KingofBattle:TheCannonDavid C.Arney,Charles ClarkIntroductionInthis section,wediscusssomeofthebasicconsiderations infiringartilleryrounds:themotionoftheprojectileandtherecoil effects ontheweaponsystem.Themodeldevelopmentforthesetwoconsiderations isdonethroughahistoricalcontext.Inessence,"thecannonproblem"hasbeenpartoftheWestPointmathematics curriculumfor centuries.Theseartillery issues were studied atWest Point byofficer candidates beforetheMilitaryAcademy was established atWestPoint in1802.GeorgeBaron,aprofessorofmathematics,taughtartilleriststhemathematicsofprojectilemotion,theproperplacementoftheweapontohandlerecoil,andotherbasicmathematicsasearlyas18o0atWestPoint.HistoricalSettingOn1July1824,a cadet,whomanyyearslaterwouldproducea controversialrifled cannon,graduatedfromtheUnited StatesMilitaryAcademy.RobertParkerParrott(1804-1877)wascommissionedupongraduationasaSecondLieutenant of theArtillery.After atour of teaching attheAcademyanda tour ofgarrison duty,hechangedhisbranchto Ordnance and servedas an inspector ofcannon duringtheperiod1834-35.In1836heresignedhiscommissionafterreaching the rank of Captain.From1836until1867heservedassuperintendentof theWestPointFoundry,also called theCold SpringFoundry,locatedat ColdSpring,NewYork ontheeastsideoftheHudsonRiveraboutonehalf milenorthof WestPoint.TheWest PointFoundrybecamesucha leadingproducerofordnanceforboththeArmyandtheNavythattheinitialsWPFandRPP,whichwereetchedonthemuzzleofParrottcannons,werewellknownthroughoutbothservices.Parrott rifles sawextensive use in field and siege as well as aboard shipthroughout the Civil War. They were praised. They were cursed. Some of thesmaller calibers weren't upto the standard of competing rifles and larger modelshad an unfortunatetendencytoburst.But considered collectively,as a rifledweaponssystem,theParrottwashardtobeat.Itwaseasytooperatebyinexperienced cannoneers.It was anextremelytough weapon.Itwasinexpensivetomanufactureandcouldbeproducedquicklyandinquantity.TheParrott riflebecame so renowned during the Civil Warthat someone suggestedtheparrot should replace the eagleas the emblematic bird of state.Inthisdiscussionofartillery,wewill firstlookatprojectilemotion,then wewillanalyze the effect of firing on thegun system itself.71
71 King of Battle: The Cannon David C. Arney, Charles Clark Introduction In this section, we discuss some of the basic considerations in firing artillery rounds: the motion of the projectile and the recoil effects on the weapon system. The model development for these two considerations is done through a historical context. In essence, “the cannon problem” has been part of the West Point mathematics curriculum for centuries. These artillery issues were studied at West Point by officer candidates before the Military Academy was established at West Point in 1802. George Baron, a professor of mathematics, taught artillerists the mathematics of projectile motion, the proper placement of the weapon to handle recoil, and other basic mathematics as early as 1800 at West Point. Historical Setting On 1 July 1824, a cadet, who many years later would produce a controversial rifled cannon, graduated from the United States Military Academy. Robert Parker Parrott (1804-1877) was commissioned upon graduation as a Second Lieutenant of the Artillery. After a tour of teaching at the Academy and a tour of garrison duty, he changed his branch to Ordnance and served as an inspector of cannon during the period 1834-35. In 1836 he resigned his commission after reaching the rank of Captain. From 1836 until 1867 he served as superintendent of the West Point Foundry, also called the Cold Spring Foundry, located at Cold Spring, New York on the east side of the Hudson River about one half mile north of West Point. The West Point Foundry became such a leading producer of ordnance for both the Army and the