Gyro Numericals Solved
February 25, 2017 | Author: captyashpal | Category: N/A
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Numerical Problems on Marine Gyrocompass
Q. In latitude 25.0 S, SA (spin axis) of a FG (free gyro) is in position S40E and horizontal. Find the tilt after 6 hours.
Solution: PZ= 65, ZX = 90, tilt after 6 hours = 90 – ZY =? In triangle PZX, Cos PX = Cos Z Sin PZ Cos PX = Cos 40 Sin 65 Cos PX = 0.694272 PX = PY = 46.031 Again, in triangle PZX, Cos P = - Cot PZ Cot PX Cos P = - Cot 65 Cot 46.031 P or angle ZPX = 116.732 Angle XPY = 6 hours = 90. Thus, angle ZPY = 26.732 In triangle PZY, Cos ZY = Cos P Sin PZ Sin PY + Cos PZ Cos PY Cos ZY = Cos 26.732 Sin 65 Sin 46.031 + Cos 65 Cos 46.031 ZY = 28.839 and tilt = 61.16 up (Ans.) Q. In latitude 25.0 N SA of a FG is tilted 55 d 20 m upwards and pointing in certain direction, Next time, when it was in the same direction, the tilt had decreased to 7 d 20 m. Find the direction of the SA.
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Numerical Problems on Marine Gyrocompass
Solution: From P drop a perpendicular PN on ZY. The perpendicular will bisect XY, so that YN = NX PZ = 65.0, ZX = 34 d 40 m, NX = 24, ZN = 58 d 40 m In triangle PZN, Sin (90-Z) = Tan ZN Tan (90-PZ) Cos Z = Tan 58.6667 Cot 65 Z = 40, hence azimuth = N40W (Ans.) Q. SA of a FG is dipping 10.0 down. Later it points N30E and is horizontal. Find the latitude.
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Numerical Problems on Marine Gyrocompass
Solution: NY = 10.0 PN = L (Lat) =? PX = PY = 10 + L In triangle PZX, Sin (90-PX) = Cos Z Cos (90-PZ) Cos PX = Cos Z Cos L Cos (10+L) = Cos Z Cos L Cos 10 Cos L – Sin 10 Sin L = Cos Z Cos L Tan L = (Cos 10 – Cos 30) Sin 10 Or Latitude = 34.374 N Q. SA of a FG(A) in 60 d 00 m latitude and 000 d 00 m long is pointing E – W with E end tilted 30.0 up. At the same time another FG (B) is lying with its SA pointing N – S and tilted 30.0 up from horizon. The two axes are parallel to each other. Find latitude and longitude of FG (B). ( Ans. 34 d 20.5 m S , 073 d 53.9 m E ).
Solution: PZA = 30, ZAX = 60, ZBX = 60, Angle P = ? and PX =? In triangle PZAX, Sin PZA = Tan ZAX Tan (90-P) Tan P = Tan ZAX Cosec PZA P = 73.899 Thus longitude = 000.00 + 73.899 = 73.899 E (Ans.) Again in triangle PZAX, Sin (90-PX) = Cos ZAX Cos PZA Cos PX = Cos 60 Cos 30 PX = 64.341
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Numerical Problems on Marine Gyrocompass
Now, PZB = PX + ZBX = 64.341 + 60.0 = 124.341 Hence Latitude = 90 - PZB = - 34.341 = 34.341 S (Ans.) Q. In lat 40.0 S, SA of a FG is pointing N and is tilted 25.0 up at 00h 00m 00s GMT. What time will it be horizontal and in what direction?
Solution: PZ = 50, NY = 25, PX = PY = PN - NY = 50 + 90 - 25 = 115, P =? , Z =? In triangle PZX, Sin (90-PX) = Cos Z Cos (90-PZ) Cos PX = Cos Z Sin PZ Cos Z = Cos 115 Sin 50 Z = 123.48 = N56.52W, the direction in which the SA will point when horizontal (Ans.) Again in triangle PZX, Sin (90-P) = - Tan (90-PX) Tan (90-PZ) Cos P = - Cot PX Cot PZ Cos P = - Cot 115 Cot 50 P = 66.966 = 4.4644 sidereal hours = 4.4644 x 23.934444 / 24 = 4.4522055 solar hours = 4h 27m 08m GMT, the time the SA will be horizontal. (Ans.)
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Numerical Problems on Marine Gyrocompass
Q. In lat 28.5 N, SA of a FG is pointing north and horizontal at 02h 12m 25s LMT. Later it was observed that maximum azimuth was reached. Find azimuth and tilt at the time of second observation. Also find the LMT time when maximum azimuth is reached.
