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The Physics of Archery (1)

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The Physics of Archery (1). Objectives. To Understand the Basic Physical Principles of Archery Through Identifying: Energy Transfers Energy Storage Trajectories. Bow Anatomy. Riser/Handle. Limbs. Grip. String. Energy Transfer. Procedure Hold up bow and put arrow on string

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slide2

Objectives

  • To Understand the Basic Physical Principles of Archery Through Identifying:
  • Energy Transfers
  • Energy Storage
  • Trajectories
slide3

Bow Anatomy

Riser/Handle

Limbs

Grip

String

slide4

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!!!)

TASK 1: Identify the stages in this energy transfer. Draw a Sankey diagram to show this.

slide5

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

slide6

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

slide7

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

0

Task 2: Calculate the energy stored in 165N bow drawn to 70cm

slide8

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

Distance (cm)

20

35

0

Task 2: Calculate the energy stored in 165N bow drawn to 70cm

slide9

Energy Storage (solution)

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 = 57.75 J

Draw length (cm)

70

0

slide10

Arrow Energy

Nock

Shaft

Fletchings

Point

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

slide11

Arrow Energy (solution)

Nock

Shaft

Fletchings

Point

Kinetic energy = 0.70 X work donebow

Kinetic energy = ½ mass X velocity2

Therefore velocity = √(2 X kinetic energy/mass)

Velocity = √(2 X 40.425 / 0.025) = √ 3234 = 56.87 ms-1

Mass = 25g

Work done = 57.75J

Efficiency = 0.7

slide12

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

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.

slide13

Trajectories (solution)

  • Split the component into vertical & horizontal:
    • v = 56.87 ms-1 for maximum range, θ = 45O
    • vh = v cos θ = 56.87 sin 45O = 40.21 ms-1
    • vv = v sin θ = 56.87 cos 45O = 40.21 ms-1
  • Taking vertical component first up to highest point:
    • u = 40.21 ms-1 a = g = -9.81 ms-2
    • v2 = u2 + 2as
    • 0 = 40.212 – 2 X 9.81 X s
    • Therefore s = 40.212 / (2 X 9.81)
    • Maximum height = 82.4m

height

θ

distance

slide14

Trajectories (solution)

    • v = u + atup
    • 0 = 40.21 - 9.81 X tup
    • Therefore tup = 40.21 / 9.81 = 4.10 s
    • Therefore tflight = 8.20 s
  • Taking the horizontal component:
    • velocity = 40.21 ms-1 time = 8.20 s
    • velocity = distance / time
    • Therefore distance = velocity X time
    • Max range = 40.21 X 8.20 = 329.7 m

height

θ

distance

slide15

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

Top view

slide16

Can humans dodge arrows?

  • Time taken for human target to dodge:
  • d2 = 32 + 0.52
  • d = 3.04 m
  • tmove = 3.04 / 10 = 0.304 s
  • tdodge = trealise + treact + tmove = 1 + 0.25 + 0.304 = 1.554 s
  • So we calculate the distance at which tflight = 1.554s
  • tup = tflight / 2 = 0.777 s
  • v = u + atup
  • u = v – atup = 0 + (9.81 X 0.777) = 7.62 ms-1 = vv
  • vv = v sin θ v = 56.87 ms-1
  • sin θ = vv / v = 7.62 / 56.87 = 0.13 therefore θ = 7.7 0
  • vh = v cos θ = 56.87 cos 7.7 = 56.36 ms-1
  • vh = d / tflight
  • s = vh X tflight = 56.36 X 1.554 = 87.58 m
  • So the human target would need to be at least 87.58 m away from the archer in order to dodge the arrow.
slide17

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
slide18

The Physics of Archery (2)

Photos from the Archery Have A Go Here!!!

slide19

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
slide20

Different Types of Bow

Longbow Crossbow Recurve Compound

slide21

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

150lbs @ 28”

12 arrows/min/archer

0.70

France

20,000-30,000 men

8000 archers

Using crossbows

75g

300lbs @ 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

slide22

Conversion Rates & Useful Formulae

1 lb = 0.45 kg

1 inch (“) = 0.025 m

Area of triangle = ½(base X height)

v = u + at s = ½ (v + u)t

k.e. = ½ mv2 v = d / t

slide23

English Longbow Range

Energy stored in bow:

Conversion: 150lb = 9.81 X 0.45 X 150 = 662.2 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 = 6489.3

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

slide24

English Longbow Range

Time taken to reach highest point:

v = u + atup

tup = (v – u) / a = 57 / 9.81 = 5.81 s

tflight = 11.62 s

Maximum range of longbow:

vh = d / tflight

d = vh X tflight = 57 X 11.62 = 662.4m

slide25

French Crossbow Range

Energy stored in bow:

Conversion: 300lb = 9.81 X 0.45 X 300 = 1324.35 N

16” = 0.4m

Work done = ½ (1324.35 X 0.4) = 264.9J

Velocity of arrow on release:

k.e. = ½ m v2

v2 = k.e. / ½ m = 0.6 X 264.9 / ½ 0.075 = 4237.9

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

slide26

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 = 431.5 m

slide27

Why did the English Win?

Terrain

Sited in a narrowing valley

Muddy rainy conditions

Timing

Hours waiting

Frequency of arrows

Class/Tradition/Organisation

French archers pushed backwards by nobility

French disorganised

Position

English archers on flanks

French multiple lines, archers behind front line

Protection

Armour

Pikes in ground

Equipment

Longer range

Greater frequency

slide29

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
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