Downwind turns equal disaster??
Hi All,
quote:
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In nil wind if I was flying thru the sky(space) heading south at 100 knots in a 206 at 1400 kg my momentum would be M X V = 140,000 kt/kg
Add a 50 kt tailwind I'm now flying thru the sky at 150 knots in a 206 at 1400 kg my momentum is 210,000kt/kg
The practical application from the pilots point of view is that in both cases he is doing 100kias. From a physics point of view the momentum of the aircraft would increase turning downwind & decrease when turning upwind and that can only be caused in this case by accelleration. Which means unballanced forces.
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As CSC rightly pointed out it always depends on the reference frame, e.g it depends on where you set your zero energy datum. Overpitched, you set yours fixed to the earth and thereby you are right in your calculations and with increased tailwind your energy will increase - with respect to your datum - the earth.
To illustrate, if you where to hit a mountain (which is bound to your reference point - the earth) you are quite right in saying that with a 50kt tailwind your impact will be a lot harder than without because your energy with regards to the reference point was increased.
In all energy systems we can chose the datum as we wish as long as we stick to this particular datum all the way. This is very important and I think the reason why you are a bit confused.
Lets look at speed vectors for a moment and to make it easier only at the speed vector along the aircrafts fligthpath at this moment in time.
I'll take your above example. You are travelling due south (although this has no relevance) at 100kts with nil wind. You are executing a rate one turn for 180 degrees to end up due North. Your speed vector is therefore changing from 100kts facing south to 100 kts facing north.
This means that in the turn, to overcome the forward momentum of the aircraft which obviously wants to keep moving in that direction (see Newton), the turning force (which is the changed Thrust vector at the rotor hub) has to decelerate the aircraft from 100 kts to zero speed (which would be at 90 degrees e.g either east or west depending on which way you go) and back to 100kts facing the other way. Remember I am only looking at the speed vector along the North-South flightpath axis and not at the aircraft airspeed! The deceleration to zero and back to 100kts equals an overall speed and therefore energy change of 200kts.
Now let's look at the 50kts tailwind scenario. Same principle applies. You start a rate one turn and this time (with reference point earth) the speed is 150kts. At 90 degrees (east or west) your speed is now not zero but 50kts going south due to the wind (still only looking at speed along North South axis) so in the first 90 degrees the speed has changed from 150kts to 50kts (100kts difference). The next 90 degrees of the turn changes the speed vector from 50kts south to 50 kts north at the end of the turn facing due north. This again equals a speed change of 100kts. So overall the speed and also energy change has been 200kts.
This shows that for the aircraft the same amount of energy is required to turn 180 degrees regardles of the wind. No extra power necessary at all. The only difference looking from our reference point (the earth) will be the shape and endpoint of the turn as we obviously will end up a lot further downwind in the second scenario.
I have recently tried this going around from a rig here in the north sea with winds of 65 kts and 75kts airspeed without any loss of height. If what you are suggesting would be true, the speed would have changed from 75kts to 10kts going downwind - below TL and the power requirement for that would have been very much higher.
Hope this helps, a drawing would probably clear things up a lot quicker. Maybe someone could do a nice graph?
Woolf
quote:
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In nil wind if I was flying thru the sky(space) heading south at 100 knots in a 206 at 1400 kg my momentum would be M X V = 140,000 kt/kg
Add a 50 kt tailwind I'm now flying thru the sky at 150 knots in a 206 at 1400 kg my momentum is 210,000kt/kg
The practical application from the pilots point of view is that in both cases he is doing 100kias. From a physics point of view the momentum of the aircraft would increase turning downwind & decrease when turning upwind and that can only be caused in this case by accelleration. Which means unballanced forces.
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As CSC rightly pointed out it always depends on the reference frame, e.g it depends on where you set your zero energy datum. Overpitched, you set yours fixed to the earth and thereby you are right in your calculations and with increased tailwind your energy will increase - with respect to your datum - the earth.
To illustrate, if you where to hit a mountain (which is bound to your reference point - the earth) you are quite right in saying that with a 50kt tailwind your impact will be a lot harder than without because your energy with regards to the reference point was increased.
In all energy systems we can chose the datum as we wish as long as we stick to this particular datum all the way. This is very important and I think the reason why you are a bit confused.
Lets look at speed vectors for a moment and to make it easier only at the speed vector along the aircrafts fligthpath at this moment in time.
I'll take your above example. You are travelling due south (although this has no relevance) at 100kts with nil wind. You are executing a rate one turn for 180 degrees to end up due North. Your speed vector is therefore changing from 100kts facing south to 100 kts facing north.
This means that in the turn, to overcome the forward momentum of the aircraft which obviously wants to keep moving in that direction (see Newton), the turning force (which is the changed Thrust vector at the rotor hub) has to decelerate the aircraft from 100 kts to zero speed (which would be at 90 degrees e.g either east or west depending on which way you go) and back to 100kts facing the other way. Remember I am only looking at the speed vector along the North-South flightpath axis and not at the aircraft airspeed! The deceleration to zero and back to 100kts equals an overall speed and therefore energy change of 200kts.
