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 Medal @ 2004 Olympics Eros – Piccadilly Circus, London, Hank, The Archer – Dungeons and Dragons

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.

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.

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.

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

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

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

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

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 = 57.75 J Draw length (cm) 70

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

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 40.425 / 0.025) = √ 3234 = 56.87 ms-1 Relate the speed to real life – ~145mph. As fast as the top speed eurostar in UK.

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.

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 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

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 See previous notes slide θ 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? 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

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. This is extended work if the other stuff gets done earlier than planned or used as homework

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

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

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.

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

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 Can flick back to the previous slide to show pictures of the longbows and crossbows

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

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

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

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

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 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

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

Why did the English Win?

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