Presentation on theme: "The Physics of Archery (1)"— Presentation transcript:
1The Physics of Archery (1) Introduce myselfIntroduce 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 RingsRobin Hood – Prince of ThievesAlison Williamson – Bronze 2004 OlympicsEros – Piccadilly Circus, London,Hank, The Archer – Dungeons and Dragons
2ObjectivesTo Understand the Basic Physical Principles of Archery Through Identifying:Energy TransfersEnergy StorageTrajectoriesOver today’s session we are going to try to build up a basic physical model for archery.
3Bow 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.
4Energy Transfer Procedure Hold up bow and put arrow on string Place fingers on string and pull string backAnchor string and hand under the chinTake aimRelease the stringArrow 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.
5Energy Transfer (solution) ProcedureHold up bow and put arrow on stringPlace fingers on string and pull string backAnchor string and hand under the chinTake aimRelease the stringArrow hits target (hopefully!!!)Main Energy transferChemical in arm to kinetic in arm, string & limbsKinetic in arm & string to elastic potential in limbsElastic potential in limbs to kinetic in string, limbs and arrowKinetic in arrow and sound in limbsKinetic in arrow to heat and sound in targetString is relatively inelastic – it’s the limbs that store the energy
6Energy Transfer (solution) Sankey Diagram Showing LossesKinetic in string; sound in limbs and string; heat in limbsChemical in armKinetic in arm, string, & limbsElastic potential in limbsKinetic in arrowHeat and sound in targetHeat and sound of arrow in flightSound in limbs; heat in arms; heat in limbs & string
7Energy Storage Draw force (N) Graph to show draw force against draw length165Work Done = Force (constant)X Displacement in direction of forceWork Done = area under the lineDraw length (cm)70Explain the bow (yellow) and arrow (purple)Describe the axes and note positive correlationExplain 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
8Task 2: Calculate the energy stored in 165N bow drawn to 70cm Energy StorageGraph to show force against distanceWork Done = Force X Displacement in direction of forceConsider an action that consists of two partspushing a 20 kg block along for 20 cmpushing 2 20kg blocks along for 15 cmArea of rectangle = height X lengthAdd the shaded boxes together!Force (N)400200Explain 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)2035Task 2: Calculate the energy stored in 165N bow drawn to 70cm
9Energy Storage (solution) Graph to show draw force against draw length Draw force (N)Graph to show draw force against draw length165Force = 165 NDistance = 70cmWork Done = ½ 165 * 0.7Work Done = JDraw length (cm)70
10Kinetic energy = ½ mass X velocity2 Arrow EnergyNockShaftFletchingsPointExplain the different parts of the arrow, and what they are forPast round an arrow so students get an idea of how it looks for realField any questions about the arrowKinetic energy = ½ mass X velocity2Task 3: Calculate the velocity of the arrow (mass 25g), assuming efficiency of energy transfer of limbs to arrow 0.70
11Arrow Energy (solution) NockShaftFletchingsPointMass = 25gWork done = 57.75JEfficiency = 0.7Kinetic energy = 0.70 X work donebowKinetic energy = ½ mass X velocity2Therefore velocity = √(2 X kinetic energy/mass)Velocity = √(2 X / 0.025) = √ 3234 = ms-1Relate the speed to real life – ~145mph. As fast as the top speed eurostar in UK.
12Trajectories 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 θv = u + at v2 = u2 + 2asv = d / theighttdistanceheighttExplain that the shape of flight is parabolic & symmetricalExplain that the vertical and horizontal components are independentUse the small graphs on left to explain the changes over time of the two componentsDraw the angles on the board with SOH CAH TOA if necessaryDescribe the equationsθdistanceTask 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.