Navy that the initials WPF and RPP, which were etched on the muzzle of Parrott cannons, were well known throughout both services. Parrott rifles saw extensive use in field and siege as well as aboard ship throughout the Civil War. They were praised. They were cursed. Some of the smaller calibers weren't up to the standard of competing rifles and larger models had an unfortunate tendency to burst. But considered collectively, as a rifled weapons system, the Parrott was hard to beat. It was easy to operate by inexperienced cannoneers. It was an extremely tough weapon. It was inexpensive to manufacture and could be produced quickly and in quantity. The Parrott rifle became so renowned during the Civil War that someone suggested the parrot should replace the eagle as the emblematic bird of state. In this discussion of artillery, we will first look at projectile motion, then we will analyze the effect of firing on the gun system itself
ProjectileMotionInordertoderiveamodel forprojectilemotion,wefirstassumethattheprojectilebehaveslikeaparticlemovinginaverticalplane(2dimensional)andthattheonlyforceacting ontheprojectileduringflight is theforceof gravity,which pointsstraight down.Withtheseassumptions,we areignoring many other effects onthemotionoftheprojectile:thegroundmovingastheearthturns,theairfrictionand wind effectsontheprojectile'smotion,andtheforceofgravitychangesastheprojectilechangesaltitude.Later,we can consider some of these effects.Let's assume that the projectile is launched from the origin at time t = O with aninitial velocity vector of vo, the first component is the the horizontal or x directionwithunitvectoriandthesecondcomponentisintheverticalorydirectionwithunit vector j. If is the angle that vmakes with the horizontal axis, then(1)。=ocoso7+|vosingj.Fromourassumptions,theprojectile's initial positionis(2)F=0i+o.Since we assume that the only force acting on the projectile during its flightcomesfromathedownwardaccelerationofgravityofmagnitudeg,wecanuseNewton's second lawtoobtaind?r(3)dt3=-gjWe find the projectile's velocity or position by solving this second orderdifferential equation (3)with the initial conditions:Equation (2)and v(0)=v。:The solution can be found through integration. We integrate (3) to obtaindr=(-gt)/ +vdtAnotherintegrationstepproduces(4)F(t)= x(t)i + y(0)j =1/2 gt? j +vot + rWe substitute the values of and from Equations (1) and (2) into (4) toobtain(5)F=(v|cose)ti+(/v/sinet-1/2gt)jor by component, x(t)=(/v/coso)t andy(t)=(/y/sin@ t-1/2gt)If wemeasure time in secondsanddistanceinmeters,g is9.8m/secand ifwemeasuretime in seconds and distance infeet,g is32ft/secExample1:Let'sseewhathappensforaspecificexample.Ifaprojectileisfired from the originoverhorizontal ground at an initial speedof 500m/sec at anangleofelevationof50degrees,let'sfindtheprojectile'slocationatt=20.We72
72 Projectile Motion In order to derive a model for projectile motion, we first assume that the projectile behaves like a particle moving in a vertical plane (2 dimensional) and that the only force acting on the projectile during flight is the force of gravity, which points straight down. With these assumptions, we are ignoring many other effects on the motion of the projectile: the ground moving as the earth turns, the air friction and wind effects on the projectile's motion, and the force of gravity changes as the projectile changes altitude. Later, we can consider some of these effects. Let’s assume that the projectile is launched from the origin at time t = 0 with an initial velocity vector of 0 v r , the first component is the the horizontal or x direction with unit vector i r and the second component is in the vertical or y direction with unit vector j r . If q is the angle that 0 v r makes with the horizontal axis, then v v i v j r r r 0 = 0 cosq + 0 sinq . (1) From our assumptions, the