Solution: PZ = 61.5, PX = PN = Latitude = 28.5, X = 90 (moment of maximum azimuth) In triangle PZX, Sin (90-PZ) = Cos PX Cos ZX Cos ZX = Cos 61.5 Sec 28.5 ZX = 57.115, thus Tilt = 32.885 (Ans.) Sin (90-Z) = Tan (90-PZ) Tan ZX Cos Z = Tan 28.5 Tan 57.115 Z = 32.885 (Ans.) Sin (90-P) = Cos ZX Cos (90-Z) Cos P = Cos 57.115 Sin 32.885 P = 72.8545 = Angle ZPX Angle NPX = 107.14553 = 7.1430352 sidereal hours = 7.123524 solar hours = 07h 07m 24.7s Thus, time of occurrence: 09h 19m 49.7s LMT (Ans.)
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Numerical Problems on Marine Gyrocompass
Q. At a place on earth’s surface, SA of a FG is pointing N30E and horizontal. Later it points N with tilt 70 d up. Find lat and tilt when it points N again.
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Numerical Problems on Marine Gyrocompass
Q. Five hours after SA of a FG was pointing N and below the horizon, it was horizontal. If the latitude was 35 N, find the azimuth at the moment. What will be the tilt, when it bears east later. Also, find the time elapsed between the two observations.
Solution: Lat = 35, PZ = 55, Angle NPX = 75, PX = PY In triangle PZX, Sin (90-PZ) = Tan Z Tan (90-P) Tan Z = Cos 55 Cot 15 Z = 64.96 (Ans.) Again in triangle PZX, Sin (90-PX) = Cos Z Cos (90-PZ) Cos PX = Cos 64.96 Sin 55 PX = 69.714 = PY Now in triangle PZY, Sin (90-PY) = Cos ZY Cos PZ Cos ZY = Cos 69.714 Sec 55 ZY = 52.81, hence tilt = 37.19 up (Ans.) Again in triangle PZY, Sin (90-P) = Tan PZ Tan (90-PX) Cos P = Tan 55 Cot 69.714 P = 58.1369 Time interval from X to Y = Angle ZPX – Angle ZPY = 105 – 58.1369 = Angle 46.8631 = 03h 07m 27.1s (Ans.)
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Numerical Problems on Marine Gyrocompass
Q. A SA of a FG pointing was horizontal pointing in direction S38W. A few hours later the other end of the axis crosses meridian with the tilt 68.5 d up. What is your latitude and tilt when next time the SA crosses meridian.
Solution: Z = 38, NU = 68.5, PN = L (latitude) Now, PU = PD = PX = NU – L = 68.5 – L In triangle PZX, Sin (90-PX) = Cos Z Cos (90-PZ) Cos PX = Cos 38 Cos L Cos (68.5 – L) = Cos 38 Cos L Tan L = Cos 38 – Cos 68.5 Sin 68.5 L (latitude) = 24.372 N (Ans.) Now, PX = NU – L = 68.5 – 24.372 = 44.128 Tilt below horizon = PD – PN = PX – L = 44.128 – 24.372 = 19.756 down (Ans.)
Q. A FG in lat 60.0 N is set with SA pointing 072 and tilted 20.0 up. Determine the tilt, when the axis is in meridian and direction when SA is horizontal next.
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Numerical Problems on Marine Gyrocompass
Solution: PZ = 30.0, Z = 72, ZY = 70.0 In triangle PZY, Cos PY = Cos Z Sin PZ Sin ZY + Cos PZ Cos ZY Cos PY = Cos 72 Sin 30 Sin 70 + Cos 30 Cos 70 PY = 63.81 = PX = PM Tilt at meridian = SM = SP – PM = SZ + PZ – PM = 90 + 30 – 63.81 = 56.19 up and pointing south. In triangle PZX, Sin (90-PX) = Cos Z Cos (90-PZ) Cos Z = Cos PX Cosec PZ Cos Z = Cos 63.81 Cosec 30 Z = 28.03, hence azimuth = N28.03W
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10 | P a g e Numerical Problems on Marine Gyrocompass Q. A FG in 60.0 S is horizontal and pointing N-S. Where will it point in ½ siderial day?
