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