projectile's initial position is r i j r r r 0 0 0 = + . (2) Since we assume that the only force acting on the projectile during its flight comes from a the downward acceleration of gravity of magnitude g, we can use Newton’s second law to obtain g j dt d r r r = - 2 2 (3) We find the projectile's velocity or position by solving this second order differential equation (3) with the initial conditions: Equation (2) and 0 v(0) v r r = . The solution can be found through integration. We integrate (3) to obtain 0 ( gt) j v dt dr r r r = - + Another integration step produces 0 0 2 r(t) x(t)i y(t) j 1/ 2 gt j v t r r r r r r r = + = + + (4) We substitute the values of 0 v r and 0 r r from Equations (1) and (2) into (4) to obtain r v t i v t gt j r r r (| |cos ) (| | sin 1/ 2 ) 2 0 0 = q + q - , (5) or by component, x(t) (| v |cos )t = 0 q and ( ) (| | sin 1/ 2 ) 2 0 y t = v q t - gt . If we measure time in seconds and distance in meters, g is 9.8 m/sec2 and if we measure time in seconds and distance in feet, g is 32 ft/sec2 . Example 1: Let’s see what happens for a specific example. If a projectile is fired from the origin over horizontal ground at an initial speed of 500 m/sec at an angle of elevation of 50 degrees, let’s find the projectile’s location at t=20. We
can use Equation (4)with /y=500,=50(元 /180)radians,g=9.8,to find theprojectile's coordinates.Thetwo equationsare:x(t) = 500cos(50元/0元/80)1and(0)=500sin(50元/元/80)1-4.9t2(6)Therefore,whent=20seconds,theprojectileis5700metershigh(ycomponent)and6428metersdown-range (xcomponent).Tofind when and where theprojectile lands when it is fired overhorizontalground,we set the vertical position (/y/sin t-1/2gt) equal to zero fromEquation (5) and solve fort. These roots are t=0 or t = 2vo sine/ .Sincet=0is the time the projectile is fred, the other root, I = 2y, sin / must be the timewhentheprojectilestrikestheground.Forourexample,thisvalueis t=78.16seconds.Tofindtheprojectile'srangeRthedistancefromtheorigintothepointof impactonhorizontal ground,wefindthevalueofx(t)when t=78.16.Forour example, R = 500 cos(50元n/)7816=25120 meters.Becarefultonote thatwearenotclaimingthis ismaximumrangeofthisweaponsystem.Forprojectileswithnoairresistance,themaximumrangeisobtainedforaspecifiedinitial velocity byfiring theprojectile at 45 degrees.Throughfurther analysis and calculation with Equation (5),we can find the highestpointofthisprojectile.Thisoccurswhenitsverticalvelocitycomponentiszero,thatiswhen /yIsin-gt=O.For ourexample, the valueof twhentheprojectile is at itshighest is t=39.08 seconds.At that time the trajectory is H=7,485 meters high.Nowlet's lookattheshapeof thetrajectoryofthisprojectile.IfweeliminatetandcombinethetwoequationsforxandyinEguation5,weobtaintheeguationg+(tane)x.(2v cos 0We see that this equation,for given values of g, ,and yo,is a parabola in xand y.Therefore,thetrajectoryofthis ideal projectile(noairresistance)isparabolic.This ideal situation isn't always accurate enough for predictingtrajectoriesofrealprojectiles,solet'srefinethemodeltoconsiderairresistanceAirResistanceAir resistance on moving projectiles can depend on many variables: the velocityoftheprojectile,the projectile's surface area,the relative densities ofthefluid73
73 can use Equation (4) with | | 500 v0 = , q = 50(p/180) radians, g = 9.8, to find the projectile's coordinates. The two equations are: x t ) t 180 50 ( ) = 500cos( p and 2 ) 4.9 180 50 y(t) = 500sin( p t - t . (6) Therefore, when t=20 seconds, the projectile is 5700 meters high (y component) and 6428 meters down-range (x component). To find when and where the projectile lands when it is fired over horizontal ground, we set the vertical position (| | sin 1/ 2 ) 2 0 v q t - gt equal to zero from Equation (5) and solve for t. These roots are t=0 or g v t 2 0 sinq = . Since t=0 is the time the projectile is fired, the