The Physics of Archery (1)

ObjectivesTo Understand the Basic Physical Principles of Archery Through Identifying:

• Energy Transfers

• Energy Storage

• Trajectories

Bow Anatomy

Limbs

Riser/Handle

String

Grip

Energy TransferProcedure

1. Hold up bow and put arrow on string

2. Place fingers on string and pull string back

3. Anchor string and hand under the chin

4. Take aim

5. Release the string

6. Arrow hits target (hopefully!!!)

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

Energy Transfer (solution)Procedure

1. Hold up bow and put arrow on string

2. Place fingers on string and pull string back

3. Anchor string and hand under the chin

4. Take aim

5. Release the string

6. 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

Chemical in arm

Energy Transfer (solution)

Kinetic in arm, string, & limbs

Sound in limbs; heat in arms; heat in limbs & string

Elastic potential in limbs

Kinetic in arrow

Heat and sound in target

Kinetic in string; sound in limbs and string; heat in limbs

Heat and sound of arrow in flight

Sankey Diagram Showing Losses

70

Energy StorageGraph to show draw force against draw lengthDraw

force (N)

Draw length (cm)

0

165

Work Done = Force (constant)

X Displacement in direction of force

Work Done = area under the line

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

20

Energy StorageGraph to show force against distance

Force (N)

Distance (cm)0

400

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!

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

200

35

70

Energy Storage (solution)Graph to show draw force against draw lengthDraw

force (N)

Draw length (cm)

0

165

Force = 165 N

Distance = 70cm

Work Done = ½ 165 * 0.7

Work Done = 57.75 J

Arrow Energy

Task 3: Calculate the velocity of the arrow (mass 25g), assuming efficiency of energy transfer of limbs to arrow 0.70

Kinetic energy = ½ mass X velocity2

Fletchings

NockShaft

Point

Arrow Energy (solution)

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

Fletchings

NockShaft

Point

Mass = 25g

Work done = 57.75J

Efficiency = 0.7

TrajectoriesParabolic 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

θ

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.

height

tdistance

t

distance

height

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 + 2as0 = 40.212 – 2 X 9.81 X sTherefore s = 40.212 / (2 X 9.81) Maximum height = 82.4m

θ

height

distance

Trajectories (solution)v = u + atup

0 = 40.21 - 9.81 X tup

Therefore tup = 40.21 / 9.81 = 4.10 sTherefore tflight = 8.20 s

Taking the horizontal component:velocity = 40.21 ms-1 time = 8.20 svelocity = distance / time Therefore distance = velocity X timeMax range = 40.21 X 8.20 = 329.7 m

θ

height

distance

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?

Top view

Target

3 m

0.5 m

Can humans dodge arrows?Time taken for human target to dodge:

d2 = 32 + 0.52

d = 3.04 mtmove = 3.04 / 10 = 0.304 stdodge = trealise + treact + tmove = 1 + 0.25 + 0.304 = 1.554 s

So we calculate the distance at which tflight = 1.554stup = tflight / 2 = 0.777 sv = 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.

Safety InformationBefore 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

The Physics of Archery (2)

Photos from the Archery Have A Go Here!!!

ObjectivesTo 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

Different Types of Bow

Longbow Crossbow Recurve Compound

The Battle of AgincourtEngland

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 rangeRemember to note down any assumptions you have made

Country:

# of Men:

# of Archers:

Style of Bow:

Mass an Arrow:

Poundage:

Release rate:

Efficiency:

1415

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

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

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

English Longbow 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

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

French Crossbow Range

TerrainSited in a narrowing valley

Muddy rainy conditions

Why did the English Win?

PositionEnglish archers on flanks

French multiple lines, archers behind front line

Class/Tradition/OrganisationFrench archers pushed backwards by nobility

French disorganised

ProtectionArmour

Pikes in ground

TimingHours waiting

Frequency of arrows

EquipmentLonger range

Greater frequency

Why did the English Win?

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.

1. Intro page

2. Equipment Anatomy & How to Shoot

3. Energy Transfers in Archery

4. Trajectories

5. The Battle of Agincourt

Poster & Presentation