The Windward Turn Theory
We are all on the same page, but just someth8ng for the poitive_negative guys to mull- you are defining positive ground speed as being in the direction the aircaft is ponting- well, if there is any cross wind, the aircraft isn' t tracking that way- so is a teack 30 degrees off the nose " moslty positove "
answering the question on a 50 knot airspeed in a 60 knot headwind- how far will the aircaft be from its starting position after one hour?
answering the question on a 50 knot airspeed in a 60 knot headwind- how far will the aircaft be from its starting position after one hour?
Said what for you?
Speed is distance over time. Velocity is distance over time with a direction component, a vector. The ground speed of this slow flying Cessna is 10 knots; it may not be making any progress towards its intended target, but its speed over the ground is 10 knots. Its velocity vector is 10 knots in the direction of the wind. There's no 'negative' speed, there's speed in the direction opposite to the intended travel. If you want to plot this on an X -Y axis then you could certainly make the direction it ends up traveling as -X, the speed (magnitude) is still 10 knots, the direction is negative.
How long is a bit of string?
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Absolute value.
Speed is distance over time. Velocity is distance over time with a direction component, a vector. The ground speed of this slow flying Cessna is 10 knots; it may not be making any progress towards its intended target, but its speed over the ground is 10 knots. Its velocity vector is 10 knots in the direction of the wind. There's no 'negative' speed, there's speed in the direction opposite to the intended travel. If you want to plot this on an X -Y axis then you could certainly make the direction it ends up traveling as -X, the speed (magnitude) is still 10 knots, the direction is negative.
Speed is distance over time. Velocity is distance over time with a direction component, a vector. The ground speed of this slow flying Cessna is 10 knots; it may not be making any progress towards its intended target, but its speed over the ground is 10 knots. Its velocity vector is 10 knots in the direction of the wind. There's no 'negative' speed, there's speed in the direction opposite to the intended travel. If you want to plot this on an X -Y axis then you could certainly make the direction it ends up traveling as -X, the speed (magnitude) is still 10 knots, the direction is negative.
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You could measure that pretty easily. Start with the the the airplane stabilized in a 100 knot cruise, dive the airplane until the speed is above 110 knots, level out, start timing when the airspeed decreases through 110 knots, and stop timing when the airplane stabilizes at 100 knots again. That's how long it takes.
Aka the speed is negative 10
When you are discussing ground speed I assume you are thinking in aviation terms, but this whole thread is not about aviation terms, it's about somehow magically gaining energy when turning windward in a steady wind environment. This seems to me to indicate that we need to treat this as a physics problem, not an interpretation of 'ground speed' in the aviation sense.
So call the Cessna's ground speed -10 knots/hr if you wish, and I'll call it 10 knots/hr at angle Θ, but I think we mean the same thing.
So, if you have a groundspeed of negative thirty knots for an hour is your displacement negative thirty miles from your origin?
This would be the same as flying at 100 kts, then quickly reducing throttle to the thrust you know will sustain 90 knots.
From experience I would say a matter of 5-10ish seconds.
https://en.wikipedia.org/wiki/Speed
in everyday use and in kinematics, the speed of an object is the magnitude of its velocity (the rate of change of its position); it is thus a scalar quantity.
A position can't have a negative rate of change.
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OK,
Scenario B2 (really fast running person inside airliner. He runs at 501 knots and can turn around on a dime.)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(501+500)-(501-500)
1001-(1)
1000
Scenario B3 (really really fast running person inside airliner. He runs at 1000 knots and can turn around on a dime.)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(1000+500)-(1000-500)
1500-(500)
1000
Holy schnike, it turns out that if you substitute any number for the airspeed, the groundspeed difference is 1000, which is twice the wind speed when going from upwind to downwind.
Unless that number is less than the wind speed, then it becomes a special case for some reason that wraps people around the axle over scalars and quantum chromodynamics and whatever else.
Scenario B2 (really fast running person inside airliner. He runs at 501 knots and can turn around on a dime.)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(501+500)-(501-500)
1001-(1)
1000
Scenario B3 (really really fast running person inside airliner. He runs at 1000 knots and can turn around on a dime.)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
or, if you like:
(walk + tailwind) - (walk - headwind)
(1000+500)-(1000-500)
1500-(500)
1000
Holy schnike, it turns out that if you substitute any number for the airspeed, the groundspeed difference is 1000, which is twice the wind speed when going from upwind to downwind.
Unless that number is less than the wind speed, then it becomes a special case for some reason that wraps people around the axle over scalars and quantum chromodynamics and whatever else.
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Your one post up is just silly- turning your back does not change your velocity.
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Do you apply that same criticism to scenario A? If not, why not? What's the difference?
Scenario A (Cessna)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20
Do you think there's an error in the math? If so, what is it?
Do you apply that same criticism to scenario A? If not, why not? What's the difference?
Scenario A (Cessna)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20
Do you apply that same criticism to scenario A? If not, why not? What's the difference?
Scenario A (Cessna)
final groundspeed minus initial groundspeed
(airspeed + tailwind) - (airspeed - headwind)
(100+10)-(100-10)
110-90
20