13Trajectories (solution) Split the component into vertical & horizontal:v = ms for maximum range, θ = 45Ovh = v cos θ = sin 45O = ms-1vv = v sin θ = cos 45O = ms-1Taking vertical component first up to highest point:u = ms a = g = ms-2v2 = u2 + 2as0 = – 2 X 9.81 X sTherefore s = / (2 X 9.81)Maximum height = 82.4mheightExplain 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 distanceθdistance
14Trajectories (solution) v = u + atup0 = X tupTherefore tup = / 9.81 = 4.10 sTherefore tflight = 8.20 sTaking the horizontal component:velocity = ms-1 time = 8.20 svelocity = distance / timeTherefore distance = velocity X timeMax range = X 8.20 = mheightSee previous notes slideθdistance
15Can humans dodge arrows? The human target would need to move outside of the area as shownAssume the archer is very accurate.Fastest human travels at 10ms-1Time for the human to realise the arrow is incoming = 1 secondHuman response time 0.25 secondsTask 5: What is the minimum distance the target needs to be before they can successfully dodge an arrow?3 m0.5 mTargetThis is extended work if the other stuff gets done earlier than planned or used as homeworkTop view
16Can humans dodge arrows? Time taken for human target to dodge:d2 =d = 3.04 mtmove = 3.04 / 10 = stdodge = trealise + treact + tmove = = sSo we calculate the distance at which tflight = 1.554stup = tflight / 2 = sv = u + atupu = v – atup = 0 + (9.81 X 0.777) = 7.62 ms-1 = vvvv = v sin θ v = ms-1sin θ = vv / v = 7.62 / = therefore θ = 7.7 0vh = v cos θ = cos 7.7 = ms-1vh = d / tflights = vh X tflight = X = mSo the human target would need to be at least 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
17FOLLOW THE INSTRUCTIONS OF THE COACH AT ALL TIMES 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 lineDo not distract anyone who is shootingOnce on the string, only ever point the arrows in the direction of the targetsIf you are not shooting stay well behind the shooting lineIf 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 targetDon’t draw and then release the bow without an arrow on it (this is called a “dry fire”) as this can damage the bowsFOLLOW THE INSTRUCTIONS OF THE COACH AT ALL TIMESOn the hour we MUST cover this before the Have a Go – also will provide handouts to take home of this
18The Physics of Archery (2) Photos from the Archery Have A Go Here!!!
19ObjectivesTo Reinforce our Understanding of the Basic Principles of Archery by:Looking at a real life application at the Battle of AgincourtCreating 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.
20Different Types of Bow Longbow Crossbow Recurve Compound Will explain the differencesLongbow Crossbow Recurve Compound
21The Battle of Agincourt 1415Country:# of Men:# of Archers:Style of Bow:Mass an Arrow:Poundage:Release rate:Efficiency:England6000 men5000 archersUsing longbows50g28”12 arrows/min/archer0.70France20,000-30,000 men8000 archersUsing crossbows75g16”4 arrows/min/archer0.60Convert the units from imperial to metricCalculate the energy stored in each type of bowCalculate the speed of the arrow on releaseCalculate the maximum rangeRemember to note down any assumptions you have madeCan flick back to the previous slide to show pictures of the longbows and crossbows
22Conversion Rates & Useful Formulae 1 lb = 0.45 kg1 inch (“) = mArea of triangle = ½(base X height)v = u + at s = ½ (v + u)tk.e. = ½ mv2 v = d / t
23English Longbow Range Energy stored in bow: Conversion: 150lb = 9.81 X 0.45 X 150 = N28” = 0.7mWork done = ½ (662.2 X 0.7) = 231.8JVelocity of arrow on release:k.e. = ½ m v2v2 = k.e. / ½ m = 0.7 X 231 / ½ 0.05 =v = 80.6 ms-1Splitting the vertical and horizontal components:vv = v sin θ vh = v cos θvv = 80.6 sin 450 = 57 ms-1 vh = 80.6 cos 450 = 57ms-1
24English Longbow Range Time taken to reach highest point: v = u + atup tup = (v – u) / a = 57 / 9.81 = 5.81 stflight = sMaximum range of longbow:vh = d / tflightd = vh X tflight = 57 X = 662.4m
25French Crossbow Range Energy stored in bow: Conversion: 300lb = 9.81 X 0.45 X 300 = N16” = 0.4mWork done = ½ ( X 0.4) = 264.9JVelocity of arrow on release:k.e. = ½ m v2v2 = k.e. / ½ m = 0.6 X / ½ =v = 65.1 ms-1Splitting the vertical and horizontal components:vv = v sin θ vh = v cos θvv = 65.1 sin 450 = 46 ms-1 vh = 65.1 cos 450 = 46ms-1
26French Crossbow Range Time taken to reach highest point: v = u + atup tup = (v – u) / a = 46 / 9.81 = 4.69 stflight = 9.38 sMaximum range of crossbow:vh = d / tflightd = vh X tflight = 46 X 9.38 = mSo the English longbow range was over 600 m, the French over 400mHowever, 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
27Why did the English Win? Terrain Timing Position Sited in a narrowing valleyMuddy rainy conditionsTimingHours waitingFrequency of arrowsPositionEnglish archers on flanksFrench multiple lines, archers behind front lineClass/Tradition/OrganisationFrench archers pushed backwards by nobilityFrench disorganisedEquipmentLonger rangeGreater frequencyProtectionArmourPikes in ground
29Poster & PresentationDesign 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 pageEquipment Anatomy & How to ShootEnergy Transfers in ArcheryTrajectoriesThe Battle of AgincourtHandouts, 1 per group