graph of Velocity with Long Time Frame.
of what changing the draw length should be.
Looking at all the figures they all have one thing in common, the time frame
is extremely short. This short time is because the model is only valid from
the time the archer releases the string to the time the arrow leaves the string.
However, it is interesting to know what happens when the model is taken out
to a relatively longer time such as 0.02 seconds.
As it can be seen in Figures 8 and 9 both the position and velocity explode
when they go past the point of the model no longer being valid. This is an
expected result because the position is constantly going forward, and the velocity
is ever increasing.
6
Other Topics of Research
The model gave a basic approximation of the interactions between a bow and
arrow. It however takes many assumptions into the model, and leaves a few
questions that need to be answered. Some of these unanswered questions are
hand shock, modeling other types of bows, and what happens when the arrow
leaves the string.
Hand shock is one of the most important aspects of shooting a bow accurately. Most archers say they would rather have a smooth shooting bow than
a fast bow. Hand shock comes from the left over energy stored in the bows
limbs and string as the arrow is fired, which is then transferred to the person
shooting the bow. This model will not take this into account, but here is a brief
explanation of how to take this into account. The easiest way would be to do
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Figure 9: Graph of Position with Long Time Frame.
force analysis on the bow arrow system, as the arrow is being shot. The force
of the bow moving forward can be figured by this, but in this model it was not
considered.
In today’s world traditional archery is not practiced by many people. The
few who do, are usually competition archers, or those who want to hunt like it
was before technical advances. Most archers, however, use compound bows, and
it does not apply to their shooting. This model can be applied to a compound
bow with a few additions. The actual bow model is accurate for a compound
bow, without the cams. The assumptions that are made actually follow the
compound bows movement just as accurately if not more. String stretch is still
minimal with the new bow strings. Compound bows are less noisy so energy
lost to noise is less. The limbs move a smaller distance so the assumption of the
hinge point matches the bow better. The changes to make the model valid for a
compound bow comes from adding the rotational effects of the cams[5]. By the
addition of the cams the model gets a lot more complicated and goes beyond
what time was allowed for this model.
When the arrow leaves the string of the bow a totally different model needs
to be established. The arrow will have the affect of gravity on it, air resistance
becomes a greater factor, a driving force is no longer applied to the arrow, and
the archers paradox has a greater affect on the arrow. All of these factors will
need to be taken into account to thoroughly model the flight of an arrow after
it leaves the bow. At the point at which the arrow leaves the bow string this
new model of the arrow trajectory will have to take over. This paper however,
just deals with the time frame of the bow interacting with the arrow.
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These other topics of research can be determined, but the model above was
not meant to discuss this. These are just mere suggestion to make a complete
archery model, or to expand the model to match other types of bows.
7
Conclusion
This model has given a basic approximation of initial velocities and positions
of an arrow as it is being shot. At the point the arrow hits the brace height
the model is no longer valid, this is due to basing the model off of conservation
of energy. The model took a different approach in modeling a bow and arrow
by using the Euler-Lagrange formula which takes the energies of the system,
and then turns the energy equations of the system into a differential equation.
This model assumed many things to simplify it in to a form the Euler Lagrange
Formula could then be used. These assumptions make the model not completely
accurate, but if the graphs are analyzed in Figures 3 and 4 the curves match
what the curves of an arrow being shot should be thought to be. So more likely
then not, the assumptions may just put the model off by a scaling factor. In
this model much was learned, and it seems to somewhat closely matches real
life.
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8
Appendix
global m m_s s m_l l B D K_l th_o L
m=0.025;
m_s=0.007;
s=0.85;
m_l=0.093;
l=0.48;
B=0.271;
D=0.75;
K_l=620;
th_o=pi/6;
L=0.40;
[t,x]=ode45(@F,[0,1],[0,0]);
i=min(find(x(:,1)>(D-l*sin(th_o))));
tr=t(1:i)
xr=x(1:i,:)
plot(tr,xr(:,1))
%plot(t,x(:,1))
xlabel(’Time (seconds)’)
ylabel(’Position (Meters)’)
figure
%plot(t,x(:,2),’r’)
plot(tr,xr(:,2),’r’)
xlabel(’Time (seconds)’)
ylabel(’Velocity (Meters per Second)’)function xprime=F(t,x)
function xprime=F(t,x)
global m m_s s m_l l B D K_l th_o L
A=(sˆ2-x(1).ˆ2+2*D*x(1)-Dˆ2-Lˆ2-lˆ2)/(2*L);
% define sin(theta) to equal the following
s_th=(2*A*(x(1)-D)+sqrt(4*Aˆ2*(x(1)-D).ˆ2-4*(Aˆ2-Lˆ2)*(Lˆ2+(x(1)-D).ˆ2)))
/(2*(Lˆ2+(x(1)-D).ˆ2));
% define cos(theta) to equal the following
c_th=(2*A*L+sqrt(4*Aˆ2*Lˆ2-4*(Aˆ2-(x(1)-D).ˆ2)*(Lˆ2+(x(1)-D).ˆ2)))
/(2*(Lˆ2+(x(1)-D).ˆ2));
% define sin(phi) to equal the following
s_phi=(D-x(1)-l*s_th)/s;
% define cos(phi) to equal the following
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c_phi=(L+l*c_th)/s;
%define phi_x to equal following
phi_x=-s_th/(s*(s_th*c_phi+s_phi*c_th));
%define theta_x to equal following
th_x=-s_phi/(l*(s_th*c_phi+s_phi*c_th));
%define phi_xx to equal following
phi_xx=(-s*(s_th*c_phi+c_th*s_phi)*c_th*th_x+s*s_th*(c_th*c_phi*th_x-s_th*s_phi*phi_xs_th*s_phi*th_x+c_th*c_phi*phi_x))/(sˆ2*(s_th*c_phi+c_th*s_phi)ˆ2);
%define theta_xx to equal following
th_xx=(-l*(s_th*c_phi+c_th*s_phi)*c_phi*phi_x+l*s_phi*(c_th*c_phi*th_x-s_th*s_phi*phi_xs_th*s_phi*th_x+c_th*c_phi*phi_x))/(lˆ2*(s_th*c_phi+c_th*s_phi)ˆ2);
% define K to equal the following
K=m_s*(1+s*phi_x*c_phi+1/3*sˆ2*phi_xˆ2)+2*m_l*B*lˆ2*th_xˆ2;
%define K_x to equal the following
K_x=m_s*(s*c_phi*phi_xx-s*phi_x*s_phi*phi_x+2/3*sˆ2*phi_x*phi_xx)+4*m_l*B*lˆ2*th_x*th_xx;
xprime=zeros(2,1);
xprime(1)=x(2);
xprime(2)=(1/2*K_x*x(2).ˆ2+K_l*s*sin(th_o)/(th_o)*(asin(s_th)*s_phi/
(s_th*c_phi+s_phi*c_th)))/(m+K);
References
[1] W. C. Marlow. Bow and Arrow Dynamics.
[2] B. W. Kooi. and J. A. Sparenberg. On the Mechanics of The Arrow:
Archer’s Paradox.
[3] wikipedia.org. Euler-Lagrange Equation.
[4] B.W. Kooi. On the Mechanics of the Modern Working-Recurve Bow.
[5] J. L. Parker. A Compound archery bow dynamic model, suggesting modifications to improve accuracy.
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