The Physics of Archery (1). Objectives To Understand the Basic Physical Principles of Archery...
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Transcript of The Physics of Archery (1). Objectives To Understand the Basic Physical Principles of Archery...
![Page 1: The Physics of Archery (1). Objectives To Understand the Basic Physical Principles of Archery Through Identifying: Energy Transfers Energy Storage Trajectories.](https://reader033.fdocuments.in/reader033/viewer/2022051614/551af2115503462e578b4a18/html5/thumbnails/1.jpg)
The Physics of Archery (1)
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ObjectivesTo Understand the Basic Physical Principles of Archery Through Identifying:
• Energy Transfers
• Energy Storage
• Trajectories
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Bow Anatomy
Limbs
Riser/Handle
String
Grip
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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.
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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
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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
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70
Energy StorageGraph to show draw force against draw length
Draw 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
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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
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70
Energy Storage (solution)Graph to show draw force against draw length
Draw force (N)
Draw length (cm)
0
165
Force = 165 N
Distance = 70cm
Work Done = ½ 165 * 0.7
Work Done = 57.75 J
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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
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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
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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
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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
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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
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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
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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.
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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
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The Physics of Archery (2)
Photos from the Archery Have A Go Here!!!
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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
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Different Types of Bow
Longbow Crossbow Recurve Compound
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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 range
Remember to note down any assumptions you have made
Country:
# of Men:
# of Archers:
Style of Bow:
Mass an Arrow:
Poundage:
Release rate:
Efficiency:
1415
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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
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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
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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
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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
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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
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Terrain
Sited in a narrowing valley
Muddy rainy conditions
Why did the English Win?
Position
English archers on flanks
French multiple lines, archers behind front line
Class/Tradition/Organisation
French archers pushed backwards by nobility
French disorganised
Protection
Armour
Pikes in ground
Timing
Hours waiting
Frequency of arrows
Equipment
Longer range
Greater frequency
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Why did the English Win?
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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