Frames of Reference: Find the speed and heading of the airplane

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SUMMARY

The discussion focuses on calculating the speed and heading of an airplane using vector addition. The key equation established is Vg = Vair + Vwing, where Vg represents ground speed, Vair is the airspeed, and Vwing is the wind speed. The participants emphasize the importance of accurately drawing velocity vectors and understanding the angles involved, particularly noting that the heading should be 45 degrees west of north for a ground speed of 400 km/h. Confusion arises regarding the terminology used for wind and airspeed, highlighting the need for clarity in vector labeling.

PREREQUISITES
  • Understanding of vector addition and subtraction
  • Familiarity with basic trigonometry, particularly angles
  • Knowledge of airspeed and ground speed concepts
  • Ability to interpret and draw vector diagrams
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  • Learn about vector addition and subtraction in physics
  • Study the graphical methods for vector representation
  • Explore the concept of wind correction angles in aviation
  • Review the differences between airspeed, ground speed, and wind speed
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Aerospace engineers, physics students, pilots, and anyone interested in understanding vector dynamics in aviation contexts.

ahira
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New user has been reminded to always show their work when posting schoolwork questions
Homework Statement
pilot is flying from City A to City B which is 300 km [NW]. If the plane will encounter a constant wind of 80 km/h from the north and the schedule insists that he complete his trip in 0.75 h, what air speed and heading should the plane have?
Relevant Equations
V=d/t
Vg= Vair +Vwing
so far what i have gotten to is that 300/0.75 = 400km/h but I dont know how to draw the diagram for this
 
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would it be like this?
diagram fof.jpg
 
ahira said:
so far what i have gotten to is that 300/0.75 = 400km/h but I dont know how to draw the diagram for this
ahira said:
would it be like this?
'like this' is a good description, but some improvement is in order:
  • heading AB is ##\pi/4##
  • you write ##v_g = v_{air} + v_{wing} ##, but you draw ##v_{wing}= =v_g - v_{air}##. That's fine, but somewhat confusing, especially if you don't label the vectors.
So far, so good; now perform the actual calculation :smile:

##\ ##
 
BvU said:
'like this' is a good description, but some improvement is in order:
  • heading AB is π/4
  • you write vg=vair+vwing, but you draw vwing==vg−vair. That's fine, but somewhat confusing, especially if you don't label the vectors.
So far, so good; now perform the actual calculation :smile:
IMG_3659.jpg

I Changed the diagram and realized that North West sits on an angle of 45 degrees so therefore the angle between the Northline and A should be 45 degrees. I'm not that sure but should the angle at B be 45 degrees as well due to alternate angles ?
 
Last edited:
City A to City B which is 300 km [NW]
So vector AB should correspond to the ground speed with a heading of 45 degrees west of north and magnitude 400 km/h. As you wrote$$v_g = v_{air} + v_{wing}$$but now your drawing shows vector AC (a.k.a.##\ \ v_{wing}\ \ ##) as ##v_g + v_{air}## !!

Lean back a little and use common sense: with a headwind your course should be aiming upwind of A !

And you can also reasonably expect that you need to make more speed than the 400 km/h ('AC should be longer than AB')

##\ ##
 
Sorry for the dumb question, but should ##v_{wing}## be ##v_{wind}## in all of the posts above (including the OP's)?
 
BvU said:
somewhat confusing, especially if you don't label the vectors.
I feel dumb for taking ##v_{air} = ## 80 km/h from the north (the speed OF the air :rolleyes:), when - most likely -@ahira perhaps meant ##v_{air} = ## the speed WRT the air.

So what about
ahira said:
Relevant Equations:
Vg= Vair +Vwing
and the picture in #3 ?

##\ ##
 

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