other root, g v t 2 0 sinq = must be the time when the projectile strikes the ground. For our example, this value is t=78.16 seconds. To find the projectile's range R, the distance from the origin to the point of impact on horizontal ground, we find the value of x(t) when t=78.16. For our example, ) 78.16 25,120 180 50 R = 500 cos( p = meters. Be careful to note that we are not claiming this is maximum range of this weapon system. For projectiles with no air resistance, the maximum range is obtained for a specified initial velocity by firing the projectile at 45 degrees. Through further analysis and calculation with Equation (5), we can find the highest point of this projectile. This occurs when its vertical velocity component is zero, that is, when | | sin 0 v0 q - gt = . For our example, the value of t when the projectile is at its highest is t=39.08 seconds. At that time the trajectory is H=7,485 meters high. Now let’s look at the shape of the trajectory of this projectile. If we eliminate t and combine the two equations for x and y in Equation 5, we obtain the equation x x v g y (tan ) 2 cos 2 2 2 0 q q + ÷ ÷ ø ö ç ç è æ = - . We see that this equation, for given values of g, q , and 0 v , is a parabola in x and y. Therefore, the trajectory of this ideal projectile (no air resistance) is parabolic. This ideal situation isn’t always accurate enough for predicting trajectories of real projectiles, so let’s refine the model to consider air resistance. Air Resistance Air resistance on moving projectiles can depend on many variables: the velocity of the projectile, the projectile’s surface area, the relative densities of the fluid
andtheprojectile,theprojectile'sshapeandsmoothness,andthefluidcompressibility.Atslowspeedsthedominanteffectprovidingresistancetomotionisthe slidingofthefluid overtheprojectile's surface (viscosity).Attheseslowspeeds,weassumethenumberofmolecules contactingthesurfaceperunit of time is proportional totheproduct of the projectile'svelocityand itssurface area.Therefore.asimplemodelforthe resistiveforce duetofrictionistheproportionalityF,=-ky,wherekisthe constantof proportionalityandvthe velocity of the projectile. The model assumes factors such as surface areaandsurfacesmoothnessareconstant.Athigherspeedsanotherair-resistancemodel maybemoreaccurate.Asthevelocityincreases,morecollisionsoftheprojectilewiththefluid'smoleculestakeplace.Eventually the effect of the collisions comestodominatethe"sliding"effect predominant atlow speeds.Themodel often used in this case andobservedexperimentallyisthattheresistiveforceFisproportionaltothesguareof the velocity.This model is written as F, =-ky?,where as before, kis theconstantofproportionalityandvthevelocityoftheprojectile.Thismakestheequationnonlinearand,therefore,moredifficulttosolveanalytically.However,graphical and numerical solution techniques may be used for the nonlinearequation case, and some of thesemethods willbe investigated later on in thetext.ProjectileMotion withViscousFrictionLet'sdevelopa modelfora projectile moving at lowvelocities,whereviscosityisthedominantresistance andgravityis still the onlyotherforce.All otherforcesare neglected. Using Newton's second law we have, ZF = F, + F, = ma.Making substitutions for a projectile with position F(t)= x(t)7 + y(t) j, we getd'r-mgj-hdrWefurther simplify this equation by writing an equationdt3dtforboththexand y componentsof motion.Ourequations are:y+m-mgdt?dt(7)d?xdx0mdtdtThese two uncoupled,linear equations can be solved separately using varioustechniques likes conjecturingandundetermined coefficients.Thetwo generalsolutions are:x(t)=C, +c,ektmgt(8)y(t)=C,+cek74