Solution: In the given situation SA can be pointing N or S. If it points N, it will point exactly N and will never rise and will remain below the horizon throughout. Hence this possibility is ruled out and the SA is then pointing initially at X coinciding with S. After half a sidereal day, it will drift through 180 degrees and be at Y. Now, PX = PY = latitude = 60.0 NY = NS – PX – PY = 180 – 60 -60 = 60.0, hence the SA will point N with upward tilt of 60.0 (Ans.) Q. Find percentage change in M.O.I, of a rotor if mass is increased by 20% and radius of gyration is decreased by 20%. Solution: MOI = Mr2 Old mass = M, new mass = 1.2M Old radius = r, new radius = 0.8r Now, %age change given by: (Old MOI – new MOI) x 100 Old MOI = [1.2M x (0.8r)2 - Mr2 ] x 100 Mr2 - 0.232 x 100 = - 23.2 % Thus the MOI of the rotor will decrease by 23.2 % (Ans.)
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11 | P a g e Numerical Problems on Marine Gyrocompass Q. At 00h 00m 00s GMT in latitude 45.0 S and longitude 000.0, SA of a FG is pointing 120 (T). At what time will it point 270 (T) on the same day?
Solution: PZ =45.0, Z = 60.0, PX = PY In triangle PZX, Sin (90-PX) = Cos Z Cos (90-PZ) Cos PX = Cos 60 Cos 45 PX = 69.295 = PY Sin (90-PZ) = Tan Z Tan (90-P) Cot P = - Tan 60 Sec 45 P = 112.21 = Angle ZPX Now in triangle PZY, Sin (90-P) = Tan PZ Tan (90-PY) Cos P = Tan 45 Cot 69.295 P = 67.792 = Angle ZPY Total time elapsed = Angle ZPX + Angle ZPY = 112.21 + 67.792 = 180.002 = 12.000133 sidereal hours = 11.967355 solar hours = 11h 59m 2.5s (Ans.)
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12 | P a g e Numerical Problems on Marine Gyrocompass Q. At equator SA of a FG is pointing N60E and is horizontal. Find tilt and direction after 3 hours. Also find the maximum tilt attained.
Solution: PZ = 90, ZX = 90, Z =60 In triangle PZX, Sin (90-PX) = Cos Z Cos (90-PZ) Sin PX = Cos Z Sin PZ PX = 60 = PY Sin (90-PZ) = Tan Z Tan (P-90) Sin 0 = Tan 60 Tan (P-90) P = 90 In triangle PZY, Z =? , ZY =? , and P = 45 Sin (90-ZY) = Cos P Cos (90-PY) Cos ZY = Cos 45 Sin 60 ZY = 52.24, hence. Tilt = 37.76 up Sin P = Tan Z Tan (90-PY) Tan Z = Sin 45 Tan 60 Z = 50.77, hence azimuth = N50.77E Max tilt attained = NM = PX = 60 up
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13 | P a g e Numerical Problems on Marine Gyrocompass Q. At equator SA of a FG is pointing N60E and tilted 20 up. Find direction and tilt after 8 hrs. Also find the max tilt attained. ( Ans.tilt: 32.260 , Az: 56.247; Max tilt 61.976 )
Solution: Angle P ZX = 60, ZX = 70 Sin (90-PX) = Cos Z Cos (90-ZX) Cos PX = Cos 60 Cos 20 PX = 61.976 = PY Sin Z = Tan P Tan (90-ZX) Tan P = Sin 60 Tan 70 P i.e. Angle ZPX = 67.204 Angle ZPY = 8 x 15 – Angle ZPX = 52.796 In triangle PZY, Sin (90-ZY) = Cos P Cos (90-PY) Cos ZY = Cos P Sin PY ZY = 57.74, hence Tilt = 32.260 (Ans.) Sin P = Tan Z Tan (90-PY) Tan Z = Sin 52.796 Tan 61.976 Z = 56.247, hence azimuth = N56.247W (Ans.) Maximum Tilt attained = PM = PX = PY = 61.976 (Ans.)
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14 | P a g e Numerical Problems on Marine Gyrocompass Q. In a certain place at 00h 00m 00s GMT SA of a FG was found to be horizontal and pointing N. Later it again pointed N with a upward tilt of 64.0. Find tilt, maximum azimuth reached during this time. Also find the GMT time at the moment and latitude of the place.
