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(1) Parallelogram method. To find the true course and ground speed. add the wind velocity vector and the air speed heading vector by the parallelogram method. The length of the resultant determines the ground speed and the direction of the resultant is the true course direction. The length of the resultant is 8.7 cm., thus the ground speed is 87 mph. The direction is 42°; thus, true course equals 42°.
(2) Triangle method.—To find the true course and ground speed, add the wind velocity vector and the air speed heading vector by the triangle method.
: (a) Draw the north line. (6) Draw the air speed heading vector at an angle of 56° from the north, making it 10 cm. long (1 cm.=10 mph.; hence 10 cm.=100 mph.).
(c) Draw the wind vector away from 110°, making it 24 cm. long (242 cm.=25 mph.) and from the head of the air speed heading vector.
(d) Air speed heading vector+wind vector=ground speed track vector (Va+Vw=V.).
(e) The angle between the ground speed track vector and the north line determines the course or track. The length of the ground speed track vector, according to the scale used, determines the ground speed.
(f) The drift angle is the angle between the air speed heading vector and the ground speed track vector. Since this is a case of left drift, the drift angle of 14° is subtracted from the heading of 56° to obtain the course 42°.
b. Exercises.-(1) On a secret maneuver, the pilot has been ordered to fly on a heading of 90° until further orders. Air speed of the airplane is 180 mph. After 1 hour of flight, motor trouble develops and a landing must be made. He is told over his radio that wind has been 40 mph. from 135o. What has been his track and ground speed?
(2) The pilot observes an enemy scouting party. To avoid being seen, he changes his heading (as point “A”) and flies on a heading of 315o with air speed 150 mph. If wind is 30 mph. from 45°, what will be his new course and ground speed? C=304°, GS=153 Answer. · (3) Because of an error in plotting his course, a pilot finds himself over unfamiliar territory. His air speed has been 160 mph., his heading has been 180°, and he verifies that wind throughout his flight has been 20 mph. from 315o. What has been his track and ground speed?
(4) Heading has been 30o and air speed of plane is 180 mph. If wind is 60 mph. from 240°, what has been the course (or track) and ground speed?
C=37°, GS=234 Answer. (5) Figure 54 was carefully drawn, but was not properly labeled. Give two different and equally reasonable sets of facts which it might be intended to represent.
46. Type II.—The type of problem next to be considered is one that must be worked out before the flight is started. As a flight is being planned, a certain course is desired. A line showing the desired course is laid off on the map and its azimuth (course angle) is measured. Wind direction and speed are accurately observed and reported at frequent intervals. The air speed of the airplane to be used is known. The heading which the airplane must have in flight in order to make good the desired course, may be determined. Heading and ground speed over the actual course may be read from a vector diagram.
a. Example.—Course to be flown is 308° Air speed of the airplane is 110 mph. and wind velocity is 15 mph. from 230o. Find the heading and ground speed. NOTE.—See back of manual for figure 57.
b. Solution: First, the desired course is indicated by a line of indefinite length drawn in the direction 308°. Since the parallelogram method is to be used to find the unknown parts of the vector diagram, the wind velocity vector originates at the same point as the course line. Wind velocity, 15 mph., is represented by a line 1.5 cm. long from the direction 230°. Since air speed, 110 mph., is known but heading is not, the compass is set with a radius of 11 cm. (11 cm.=110 mph.), and, using the head of the wind velocity vector as center, strike an arc which will intersect the course line. The line drawn from this point to the head of the wind vector has the length and direction
Scole : 1 Inch = 40 mph
of the air speed heading vector. The segment of the course line cut off by the intersecting arc is the length representing the ground speed. The parallelogram is completed in such a way that all three vectors originate from the origin point with the wind velocity vector and the air speed heading vector as sides and the ground speed track vector as the diagonal.
c. The solution of the same problem by the triangle method is given next.
(1) Draw the north line. (2) Draw the wind velocity vector from 230°, making it 1.5 cm. long.
(3) Draw a line of indefinite length in the direction 308°, indicating the desired course.
(4) Draw the air speed heading vector by placing one end of a ruler on the head of the wind vector and mark where a vector 11 cm. long will touch the course line.
(5) The segment of the course line cut off by the head of the air speed heading vector represents the ground speed.
(6) The angle between the air speed heading vector and the north line measured clockwise determines the heading.
d. Exercises.—(1) Course to be flown is 270°. Air speed of the airplane is 120 mph., and wind is 40 mph. from 45°. Find the required heading and the ground speed.
(2) Wind is 30 mph. from 180°. Desired course is 45o, and air speed of the airplane is 140 mph. What should be the heading, and what will be the ground speed?
H=54°, GS=160 Answer. (3) Air speed of the airplane is 125 mph. What will be the required heading to fly a course 135° if wind is blowing 25 mph. from 225°? What will be the ground speed? .
(4) Course to be flown is 225°, and wind is 30 mph. from 90°. If air speed of the airplane is 160 mph., find the required heading and the ground speed.
H=217°, GS=180 Answer. (5) Wind is 45 mph. from 10o, and air speed of the airplane is 165 mph. If the course to be flown is 150°, what will be the required heading and what will be the ground speed?
47. Type III.-If the air speed, heading, ground speed, and track (angle of actual course flown) are known, the wind velocity can be found. These cases arise when the first two factors are obtained from instrument reading and the last two are computed by timing the flight between two landmarks on a map.
a. Plot both the air speed heading vector and the ground speed track vector from the given data.
b. Draw a vector with tail at the arrow end of the air vector and arrow at the arrow end of the ground vector. This is the wind vector.
c. By drawing a north line through the tail of the wind vector, one can measure the azimuth of the wind vector, that is, get the wind direction. The length represents the wind speed.
d. Note that the sum of the wind vector and the air vector is the ground vector; that is, Vw+VA=VG.
e. Example: An airplane's air speed is 155 mph, and its heading is 240°. By computation from a chart, it is found that the ground speed is 170 mph, and the track is 251°. What is the wind velocity? From the vector diagram the wind velocity is 33 mph. from 129o.
Answer. NOTE.-See back of manual for figure 59.
48. Summary.-In any vector triangle, there are six quantities involved: the length and the direction of each of the three vectors.
a. It can be seen from the preceding examples that when any four of the six quantities are given, the other two can always be found. However, the triangle cannot be solved if less than four of the quantities are given.
6. The two unknown quantities can always be found by using Vw+VA=Va, or briefly W A G.
c. Exercises.—(1) If an airplane with an air speed of 180 mph. flies with a heading of 160° in a 20-mph. wind blowing from 215°, what are its track and ground speed?