74 and the projectile, the projectile's shape and smoothness, and the fluid compressibility. At slow speeds the dominant effect providing resistance to motion is the sliding of the fluid over the projectile's surface (viscosity). At these slow speeds, we assume the number of molecules contacting the surface per unit of time is proportional to the product of the projectile's velocity and its surface area. Therefore, a simple model for the resistive force due to friction is the proportionality F kv r = - , where k is the constant of proportionality and v the velocity of the projectile. The model assumes factors such as surface area and surface smoothness are constant. At higher speeds another air-resistance model may be more accurate. As the velocity increases, more collisions of the projectile with the fluid's molecules take place. Eventually the effect of the collisions comes to dominate the "sliding" effect predominant at low speeds. The model often used in this case and observed experimentally is that the resistive force F is proportional to the square of the velocity. This model is written as 2 F kv r = - , where as before, k is the constant of proportionality and v the velocity of the projectile. This makes the equation nonlinear and, therefore, more difficult to solve analytically. However, graphical and numerical solution techniques may be used for the nonlinear equation case, and some of these methods will be investigated later on in the text. Projectile Motion with Viscous Friction Let’s develop a model for a projectile moving at low velocities, where viscosity is the dominant resistance and gravity is still the only other force. All other forces are neglected. Using Newton's second law we have, F F F ma. å = g + r = Making substitutions for a projectile with position r t x t i y t j r r r ( ) = ( ) + ( ) , we get dt dr mg j k dt d r m r r r = - - 2 2 . We further simplify this equation by writing an equation for both the x and y components of motion. Our equations are: 0 2 2 2 2 + = + = - dt dx k dt d x m mg dt dy k dt d y m . (7) These two uncoupled, linear equations can be solved separately using various techniques likes conjecturing and undetermined coefficients. The two general solutions are: m kt x t c c e - = + 1 2 ( ) k mgt y t c c e m kt = + - - 3 4 ( ) . (8)
The constants cy, Ca, C,,and cy are determined from substitution of the initialconditions: v=vocosoi+vosinoj and =0i +oj.These calculationsproduceG, =m/ro /cosokm|vo/cosekm(I v/sine+mgSkk"(。 sin +m)kkTherefore, the solution isemlvolcose-x(0) = m/ o /cosokk(9)"( o sin +m)-( sin +m)emgt(t) =k?kkkkExample2:Let'sseehowaddingthisairresistancetermchangesourresultsfromExample1.Usingthesamedatathat/y500=50(元/180)radians,andg=9.8, alongwith the newdata that the mass of theprojectile is m=3slugs andtheproportionalityconstantisk=0.05.WesubstitutethesevaluesintoEq(9)toobtain:x(t) = 19284 -19284e-0.01671(10)y(t)= 58261-588t-58261le-0.0167WedeterminemaximumrangeandmaximumprojectileheightusingthesametechniguesasusedinExample1.WeobtainthemaximumrangeasR=12.888meters (att=66.225seconds)andthemaximumheightof H=5,284meters(att=30.1seconds)Thesevaluesaresignificantlydifferent(less)thanthosefound inExample1Therefore,weseethathismuchairresistancedoeseffectthetrajectoryoftheprojectile.WeplotEqn(5)and(10)onthesameaxestoseetheeffects.Figure1showstheeffectsofaddingairresistance75
75 The constants 1 c , 2 c , 3 c , and 4 c are determined from substitution of the initial conditions: v v i v j r r r 0 = 0 cosq + 0 sinq and r i j r r r 0 0 0 = + . These calculations produce k m v c | 0 | cosq 1 = , k m v c | 0 | cosq 2 = - , (| | sin ) 3 0 k mg v k m c = q + , (| | sin ) 4 0 k mg v k m c = - q + . Therefore, the solution is m kt e k m v k m v x t - = - | | cosq | | cosq ( ) 0 0 (9) k mgt e k mg v k m k mg v k m y t m kt = + - + - - ( ) (| | sin ) (| |sin ) 0 q 0 q . Example 2: Let’s see how adding this air resistance term changes our results from Example 1. Using the same data that | | 500 v0 = , q = 50(p/180) radians, and g = 9.8, along with the