1 / 29

The Physics of Archery (1)

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

sherlock_clovis
Download Presentation

The Physics of Archery (1)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Physics of Archery (1)

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

  3. Bow Anatomy Riser/Handle Limbs Grip String

  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!!!) 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

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

  8. 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

  9. 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

  10. 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

  11. 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

  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 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 = 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

  14. 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

  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 Top view

  16. 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.

  17. 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

  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

  20. Different Types of Bow 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 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

  22. 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

  23. 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

  24. 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

  25. 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

  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 = 431.5 m

  27. 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

  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

More Related