An airplane is heading at a bearing of N23°E at a speed of 700 km/hr, but there is a wind blowing from the west at 60 km/hr. If the plane does not make any correction, what will be the bearing of the plane? What is its speed relative to the ground (ground speed = airspeed + wind speed). In order for the plane to fly to its original bearing, what correction needs to be made?
craft travel on a heading, not a bearing.
700 @ N23°E = (273.51,644.35)
60 @ E = (60,0)
Add them up and you have
(333.51,644.35) = 725.55 @ N27.36°E
Can you explain how to get these from here:
700 @ N23°E = (273.51,644.35)
(333.51,644.35) = 725.55 @ N27.36°E
700 sin 23 = 273.51 = x speed (east)
700 cos 23 = 644.35 = y speed (north)
If x = 333.51
and y = 644.35
then resultant = sqrt (x^2+y^2)
and for angle east of north
tan angle = x/y
Vpw = Vp + Vw = 700[67o] + 60[0o] =
700*Cos67 + 700*sin67 + 60 =
273.5 + 644.4i + 60 = 333.5 + 644.4i =
725.6km/h[62.64o].
Vpw = Vp + Vw = 700[67o].
Vp + 60 = 700[67].
Vp = 273.5 + 644.4i - 60=213.5 + 644.4i
= 679km/h[71.7o] N. of E. = 18.3o E. of
N.
Hi Henry, can you explain the second part of your answer please?
Is that the answer to the "What correction needs to be made" part?
What is that "i"?
To determine the bearing of the plane if it doesn't make any correction, we can use vector addition.
First, let's convert the given bearing of N23°E into components. Considering North as the y-axis and East as the x-axis, N23°E can be split into N(23° - 0°)E(90° - 23°).
This can be simplified to N23°E = N67°E.
Now, let's assume the plane's airspeed vector to be 700 km/hr at N67°E and the wind's velocity vector to be 60 km/hr towards west, or W90°.
We can add these two vectors to determine the resultant vector or the velocity of the plane relative to the ground.
Using vector addition, we find:
Resultant vector = N67°E + W90°
To simplify vector addition, we need to convert the resultant vector into its North and East components.
Using trigonometry, we can find the x and y components of the resultant vector:
x-component = 700 km/hr * cos(67°) = 322.59 km/hr towards East
y-component = 60 km/hr - 700 km/hr * sin(67°) = -678.58 km/hr towards South
Using these components, we can determine the bearing of the plane if it doesn't make any correction:
tan(bearing) = (y-component / x-component)
tan(bearing) = (-678.58 km/hr / 322.59 km/hr)
bearing = tan^(-1)(-678.58 km/hr / 322.59 km/hr)
Calculating this value gives us a bearing of approximately -65.07° or S65°W.
To determine the ground speed, we can use the formula:
ground speed = airspeed + wind speed.
ground speed = 700 km/hr + 60 km/hr
ground speed = 760 km/hr.
Finally, to fly back to its original bearing of N23°E, the plane needs to make a correction equal to the wind vector (opposite in direction) since the wind is pushing it off course.
Therefore, the correction needed is a velocity vector of 60 km/hr at E90° or simply W90°.