new data that the mass of the projectile is m=3 slugs and the proportionality constant is k=0.05. We substitute these values into Eq (9) to obtain: t x t e 0.0167 ( ) 19284 19284 - = - (10) t y t t e 0.0167 ( ) 58261 588 58261 - = - - . We determine maximum range and maximum projectile height using the same techniques as used in Example 1. We obtain the maximum range as R=12,888 meters (at t=66.225 seconds) and the maximum height of H= 5,284 meters (at t=30.1 seconds). These values are significantly different (less) than those found in Example 1. Therefore, we see that his much air resistance does effect the trajectory of the projectile. We plot Eqn (5) and (10) on the same axes to see the effects. Figure 1 shows the effects of adding air resistance
altitudeProjectileMotion6000500040003000No Air Resistance2000AirResistance1000feet250005000100001500020000Figure1:Trajectoriesofprojectilewithoutairresistance(Eqn(5))andwithairresistance (Eqn (10)GunSystemModelLet'sgobackand analyzetheParrottGun itself.Anewexperimental 4.2inchParrott is to be testfiredwhile mounted on a two-wheel, brackettrail guncarriage.To determinethe gas pressureforcegenerated uponfiring,thecannon was first testfiredat zero degrees elevation froma fixed casemate.Witha30 pound projectileand a3.25pound charge,thegaspressureforcegenerated was determinedtobe4500pounds.Withthisprojectileandcharge ittakes one-fifth of a second for theroundtoleavethe cannontube (i.e.thegaspressureforce lastsforone-fifthofa second).Thecannonand carriageofourParrotttogetherweigh48oo pounds.Thedampingforce (thatforcecreated bythetrails scrapingalongthegroundandretardingtherearwardrecoil)actshorizontallyonthecannonandcarriageandis known to be numerically equal to 600 /2 times the instantaneous velocity ofthecannonandcarriage.Theexperimental Parrottwill betestfiredwithzerodegrees elevation on a rangewhich hasa safetywall constructedfivefeetbehind the end of the carriage trails as shown in Figure 2. We produce both afirst order and second ordermodel just to compare the results.-Figure2:ParrottGun76
76 5000 10000 15000 20000 25000 feet 1000 2000 3000 4000 5000 6000 altitude Projectile Motion Air Resistance No Air Resistance Figure 1: Trajectories of projectile without air resistance (Eqn (5)) and with air resistance (Eqn (10)). Gun System Model Let’s go back and analyze the Parrott Gun itself. A new experimental 4.2 inch Parrott is to be test fired while mounted on a two-wheel, bracket trail gun carriage. To determine the gas pressure force generated upon firing, the cannon was first test fired at zero degrees elevation from a fixed casemate. With a 30 pound projectile and a 3.25 pound charge, the gas pressure force generated was determined to be 4500 pounds. With this projectile and charge it takes one-fifth of a second for the round to leave the cannon tube (i.e. the gas pressure force lasts for one-fifth of a second). The cannon and carriage of our Parrott together weigh 4800 pounds. The damping force (that force created by the trails scraping along the ground and retarding the rearward recoil) acts horizontally on the cannon and carriage and is known to be numerically equal to 600 2 times the instantaneous velocity of the cannon and carriage. The experimental Parrott will be test fired with zero degrees elevation on a range which has a safety wall constructed five feet behind the end of the carriage trails as shown in Figure 2. We produce both a first order and second order model just to compare the results. Figure 2: Parrott Gun
