1. The Physics of Archery (1)
Introduce myself
Introduce what I’ll be doing (i.e. session now, wed have a go, session after and Friday)
ICE BREAKER - Guessing who the pictures are!!!
Legolas – Lord of the Rings
Robin Hood – Prince of Thieves
Alison Williamson – Bronze 2004 Olympics
Eros – Piccadilly Circus, London,
Hank, The Archer – Dungeons and Dragons

2. Objectives
To Understand the Basic Physical Principles of Archery Through Identifying:
Energy Transfers
Energy Storage
Trajectories
Over today’s session we are going to try to build up a basic physical model for archery.

3. Bow Anatomy Riser/Handle Limbs Grip String
Over today’s session we are going to try to build up a basic physical model for archery. Explain the main parts of a basic bow. Can use my bow as a prop also. Answer any questions about the other parts of equipment if necessary.

4. Energy Transfer Procedure Hold up bow and put arrow on string
Place fingers on string and pull string back
Anchor string and hand under the chin
Take aim
Release the string
Arrow hits target (hopefully!!!)
Ask if anyone has done archery before.
I will bring in my bow and demonstrate how to pull the string back as a visual prop!!
TASK 1: Identify the stages in this energy transfer. Draw a Sankey diagram to show this.

5. Energy Transfer (solution)
Procedure
Hold up bow and put arrow on string
Place fingers on string and pull string back
Anchor string and hand under the chin
Take aim
Release the string
Arrow hits target (hopefully!!!)
Main Energy transfer
Chemical in arm to kinetic in arm, string & limbs
Kinetic in arm & string to elastic potential in limbs
Elastic potential in limbs to kinetic in string, limbs and arrow
Kinetic in arrow and sound in limbs
Kinetic in arrow to heat and sound in target
String is relatively inelastic – it’s the limbs that store the energy

6. Energy Transfer (solution)
Sankey Diagram Showing Losses
Kinetic in string; sound in limbs and string; heat in limbs
Chemical in arm
Kinetic in arm, string, & limbs
Elastic potential in limbs
Kinetic in arrow
Heat and sound in target
Heat and sound of arrow in flight
Sound in limbs; heat in arms; heat in limbs & string

7. Energy Storage Draw force (N)
Graph to show draw force against draw length
165
Work Done = Force (constant)
X Displacement in direction of force
Work Done = area under the line
Draw length (cm) 70
Explain the bow (yellow) and arrow (purple)
Describe the axes and note positive correlation
Explain why it is NOT just max force X max distance. Hence constant. Can explain through drawing multiple rectangles instead of going into calculus (see next slide)
Task 2: Calculate the energy stored in 165N bow drawn to 70cm

8. Task 2: Calculate the energy stored in 165N bow drawn to 70cm
Energy Storage
Graph to show force against distance
Work Done = Force X Displacement in direction of force
Consider an action that consists of two parts
pushing a 20 kg block along for 20 cm
pushing 2 20kg blocks along for 15 cm
Area of rectangle = height X length
Add the shaded boxes together!
Force (N)
400
200
Explain why it is NOT just max force X max distance. Hence constant. Can explain through drawing multiple rectangles instead of going into calculus.
Distance (cm) 20 35
Task 2: Calculate the energy stored in 165N bow drawn to 70cm

9. Energy Storage (solution) Graph to show draw force against draw length
Draw force (N)
Graph to show draw force against draw length
165
Force = 165 N
Distance = 70cm
Work Done = ½ 165 * 0.7
Work Done = J
Draw length (cm) 70

10. Kinetic energy = ½ mass X velocity2
Arrow Energy
Nock
Shaft
Fletchings
Point
Explain the different parts of the arrow, and what they are for
Past round an arrow so students get an idea of how it looks for real
Field any questions about the arrow
Kinetic energy = ½ mass X velocity2
Task 3: Calculate the velocity of the arrow (mass 25g), assuming efficiency of energy transfer of limbs to arrow 0.70

11. Arrow Energy (solution)
Nock
Shaft
Fletchings
Point
Mass = 25g
Work done = 57.75J
Efficiency = 0.7
Kinetic energy = 0.70 X work donebow
Kinetic energy = ½ mass X velocity2
Therefore velocity = √(2 X kinetic energy/mass)
Velocity = √(2 X / 0.025) = √ 3234 = ms-1
Relate the speed to real life – ~145mph. As fast as the top speed eurostar in UK.

12. Trajectories Parabolic shape of arrow flight
Can consider the vertical and horizontal components of the flight separately. Think SOH CAH TOA!!!
vh = v cos θ vv = v sin θ
v = u + at v2 = u2 + 2as
v = d / t
height t
distance
height t
Explain that the shape of flight is parabolic & symmetrical
Explain that the vertical and horizontal components are independent
Use the small graphs on left to explain the changes over time of the two components
Draw the angles on the board with SOH CAH TOA if necessary
Describe the equations θ
distance
Task 4: Split the components of the arrow velocity up and calculate the max range and the max height at that range . Assume air resistance is negligible.

13. Trajectories (solution)
Split the component into vertical & horizontal:
v = ms for maximum range, θ = 45O
vh = v cos θ = sin 45O = ms-1
vv = v sin θ = cos 45O = ms-1
Taking vertical component first up to highest point:
u = ms a = g = ms-2
v2 = u2 + 2as
0 = – 2 X 9.81 X s
Therefore s = / (2 X 9.81)
Maximum height = 82.4m
height
Explain the 4 steps required:
-split the component parts up
-take the vertical part up to highest point to ascertain max height
-ascertain the time taken as this is constant for the flight regardless of whether the vertical or horizontal component is being addressed.
-ascertain the max distance θ
distance

14. Trajectories (solution)
v = u + atup
0 = X tup
Therefore tup = / 9.81 = 4.10 s
Therefore tflight = 8.20 s
Taking the horizontal component:
velocity = ms-1 time = 8.20 s
velocity = distance / time
Therefore distance = velocity X time
Max range = X 8.20 = m
height
See previous notes slide θ
distance

15. Can humans dodge arrows?
The human target would need to move outside of the area as shown
Assume the archer is very accurate.
Fastest human travels at 10ms-1
Time for the human to realise the arrow is incoming = 1 second
Human response time 0.25 seconds
Task 5: What is the minimum distance the target needs to be before they can successfully dodge an arrow?
3 m
0.5 m
Target
This is extended work if the other stuff gets done earlier than planned or used as homework
Top view

16. Can humans dodge arrows?
Time taken for human target to dodge:
d2 =
d = 3.04 m
tmove = 3.04 / 10 = s
tdodge = trealise + treact + tmove = = s
So we calculate the distance at which tflight = 1.554s
tup = tflight / 2 = s
v = u + atup
u = v – atup = 0 + (9.81 X 0.777) = 7.62 ms-1 = vv
vv = v sin θ v = ms-1
sin θ = vv / v = 7.62 / = therefore θ = 7.7 0
vh = v cos θ = cos 7.7 = ms-1
vh = d / tflight
s = vh X tflight = X = m
So the human target would need to be at least m away from the archer in order to dodge the arrow.
This is extended work if the other stuff gets done earlier than planned or used as homework

17. FOLLOW THE INSTRUCTIONS OF THE COACH AT ALL TIMES
Safety Information
Before taking part in archery you need to understand certain safety rules!!!
Do not put the arrow on the string until you are standing on the shooting line
Do not distract anyone who is shooting
Once on the string, only ever point the arrows in the direction of the targets
If you are not shooting stay well behind the shooting line
If you see any possible hazard or danger (e.g. someone is walking behind the targets) then shout the word “FAST”. If you hear the word “FAST”, then do not shoot any arrows under any circumstances.
One whistle means shooting can start, two whistles means that you can collect your arrows from the target
Don’t draw and then release the bow without an arrow on it (this is called a “dry fire”) as this can damage the bows
FOLLOW THE INSTRUCTIONS OF THE COACH AT ALL TIMES
On the hour we MUST cover this before the Have a Go – also will provide handouts to take home of this

18. The Physics of Archery (2) Photos from the Archery Have A Go Here!!!

19. Objectives
To Reinforce our Understanding of the Basic Principles of Archery by:
Looking at a real life application at the Battle of Agincourt
Creating a poster and presenting on an area of what has been learnt
-Look at the specification of the bows on the English and French sides. Decide which has the advantage.
-Look at what really happened at the Battle of Agincourt and how the English won
-Split 5 groups across the 2 sets. Each group to write one page. If they haven’t done websites before, I will create the basic pages with hyperlinks, and they can create the content and source the diagrams using MS FrontPage.
Homework will be to finish off the web pages.

