Standard Trigonometric Functions

Transcription

1 CRASH KINEMATICS For ngle A: opposite sine A = = hypotenuse djent osine A = = hypotenuse opposite tngent A = = djent For ngle B: opposite sine B = = hypotenuse djent osine B = = hypotenuse opposite tngent B = = djent Stndrd Trigonometri Funtions A C = Sin A = Cos A = Tn A = Sin B = Cos B = Sin B B A= Sin -1 ( ) A= Cos -1 ( ) A= Tn -1 ( ) B= Sin -1 ( ) B= Cos -1 ( ) B= Tn -1 ( ) A = mesured y surveyor = Distne from surveyor to tree Sin A = = sin A + Surveyor height = Height of tree A 1

2 Crsh Fore Computtion Terminology Horizontl Flight Pth Terrin Impt Term Desription Notes The ngle etween the impt surfe nd the horizontl, mesured in the vertil plne of the flight pth. the diretion of the flight is downhill. Terrin Flight Pth Impt The ngle etween the irrft s flight pth nd the horizontl t the moment of impt. The ngle etween the flight pth nd the terrin, mesured in the vertil plne of the flight pth. The lgeri sign of the terrin ngle is positive when the diretion of flight is uphill & negtive when The lgeri sign of the flight pth ngle is positive if the irrft is moving downwrd immeditely prior to impt. The sign is negtive if impt ours while the irrft is moving upwrd. The impt ngle is the lgeri sum of the flight pth ngle plus the terrin ngle. [ Impt = Flight Pth + Terrin ] Airrft Pith Crsh Fore Resultnt Crsh Fore G R Resultnt Crsh Fore G H Horizontl Crsh Fore G V Vertil Crsh Fore Resultnt Term Desription Notes The geometri sum of the horizontl nd vertil rsh fores. Horizontl nd vertil rsh fores re determined on the sis of horizontl nd vertil veloity omponents t impt nd horizontl nd vertil stopping distnes. Attitude t Impt Crsh Fore The irrft ttitude t moment of impt. Pith Nose ove or elow horizon Yw Nose right or left Roll Right or left nk The ngle etween the resultnt rsh fore nd the longitudinl xis of the irrft. Crsh* Resultnt Fore = + * Resultnt ngle mesured from horizontl. Other solutions re possile. Pith The lgeri sign of the Crsh Fore Resultnt ngle is positive when the line of tion of the resultnt is ove the horizontl nd is negtive if the line of tion is elow the horizontl. The lgeri sign of the irrft pith ngle is negtive when the nose of the irrft points elow the horizon nd positive when the nose points ove the horizon. For impts with little lterl omponent of fore the rsh fore ngle is the lgeri sum of the rsh fore resultnt ngle plus the irrft pith ngle. 2

3 V l V l l I V FP Down Hill R = Resultnt CF G ll = G Prllel G l = G Perpendiulr G R = G Resultnt FP T Horizon I = Impt FP = Flight Pth T = Terrin V FP = Veloity Flight Pth = Veloity Vertil V ll = Veloity Prllel = Veloity Perpendiulr V i G l G l l G R R V FP = sin I Tn FP = V l V ll G R = V l = sin I V FP V ll = os I V FP G l = G l sin R Tn R = V l SD G ll = G l G ll V ll SD MPH = Ft/Se KTS 1.69 = Ft/Se is mesured perpendiulr to the Horizon, not the Terrin. V FP Up Hill i fp Horizon V l t V l l G l G R R G l l 3

4 Solve for: Opposite A Sine A = Hypotenuse = B Adjent Cosine A = = Hypotenuse Opposite Tngent A = = Adjent Opposite Sine B = Hypotenuse = Adjent Cosine B = = Hypotenuse Opposite Tngent B = = Adjent Sine of A =.3420 Length of side is 10 feet long Solve for length of side. A = 2 C = 90º Exmple: Solve for length of sides nd if ngle A is 20º nd side is 10 feet long Sine A = = 10 = (0.3420)(10) = feet long Mth Key Cosine of A =.9397 Length of side is 10 feet long Solve for length of side. C B Cosine A = = 10 = (0.9397)(10) = feet long Chek ury of omputtions: = 2 (3.420) 2 + (9.397) 2 = = V = Veloity in feet/seond (f/s) = Vertil Veloity V H = Horizontl Veloity V FP = Veloity of flight pth V(MPH) = V f/s V(KTS) 1.69 = V f/s Grvity (g) = GV = G Lod Vertil GH = G Lod Horizontl SD = Stopping Distne in feet SDV = Stopping Distne Vertil SDH = Stopping Distne Horizontl KE = Keneti Energy in foot/pounds (ft/ls) W = Weight of Ojet h = Height CF = Centrifugl Fore in pounds Rdius = 1/2 dimeter Solve for impt ngle: Impt Opposite Tn A = Adjent = G V = Solve for vertil impt G lod: ( ) 2 g SD V CF = Solve for horizontl imlne: W Rdius (RPM) Solve for vertil veloity, given Solve for horizontl impt G lod: Solve for kineti energy: flight pth veloity (f/s): Sine A = G H = ( ) 2 W KE = 1/2 V 2 V g x SD g H FP (V in fps, nswer in foot/pounds) Solve for horizontl veloity given flight Solve for Veloity: pth veloity (f/s) V H Cosine A = V = 8 h V FP Stopping Distne V2 SD = G s (Ft needed to survive x mount of G s) Sovle for veloity: V 2 = K.E. 1 weight G s = Solve for G s: V SD MPH = Feet/Seond KTS 1.69 = Feet/Seond 4