First-orderModelLet's analyze this situation. We model the given interaction as a first orderdifferential equation with appropriate initial values. We use velocity of thecannon v(t) as the dependent variable and time t as the independent variable.Weassumethefollowing:1. The weight of the projectile is negligible in relation to the weight of thecannon.2.Thereis no friction in the cannontube.3.Att=0gaspressure inthetubeis immediately4500Ibs.4. The forces are projected only in the horizontal direction.5. We use the following notation: F, = gas pressure force, F, = damping forceWeareconcerned withthehorizontal displacement ofthe cannon,andthere isno horizontal component of the cannon's weight.First, we calculate the mass of the cannon with the fomula mg = weight herefore m 2 4800lbsormsec?dyfor the=15oslugs.ThenweapplyNewton'sSecondLaw,usinga=dtacceleration of the gun and v(t) as the velocity of the gun. We separate the11andt≥The followingcalculations into the twoperiods of time, 0≤t<55calculations are used toformulate themodeldifferential equation:110≤<t≥S5ZF=maZF= maF-F=ma=mo4500-600/2v= 150dy600dtdt30-4/2v=hdy-4V2vdtdtorord +4/2v =30dv2+4/2v=0dtdt77
77 First-order Model Let’s analyze this situation. We model the given interaction as a first order differential equation with appropriate initial values. We use velocity of the cannon v(t) as the dependent variable and time t as the independent variable. We assume the following: 1. The weight of the projectile is negligible in relation to the weight of the cannon. 2. There is no friction in the cannon tube. 3. At t = 0 gas pressure in the tube is immediately 4500 lbs. 4. The forces are projected only in the horizontal direction. 5. We use the following notation: F1 = gas pressure force, F2 = damping force We are concerned with the horizontal displacement of the cannon, and there is no horizontal component of the cannon’s weight. First, we calculate the mass of the cannon with the formula mg = weight; therefore, lbs ft m 4800 sec 32 2 =ú û ù ê ë é or m = 150 slugs. Then we apply Newton’s Second Law, using a dv dt = for the acceleration of the gun and v(t) as the velocity of the gun. We separate the calculations into the two periods of time, 5 1 0 £t < and t ³ 1 5 . The following calculations are used to formulate the model differential equation: 5 1 0 £t < : t ³ 1 5 : å F = ma å F = ma F F ma 1 2 - = - F = ma 2 4500 - 600 2v = 150 dv dt dt dv - 600 2v = 150 30 - 4 2v = dv dt - 4 2v = dv dt or or + 4 2v = 30 dt dv + 4 2v = 0 dt dv
Therefore,thedifferentialeguationmodelis:30元0<tdv+4/2v=(11)dt0,15Wealsoneedaninitial conditionforthiseguation.Wenotethatatt=othevelocityanddisplacementarebothzero.Nowwecansolvethisinitialvalueproblemto determine whether ornot the carriagetrailswill strikethe safetywallafter the cannon is fired.First-orderSolution+4/2v=30 on the interval 0<1<1The solution ofisasfollows:dtThisisaseparableequation,soweseparatevariablesand integratebothsidesdy[dt.Lettingu=30-4/2v,wesubstituteof the equation as shown:30-4V2vduto get the integral=t+C,.Thefollowingstepsperformthe4V2Juintegrationandback substitutionforthisintegral:nu=t+C,4/2n|30-4/2=1+C,4/2"In|30-4/2v=-4/2t+C,30-4/2v=Ce-4/230+Ce-4/274V2Weapplythe initial conditionto gettheparticular solution.Theequation3030v(0)= 0=+C is solved for C to get C-Thisissubstitutedtoproduce4/24V2the solution:15(12)v(t):2/278
78 Therefore, the differential equation model is: ï þ ï ý ü ï î ï í ì ³ < < + = 5 1 0, 5 1 30, 0 4 2 t t v dt dv . (11) We also need an initial condition for this equation. We note that at t = 0 the velocity and displacement are both zero. Now we can solve this initial value problem to determine whether or not the carriage trails will strike the safety wall after the cannon is fired. First-order Solution The solution of dv dt + 4 2v = 30 on the interval 0 1 5 < t < is as follows: This is a separable equation, so we separate variables and integrate both sides of the equation as shown: ò = ò - dt v dv 30 4 2 . Letting u = 30 - 4 2v , we substitute to get the integral - ò = + . 4 2 1 C3 t u du The following steps perform the integration and back substitution for this integral: t t v Ce v C e v t C v t C u t C 4 2 4 2 1 2 3 3 4 2 30 30 4 2 ln 30 4 2 4 2 ln 30 4 2 4 2 1 ln 4 2 1 - - = + - = - = - + - - = + - = + We apply the initial condition to get the particular solution. The equation v(0) 0 C 30 4 2 = = + is solved for C to get C = - 30 4 2 . This is substituted to produce the solution: v t ( e ) t ( ) = - 15 - 2 2 1 4 2 . (12)