Solution: Upper tilt = NM = 64.0m Lower tilt = 0, Latitude = PN = PX =? , ZX =? , Z =? , P =? From diagram it is obvious that PM = PN = MN/2 = 64/2 = 32 Hence Latitude = 32.0 N and also PX = 32 (Ans.) In triangle PZX: Sin (90-P) = Tan PX Tan (90-PZ) Cos P = Tan 32 Tan 32 P = Angle ZPX = 67.016769 degrees Thus angle NPX = 180 - 67.017 = 112.98323 degrees = 7.5322153 sidereal hours = 7.5116411 solar hours = 07h 30m 41.9s Thus time at occurrence of maximum azimuth = 07 h 30 m 41.9 s GMT (Ans.) Again in triangle PZX: Sin (90-PZ) = Cos PX Cos ZX Cos ZX = Sin 32 Sec 32 ZX = 51.327, hence tilt = 38.673 degrees (Ans.) Sin PX = Cos (90-Z) Cos (90-PZ) Sin Z = Sin 32 Sec 32 Z = 38.673, hence azimuth = N38.673E (Ans.)
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15 | P a g e Numerical Problems on Marine Gyrocompass Q. In a certain latitude at 00h 00m 00s GMT, SA of a FG was pointing N and tilted 15 down. Later it again pointed N with an upward tilt of 45. Find time and direction when SA was horizontal during this time. Also find the lat.
Solution: NU = 45, ND = 15, PD = PU = PX, Latitude = PN = L =? Z =? P =? ND = 15 = PX – L --------- (1) NU = 45 = PX + L --------- (2) Subtracting (1) from (2), 30 = 2L, hence L = latitude = 15.0 N (Ans.) PX = 45 – L = 45 – 15 = 30 degrees In triangle PZX: Sin (90-PX) = Cos Z Cos (90-PZ) Cos Z = Cos 30 Sec 15 Z = 26.288 degrees, hence azimuth = N26.288E (Ans.) Sin (P-90) = Tan (90-PX) Tan (90-PZ) Cos P = Tan 60 Tan 15 P = Angle ZPX = 117.652 degrees Therefore time taken = Angle DPX =180 – 117.652 = 62.348 degrees = 4.1565333 Sidereal hours = 4.1451798 Solar hours = 04 h 08 m 42.6 s GMT, the time when SA was horizontal during this time.
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16 | P a g e Numerical Problems on Marine Gyrocompass Q. In a certain latitude at 00h 00m 00s GMT, SA of a FG was tilted 20 down and pointing N. Later it was found to be pointing to zenith. Find the direction and time when it became horizontal during this time. Also find the latitude.
Solution: SZ = 90, SD = 20, PD = PZ = PX, Latitude = PS = L =? Z =? P =? SD = 20 = PX – L --------- (1) SZ = 90 = PX + L --------- (2) Subtracting (1) from (2), 70 = 2L, hence L = latitude = 35.0 S (Ans.) PX = 90 – L = 90 – 35 = 55 degrees = PZ In triangle PZX: Sin (90-PX) = Cos Z Cos (90-PZ) Cos Z = Cos 55 Sec 35 Z = 45.556, hence azimuth = S45.556E (Ans.) Sin (90-P) = -Tan (90-PX) Tan (90-PZ) Cos P = -Tan 35 Tan 35 P = 119.35968 degrees = Angle ZPX Angle SPX = 180 - 119.3597 = 60.640317 degrees = 4.0316452 Solar hours = 04 h 01 m 53.9 s GMT, the time SA became horizontal (Ans.)
19. In a certain place at 00h 00m 00s GMT, SA of a FG pointed S with upward tilt of 75.0 degrees and it was tilted down 15.0 degrees while pointing N. Find: a) Tilt and time when SA was pointing E-W. b) Direction and time when SA was horizontal during this time. Also find the latitude of the place.
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17 | P a g e Gyrocompass
Numerical Problems on Marine
Solution: SU = 75, ND = 15, PU = PD = PX = PY, Latitude = PN = L =? Z =? ZX =? P =? ND = 15 = PX – L --------- (1) NU = 105 = PX + L --------- (2) Subtracting (1) from (2), 90 = 2L, hence L = latitude = 45.0 N (Ans.) and PZ = 45 PX = 105 – L = 105 – 45 = 60 degrees = PY In triangle PZX: Sin (90-PX) = Cos ZX Cos PZ Cos ZX = Cos 60 Sec 45 ZX = 45, hence tilt = 45.0 degrees up (Ans.) Sin (90-P) = Tan PZ Tan (90-PX) Cos P = Tan 45 Tan 30 P = angle ZPX = 54.73561 degrees = 03h 38m 20.7s GMT (Ans.) Tri PZY: Sin (90-PY) = Cos Z Cos (90- PZ) Cos Z = Cos 60 Sec 45 Z = 45, hence azimuth = N45W (Ans.) Sin (90-P) = - Tan (90-PZ) Tan (90-PY) Cos P = - Tan 45 Tan 30 P = angle ZPY = 125.26439 degrees = 08h 19m 41.3s GMT (Ans.)
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