5 Prolem #1 An irplne impts on level ground fter pssing through the top rnhes of tree. By mesurement, you determine tht the irplne struk the tree 70 feet ove the ground t point 200 feet horizontlly from the point of impt. Find: 1. The ngle of impt. 2. The horizontl nd vertil veloities t impt if the flight pth veloity is 150 feet per seond. 70 A Find the ngle of impt Tn A = A = tn -1 (70 200) = 19.29º 2 The horizontl nd vertil veloities t impt if the flight pth veloity is 150 feet per seond.. Drw sketh of irplne impt ngle A = 19.29º = A = V H. Find nd V H using sine nd osine =, V H = = (sin A = ) sin A = sin (150) = fps V H = (os A = ) os A = os (150) = fps 5

6 Prolem #2 An irrft rshes in level open field. Flight pth ngle is 10 degrees nd the true irspeed is 85 mph. Initil impt ours with the fuselge level (zero pith ngle). The impt uses 2-foot-deep gouge, nd the irrft omes to rest 25 feet from initil impt. The fuselge is rushed 12 inhes vertilly nd 5 feet horizontlly. Find: 1. The irrft ground speed (V H ) nd vertil veloity ( ) in feet per seond. 2. The men vertil nd horizontl elertions, in G s. 3. The mgnitude nd diretion of the men rsh resultnt. 1 To find V H nd. Sketh onditions t impt. Convert 85 mph to ft/se (85 mph)(1.467) = ft/se V H 10 o Answer rounded to ft/se. Determine vertil nd horizontl veloities (Round nswers to nerest 10th for next step) sin 10 o = V FP = V FP sin 10 o = (124.7)(0.174) = ft/se os 10 o = V H V FP V H = V FP os 10 o V H = (124.7)(0.985) = ft/se 2 To find G V nd G H. Determine vertil nd horizontl stopping distnes Vertil stopping distne (S V ) = (2 ft) gouge + (1 ft) struture = 3 ft Horizontl stopping distne (S H ) = (25 ft) gouge + (5 ft) struture = 30 ft. Determine vertil nd horizontl elertions (G s) G = V 2 64S 3 To find R nd G R G V = 2.45G G H = 7.85G G V = (21.7) 2 (64 3) = 2.45 G G H = (122.8) 2 (64 30) = 7.85 G. Sketh the vetor digrm of the impt elertions. Use tngent trig funtion to find diretion of resultnt elertion tn R = G V 2.45 = G H 7.85 = R. Use the osine funtion to find the mgnitude of the resultnt elertion os 17.3º = G R = = 8.22 G G R os 17.3º R = r tn = 17.3º 6