We determine the displacement x at t :sec. This involves the integration ofSthevelocityfunction overthe interval:[v(t)dt. Using Equation (12), we-e-4/2.)dt: The integral evaluates to get.4256ft.throughsubstitutiontogetWe can also calculatevelocity at t=5-4/5F(13)3.5925sec-Wecontinuebysolving the equation on the second interval of themodel in Eqn(11), t ≥,usingsimilarstepsandEquation(13)astheinitialcondition:5+4/2v=0,= 3.5925. Separating variables and integrating produces:dt-[4/2dt +C, and Inv=-4/2t+C,.Therefore, the general solution is=3.5925= Ce-42/5 and solve to findv= Ce+/2t. We apply the initial condition111378= C. Therefore, the particular solution is v(t)=111378e--/2Todeterminetotaldisplacementwecombinethedisplacementduringo579
79 We determine the displacement x at t = 1 5 sec. This involves the integration of the velocity function over the interval: x v t dt 1 5 0 1 5 æ è ç ö ø ÷ = ò ( ) / . Using Equation (12), we get x ( e )dt 1 t 5 15 2 2 1 4 2 0 1 5 æ è ç ö ø ÷ = - - ò / . The integral evaluates to x ft 1 5 4256 æ è ç ö ø ÷ =. . We can also calculate velocity at t = 1 5 through substitution to get v e 1 5 15 2 2 1 4 2 5 æ è ç ö ø ÷ = - æ è ç ç ö ø ÷ ÷ - = 3.5925 sec ft . (13) We continue by solving the equation on the second interval of the model in Eqn (11), t ³ 1 5 , using similar steps and Equation (13) as the initial condition: dv dt + 4 2v = 0; v 1 5 35925 æ è ç ö ø ÷ = . . Separating variables and integrating produces: dv v ò = - ò4 2dt + C1 and lnv = - 4 2t + C1 . Therefore, the general solution is v = Ce t - 4 2 . We apply the initial condition v Ce 1 5 35925 æ 4 2 5 è ç ö ø ÷ = = - . / and solve to find 11.1378 = C . Therefore, the particular solution is v t e t ( ) = . - 111378 4 2 . To determine total displacement we combine the displacement during 0 1 5 < t < and the displacement during t = 1 5 . These calculations produce an improper integral: x e dt t = + - ¥ .4256 ò11.1378 4 2 1 5 . Integration gives the total displacement as 1.06 feet. Since this is far less than the 5 feet to the wall, the cannon will not hit the wall. Second-order Model We now model the same situation as second order differential equation with initial values. With this model we use displacement (instead of velocity) as the dependent variable and time as the independent variable. Making the same assumptions as before, we build the following model: 0 1 5 < t < :: t ³ 1 5
F=maJ=maF -F, =150a-F, = 150a150dr2dx4500-600/-600V2dr_150dt?dt?dtdt30-4/2x"=x"-4/2x=x"1300≤t5x"+4/210.t≥5We know that at t=o, both the displacement and velocity are zero. Thus, thesecond-orderinitialvalueproblemiswrittenas:130,0≤t0 and =0,t<0. Therefore, x"+4/2x'=30-30U(t-).Weconvert5thisto theLaplace spacetoformanalgebraicequation:3030s°X(s)-sx(0)- x(0)+4/2 s X(s)+4/2 x(0)= -115s30305/5(s? + 4/2s)X(s)= 3ss30 e~s/530X(s) =s(s+4/2)s(s+4/2s)Using partial fractions to separate the first term on the right-hand side, weobtain:80
80 å F = ma å F = ma F F 150a 1 - 2 = - F = a 2 150 4500 600 2 150 2 2 - = dx dt d x dt - 600 2 = 150 2 2 dx dt dx dt 30 - 4 2x¢= x¢¢ - 4 2x¢= x¢¢ ï þ ï ý ü ï î ï í ì ³ £ 0 and = 0,t < 0. Therefore, ). 5 1 x¢¢+ 4 2 x¢= 30 - 30U(t - We convert this to the Laplace space to form an algebraic equation: ( 4 2 ) 30 ( 4 2) 30 ( ) 30 30 ( 4 2 ) ( ) 30 30 ( ) (0) (0) 4 2 ( ) 4 2 (0) 2 / 5 2 2 / 5 2 1/ 5 s s s s e s s X s e s s s s X s e s s s X s sx x s X s x s s + - + = + = - - - ¢ + + = - - - - Using partial fractions to separate the first term on the right-hand side, we obtain: