Question
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?
Answers
Steve
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
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
Alexa
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 @ N23°E = (273.51,644.35)
(333.51,644.35) = 725.55 @ N27.36°E
Damon
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
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
Henry
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.
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.
Ryan
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"?
Is that the answer to the "What correction needs to be made" part?
What is that "i"?