7 Prolem #3 An irrft rshes on level ground t n irspeed of 140 knots. Aident investigtors disover tht the irplne struk the top of tree t point 60 feet ove the ground nd rshed 100 feet from the se of the tree. The irrft me to rest t the end of gouge 32 feet long nd 3 feet deep. Mesurements show tht the irplne ws rushed 60 inhes longitudinlly nd 12 inhes vertilly. Find: 1. Horizontl nd vertil veloities, in feet per seond. 2. Men vertil nd horizontl elertions, in G s. 3. Mgnitude nd diretion of the rsh fore resultnt. 1 To find V H nd. Sketh onditions t impt. Convert 140 knots to ft/se (140 kts)(1.69) = ft/se = 60 = 100 V H d. Determine vertil nd horizontl veloities I. Determine impt ngle tn I = I = tn ( ) = 30.96º 100 (Answer rounded to 31º for next omputtion) (Round nswers to nerest 10th) sin 31 o = V FP = (236.6)(sin 31 o ) = (236.6)(0.515) = ft/se os 31 o = V H V FP V H = (236.6)(os 31 o ) V H = (236.6)(0.857) = ft/se 2 To find G V nd G H. Determine vertil nd horizontl stopping distnes Vertil stopping distne (S V ) = (3 ft) gouge + (1 ft) struture = 4 ft Horizontl stopping distne (S H ) = (32 ft) gouge + (5 ft) struture = 37 ft. Determine vertil nd horizontl elertions (G s) G = V 2 64S 3 To find R nd G R G V = (121.8) 2 (64 4) = G G H = (202.8) 2 (64 37) = G (Round nswers to nerest 10th). Sketh the vetor digrm of the impt elertions. Use tngent trig funtion to find diretion of resultnt elertion G tn R = V 58.0 = = 3.33 G H 17.4 R = tn -1 (3.33) = 73.28º (Round nswer to nerest 10th). Use the osine funtion to find the mgnitude of the resultnt elertion R G H = 17.4 G G sin 73.3º = V = G R = = G G R G R sin73.3º 7

8 Prolem #4 An irrft rshes ginst 10-degree uphill slope. The impt ngle is 20 degrees. At the time of the impt, the irrft vertil veloity ws 1,800 feet per minute. The irplne me to rest fter sliding 80 feet. Mximum depth of the gouge ws 1 foot, nd inspetion reveled tht the irplne struture ws rushed 1 foot vertilly. There ws no horizontl rushing of the struture. Find: 1. Flight pth veloity. 2. Men longitudinl nd vertil elertion, in G s, with respet to the fe of the hill. 3. Men rsh fore resultnt mgnitude nd diretion.. 1 To find flight pth veloity. Sketh onditions t impt. Convert to feet/se 1,800 ft/min 60 = 30 ft/se FP = 10º 10º I = 20º 2 To find G l nd G ll. Determine flight pth veloity sin FP = sin 10º = = V FP = 30 sin 10º V FP 30 V FP = ft/se (Round nswer to nerest 10th). Sketh the known onditions, with respet to the fe of the hill. (This my e inluded in the erlier sketh). I = 20º. Determine veloity prllel to fe of hill nd veloity perpendiulr to fe of hill. (Round nswers to nerest 10th) (1) os 20º = V ll V FP V ll = (V FP )(os 20º) = (172.8)(os 20º) = ft/se (2) sin 20º = V l V FP V l = (V FP )(sin 20º) = (172.8)(sin 20º) = ft/se. Determine stopping distnes prllel to the fe of the hill nd perpendiulr to the fe of the hill. (1) S ll = 80 feet (2) S l = (1 foot) + (1 foot) = 2 feet d. Determine elertions prllel to the slope nd perpendiulr to the slope. G = V 2 64S (1) G ll = (162.4) 2 (64 80) = 5.15 G s (2) G l = (59.1) 2 (64 2) = G s 8

9 Prolem #4 ontinued: 3 To find R nd G R. Sketh the vetor digrm of the men elertions.. Determine R. tn R = G l G = ll 5.15 = 5.29 R = tn -1 (5.29) = 79.29º (Round nswer to nerest 10th). Determine mgnitude of rsh fore resultnt. G ll = 5.15 G G l sin 79.3 = = G R G R = sin 79.3º = G G R 9

10 Prolem #5 (Bsed on Prolem #4: An irrft rshes ginst 10-degree uphill slope. The impt ngle is 20 degrees. At the time of the impt, the irrft vertil veloity ws 1,800 feet per minute. The irplne me to rest fter sliding 80 feet. Mximum depth of the gouge ws 1 foot, nd inspetion reveled tht the irplne struture ws rushed 1 foot vertilly. There ws no horizontl rushing of the struture.) ADD: The longitudinl xis of the irplne is prllel to the flight pth. Find: Longitudinl nd vertil elertions with respet to the irrft xes. 1 Determine longitudinl nd vertil elertions with respet to the irrft xes.. Sketh onditions t impt Airrft Vertil Axis Horizontl Referene 20º. Determine from the sketh tht the G rsh fore resultnt ts t n ngle of 59.3 degrees from the longitudinl xis of the irplne (79.3º - 20º = 59.3º) Therefore, the men longitudinl elertion of the irplne is: G longitudinl = (G R )(os 59.3º) = (27.28)(os 59.3º ) = ft/se And the men vertil elertion of the irplne is: G vertil = (G R )(sin 59.2º) = (27.28)(sin 59.3º ) = ft/se 10