I have a question about an airspeed/groundspeed word problem

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SUMMARY

The discussion revolves around calculating the ground speed of a plane climbing at 60 km/hr with an airspeed of 540 km/hr while facing a northwest wind of 110 km/hr. Participants emphasize the need to resolve the plane's motion into orthogonal components, specifically separating the vertical and horizontal velocities. The correct approach involves using vector addition to combine the airspeed and wind components, discarding the vertical component for ground speed calculations. The consensus is that the plane's heading is due south, and the wind's northwest direction must be accurately accounted for in the calculations.

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Homework Statement



A plane is heading due south and climbing at the rate of 60 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 110 km/hr to the northwest, what is the ground speed of the plane?

Homework Equations



Pythagorean theorem

The Attempt at a Solution



√(540^2+110^2). I tried adding different vectors using this method, but none of the results are correct.
 
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South and northwest do not make a right angle. you need to resolve each motion into suitable orthogonal components and add the components separately. Don't forget that some of the airspeed is upwards, and this won't count at all towards groundspeed.
 
Thanks for the reply. Now what?

I'm still confused about exactly how I break down the components. Am I at least using the right formula? Please no vague answers.
 
Well, if there are no other respondents:

It says the plane is "heading" south. In aircraft parlance that would mean that it's pointed south but not necessarily flying due south.

I see the problem solvable only if we assune the direction of the flight is due south.

So: let x = east, y = north, z = up. What is the x component of indicated airspeed IAS, given that the plane's x ground speed is v_px? Recall there is a wind blowing to the northwest which his IAS indicator will register.
and same question for indicated y airspeed if its y ground speed is v_py?
The how about combining those velocity components along with v_pz with the total airspeed to solve for v_px and v_py?

(Hint: what is v_px right off the bat?)
 
I don't know what the x component of the airspeed is. How would I find it?
 
PatrickE said:
I don't know what the x component of the airspeed is. How would I find it?

It's going to be the vector sum of the plane's airspeed without any wind, plus you have to add the x component of wind, right?
 
You mean the vector sum of 540 km/hr South? Since there are no degree measures in the problem, should I assume that it means exactly Northwest at 45 degrees?
 
It would help if I had any idea how to draw this scenario.
 
rude man said:
It says the plane is "heading" south. In aircraft parlance that would mean that it's pointed south but not necessarily flying due south.

I see the problem solvable only if we assune the direction of the flight is due south.
I agree that south is the direction relative to the wind. Not sure if that's what you meant by the last sentence above. But it's solvable with either interpretation.
The wind velocity is one vector, the velocity of the plane relative to the wind another. Just add the two (but discard the vertical component).
PatrickE: I don't know what the x component of the airspeed is. How would I find it?
X component of the windspeed. Airspeed means the speed of the plane relative to the air. If something is moving northwest at 1kph, how much further west is it after one hour?
 
  • #10
haruspex said:
I agree that south is the direction relative to the wind. Not sure if that's what you meant by the last sentence above. But it's solvable with either interpretation.

'Heading' is not direction wrt a wind direction. It's the direction of flight plus/minus the crab angle. The crab angle varies with aircraft aerodynamical properties, speed, etc.
 
  • #11
If you go northwest for an hour at 1 kph you should go about 0.7 km west, right?
 
  • #12
PatrickE said:
If you go northwest for an hour at 1 kph you should go about 0.7 km west, right?
Yes, or more precisely 1/√2 km. so if The wind is blowing at 110kph NW, what components does that velocity have in the north and west directions?
 
  • #13
rude man said:
'Heading' is not direction wrt a wind direction.

I stand corrected. But do we agree that that's probably how it's intended in the questioin?
 
  • #14
PatrickE said:
If you go northwest for an hour at 1 kph you should go about 0.7 km west, right?

Well, you're not really 'going' NW. That's the direction of the wind. You're 'going' south. But your IAS indicator registers wind as though it were the airplane flying; it can't tell the difference between wind and aircraft motion. They're both 'wind' to it.

For the IAS indicator, yes, if it's going south and the wind is nw then it will register +0.707 times the actual windspeed of 110 km/h, pplus the IAS due to the plane's own motion.

What you're trying to do is combine wind and aircraft motion to determine the aircraft ground speed.
 
Last edited:
  • #15
haruspex said:
I stand corrected. But do we agree that that's probably how it's intended in the questioin?

I don't know. I am assuming the plane continues on a due-south path. Like I said - could be wrong!

I do know this - an aircraft is not subject to wind motion the way a canoe is to moving water. So you can't just add wind and aircraft velocities the way you can with a boat and a moving river. In other words, there is no 'slippage' assumed with a boat but there is with a plane.
 
  • #16
rude man said:
Well, you're not really 'going' NW. That's the direction of the wind. You're 'going' south.
I was just using that to get PatrickE to figure out how to resolve the windspeed into components.
I don't know. I am assuming the plane continues on a due-south path. Like I said - could be wrong!

I do know this - an aircraft is not subject to wind motion the way a canoe is to moving water. So you can't just add wind and aircraft velocities the way you can with a boat and a moving river. In other words, there is no 'slippage' assumed with a boat but there is with a plane.
As I understand it, an airspeed indicator measures just that - the speed (and direction) of the plane relative to the air. I don't see how it could do anything else. Note that this is nothing to do with how the plane creates that motion. It could be a rocket; could be a glider pulled along by cable by a truck on the ground.
 
  • #17
haruspex said:
I was just using that to get PatrickE to figure out how to resolve the windspeed into components.

As I understand it, an airspeed indicator measures just that - the speed (and direction) of the plane relative to the air. .

Speed, yes. Direction, no.

It measures pitot vs. static pressure and determines the indicated airspeed from that.
 

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