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Originally Posted by Wizofoz
(Post 10216520)
Point being, I would say his ground speed has gone from 2kts to 7 knots. His ground velocity frim 2cknots North to 7 knots South (or whatever).
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Originally Posted by Meikleour
(Post 10203933)
Wizofoz: I agree with you however I think you will find that I never claimed lift was proportional to groundspeed! For the windshear analogy to apply the momentum must be related to the groundspeed ie. relative to the earth. Why is it therefore that when operating in the "bubble of air" concept then reference to the earth is discounted? Likewise can you help me understand the often observed phenomenon that I cited.
Because we measure wind velocity relative to the Earth! e.g. if you are heading 270 and you fly through a gradient from 270/10 kts to 270/30 kts then we have a +20 knot effect. The same as if you went from 090/5 to 270/15. It really doesn't matter what the absolute values are relative to the Earth, only what the change is. But, if you wanted to calculate it based on reported wind values,we measure wind relative to a stationary point on the ground. This might give the impression that the relationship to the ground is fundamentally important rather than being a book keeping convenience. |
Originally Posted by Jet_Fan
(Post 10216268)
no, it's -499 because he is facing the tail. If he stops it goes back to -500. If he turns to face the cockpit while stationary it goes from -500 to +500 almost instantly.
Groundspeed is a Scalar. An objects ground vector is comprised of its ground speed and its track. When an aircraft is heading directly into a wind that exceeds the TAS then it's track is the reciprocal of the heading. If you try and call that a negative ground speed you are taking account of the same thing twice. |
Originally Posted by Brercrow
(Post 10216330)
The difference between these two examples is that in one the windspeed is less than the airspeed and in the other the windspeed is greater than the airspeed.
In both cases airspeed is positive. In the first case the groundspeed reverses direction and in the second case the groundspeed is in the same direction. That means that in the first case you are entitled to take the difference between the groundspeeds but in the second case you have to take the sum. |
Originally Posted by Vessbot
(Post 10216463)
If groundspeed is only positive, then airspeed is always positive, and therefore airplane should fly just the same in a tailslide as they do going forward. True or untrue? +ve airspeed and Alpha(plus Beta for 3d) gives us a system that covers all the bases for where the air is coming from. |
Originally Posted by Capt Pit Bull
(Post 10216548)
This, and everything else resting upon it, is total bollocks.
Groundspeed is a Scalar. An objects ground vector is comprised of its ground speed and its track. When an aircraft is heading directly into a wind that exceeds the TAS then it's track is the reciprocal of the heading. If you try and call that a negative ground speed you are taking account of the same thing twice. And don't act as if this is esoteric or meaningless, as we, conventionally, do exactly that with airspeed as I've pointed out a few times by now. |
Originally Posted by Capt Pit Bull
(Post 10216556)
Well, AOA will be cracking on for 180 degrees. That might have a small part to play.....
+ve airspeed and Alpha(plus Beta for 3d) gives us a system that covers all the bases for where the air is coming from. |
Originally Posted by Capt Pit Bull
(Post 10216548)
This, and everything else resting upon it, is total bollocks.
Groundspeed is a Scalar. An objects ground vector is comprised of its ground speed and its track. When an aircraft is heading directly into a wind that exceeds the TAS then it's track is the reciprocal of the heading. If you try and call that a negative ground speed you are taking account of the same thing twice. |
Jet Fan, I think it's time me and you resign from this part of the project, and think up a new and different counterexample to demonstrate that initial and final ground velocity is irrelevant to accelerations that are measured and enacted only by reference to the uniform airmass... :ugh::{
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Originally Posted by Vessbot
(Post 10216574)
Jet Fan, I think it's time me and you resign from this part of the project, and think up a new and different counterexample to demonstrate that initial and final ground velocity is irrelevant to accelerations that are measured and enacted only by reference to the uniform airmass... :ugh::{
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Now it's your turn again to deal with the Cessna flying at 50 knots into a 60 knot headwind. What's his groundspeed? The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion (e.g. 60 km/h to the north) I'm a dumb winemaker and even I understand this. |
Originally Posted by Winemaker
(Post 10216598)
His ground speed is 10 knots. His ground velocity is 10 knots in the direction of the wind. Speed and velocity are two different things. A good definition of velocity is from Wikipedia, which I use for its clarity:
Please note that velocity must have a frame of reference and that's what's leading to all this BS. Define your frame of reference then do all the vector sums from that frame. You can't mix both the 'air' reference and the 'ground' reference. Use one or the other. There is no energy generated when a plane turns constant circles in a constantly moving air stream. The velocity of the reference air stream over the ground doesn't make one whit of difference to the aircraft. I'm a dumb winemaker and even I understand this. Ground speed can be determined by the vector sum of the aircraft's true airspeed and the current wind speed and direction; a headwind subtracts from the ground speed, while a tailwind adds to it. Winds at other angles to the heading will have components of either headwind or tailwind as well as a crosswind component. |
Originally Posted by Jet_Fan
(Post 10216611)
What is 50 - 60?
Ever since the 3rd grade I always wondered when in the world you'd ever use the | | symbol to strip out a negative, and I've finally found it! |
Originally Posted by Vessbot
(Post 10216622)
Apparently groundspeed = | airspeed - headwind |
Ever since the 3rd grade I always wondered when in the world you'd ever use the | | symbol to strip out a negative, and I've finally found it! This might help some in here: |
Thanks Vessbot, you said it for me.
Race Fan, I didn't say I loved Wiki, I said their definition was clear. |
Originally Posted by Jet_Fan
(Post 10216559)
...is just another way of saying its GS has become negative.
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Originally Posted by Capt Pit Bull
(Post 10216651)
NO. It really isn't. It might look like it, but if you just choose to invert the sign of something on a whim, expect difficulties if you apply that thought process to a navigation or reference system.
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Originally Posted by Winemaker
(Post 10216644)
Thanks Vessbot, you said it for me.
Race Fan, I didn't say I loved Wiki, I said their definition was clear. |
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? |
Can I propose a gedanken (thought experiment)?
A Cessna is flying into a 10 kt headwind and has ground speed 90 kts. Instantaneously the headwind increases to 20 kts. How long does it take for the Cessna's ground speed to attain 80 kts? I've always wanted to know. |
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