20. Different Types of Bow Longbow Crossbow Recurve Compound
Will explain the differences
Longbow Crossbow Recurve Compound

21. The Battle of Agincourt
1415
Country:
# of Men:
# of Archers:
Style of Bow:
Mass an Arrow:
Poundage:
Release rate:
Efficiency:
England
6000 men
5000 archers
Using longbows
50g
28”
12 arrows/min/archer
0.70
France
20,000-30,000 men
8000 archers
Using crossbows
75g
16”
4 arrows/min/archer
0.60
Convert the units from imperial to metric
Calculate the energy stored in each type of bow
Calculate the speed of the arrow on release
Calculate the maximum range
Remember to note down any assumptions you have made
Can flick back to the previous slide to show pictures of the longbows and crossbows

22. Conversion Rates & Useful Formulae
1 lb = 0.45 kg
1 inch (“) = m
Area of triangle = ½(base X height)
v = u + at s = ½ (v + u)t
k.e. = ½ mv2 v = d / t

23. English Longbow Range Energy stored in bow:
Conversion: 150lb = 9.81 X 0.45 X 150 = N
28” = 0.7m
Work done = ½ (662.2 X 0.7) = 231.8J
Velocity of arrow on release:
k.e. = ½ m v2
v2 = k.e. / ½ m = 0.7 X 231 / ½ 0.05 =
v = 80.6 ms-1
Splitting the vertical and horizontal components:
vv = v sin θ vh = v cos θ
vv = 80.6 sin 450 = 57 ms-1 vh = 80.6 cos 450 = 57ms-1

24. English Longbow Range Time taken to reach highest point: v = u + atup
tup = (v – u) / a = 57 / 9.81 = 5.81 s
tflight = s
Maximum range of longbow:
vh = d / tflight
d = vh X tflight = 57 X = 662.4m

25. French Crossbow Range Energy stored in bow:
Conversion: 300lb = 9.81 X 0.45 X 300 = N
16” = 0.4m
Work done = ½ ( X 0.4) = 264.9J
Velocity of arrow on release:
k.e. = ½ m v2
v2 = k.e. / ½ m = 0.6 X / ½ =
v = 65.1 ms-1
Splitting the vertical and horizontal components:
vv = v sin θ vh = v cos θ
vv = 65.1 sin 450 = 46 ms-1 vh = 65.1 cos 450 = 46ms-1

26. French Crossbow Range Time taken to reach highest point: v = u + atup
tup = (v – u) / a = 46 / 9.81 = 4.69 s
tflight = 9.38 s
Maximum range of crossbow:
vh = d / tflight
d = vh X tflight = 46 X 9.38 = m
So the English longbow range was over 600 m, the French over 400m
However, in order to be accurate, the distance between the warring lines was far shorter (250 yards = ~230m). So how did the English win? – next slide

27. Why did the English Win? Terrain Timing Position
Sited in a narrowing valley
Muddy rainy conditions
Timing
Hours waiting
Frequency of arrows
Position
English archers on flanks
French multiple lines, archers behind front line
Class/Tradition/Organisation
French archers pushed backwards by nobility
French disorganised
Equipment
Longer range
Greater frequency
Protection
Armour
Pikes in ground

28. Why did the English Win?

29. Poster & Presentation
Design a Poster. Presentations to be given at Friday’s lesson.
Split into groups – each responsible for one area.
Poster options: Physical A1 poster. PowerPoint poster. Web page poster.
Intro page
Equipment Anatomy & How to Shoot
Energy Transfers in Archery
Trajectories
The Battle of Agincourt
Handouts, 1 per group