Now let's look at the 50kts tailwind scenario. Same principle applies. You start a rate one turn and this time (with reference point earth) the speed is 150kts. At 90 degrees (east or west) your speed is now not zero but 50kts going south due to the wind (still only looking at speed along North South axis) so in the first 90 degrees the speed has changed from 150kts to 50kts (100kts difference). The next 90 degrees of the turn changes the speed vector from 50kts south to 50 kts north at the end of the turn facing due north. This again equals a speed change of 100kts. So overall the speed and also energy change has been 200kts.
This shows that for the aircraft the same amount of energy is required to turn 180 degrees regardles of the wind. No extra power necessary at all. The only difference looking from our reference point (the earth) will be the shape and endpoint of the turn as we obviously will end up a lot further downwind in the second scenario.
I have recently tried this going around from a rig here in the north sea with winds of 65 kts and 75kts airspeed without any loss of height. If what you are suggesting would be true, the speed would have changed from 75kts to 10kts going downwind - below TL and the power requirement for that would have been very much higher.
Hope this helps, a drawing would probably clear things up a lot quicker. Maybe someone could do a nice graph?
Woolf
Flungdung - I agree, I think we've done this one to death now; unless overpitched is still not convinced!
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one more go - I'm pretty sure this won't work either...
Say you are in a C-5A flying along at 350 KT. What is the airspeed INSIDE the aircraft? If you were flying a little R/C helo inside the airplane, would it have a tailwind? Of course not. It would fly exactly the same inside the moving C-5A as it would inside a parked C-5A.
Wow - I don't really even get that one
Oh well, I'm done.
Say you are in a C-5A flying along at 350 KT. What is the airspeed INSIDE the aircraft? If you were flying a little R/C helo inside the airplane, would it have a tailwind? Of course not. It would fly exactly the same inside the moving C-5A as it would inside a parked C-5A.
Wow - I don't really even get that one
Oh well, I'm done.
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You're right flingwing.. not even close to working. But nice try
Crab. As much fun as this has been I think it was finished a couple of days ago However I would like to say in closing....
I feel my arguements were logical, well researched, well presented and also very convincing and if it was anybody else other than Nick lappos whom everybody knows ....
1. Is a guru &
2. Knows heaps more about this crap than me &
3. Has a cooler job.
Then I feel I would have won many supporters
And if you really haven't had enough try the plane and the wind
http://www.physicsclassroom.com/Clas...ors/u3l1f.html
Crab. As much fun as this has been I think it was finished a couple of days ago However I would like to say in closing....
I feel my arguements were logical, well researched, well presented and also very convincing and if it was anybody else other than Nick lappos whom everybody knows ....
1. Is a guru &
2. Knows heaps more about this crap than me &
3. Has a cooler job.
Then I feel I would have won many supporters
And if you really haven't had enough try the plane and the wind
http://www.physicsclassroom.com/Clas...ors/u3l1f.html
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...do you notice that in the website you posted, the velocities are always described as "relative to the ground", "velocity of the plane relative to an observer on the ground", "its speed with respect to an observer on the shore", and so on? No mention of velocity with respect to the air rather than airspeed, always referred to an observer "on the ground".
Any bells going off?
Anything?... ...Anybody?... ...Beuler? Ferris Beuler?
Any bells going off?
Anything?... ...Anybody?... ...Beuler? Ferris Beuler?
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Yeah you're right flingwing, as a matter of fact I took my boat out for a sail this morning and made the silly mistake of turning down wind and down current at the same time and as you pointed out the boat maintained its speed relative to the water and the sails maintained their speed relative to the wind. If only the current and wind were exactly the same I guess the sails would still be on the deck ????
Ever think that the site refers to an observer in a position fixed in space separate to the system you are observing because its relevant !
Ever think that the site refers to an observer in a position fixed in space separate to the system you are observing because its relevant !
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When you are IMC you do not notice the wind (unless it is gusty or turbulent) - it is only when you can see a fixed point that you notice the wind. The helicopter still flies in IMC / at night and its characteristics dont change the moment you enter a cloud. Therefore using the ground as a reference for aircraft aerodynamics is irrelevant, in the same way that the seabed is irrelevant in ship design.
overpitched:
"Ever think that the site refers to an observer in a position fixed in space separate to the system you are observing because its relevant !"
You need to understand that in the circumstances we originally started with - downwidn turns - it IS NOT relevant. Once you've got that hurdle over with, the rest is easy.
"Ever think that the site refers to an observer in a position fixed in space separate to the system you are observing because its relevant !"
You need to understand that in the circumstances we originally started with - downwidn turns - it IS NOT relevant. Once you've got that hurdle over with, the rest is easy.
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overpitched,
This thing about reference frames has you boggled, but let me try this:
In this case, there is no absolute energy or momentum. It is all relative to the reference frame you chose. Since you calculate the energy/momentum in one frame, you get confused and wonder where the extra energy/momentum came from when downwind vs upwind. You forget that the wind is a big contributer to the energy/momentum of the situation, perhaps because the air is so clear it seems to not be there.
If you are in a balloon drifting in a 50 knot wind, and you bump into a traffic sign, the energy/momentum is enormous, and you might be killed. If you bump into another balloon, the energy/momentum is small, and you say "Pardon me!". Why? Because these quantities are relative, not absolute, and are dependant on the reference frame you decide to use.
You have decided to stay in the earth reference frame, and calculate all energy/momentum that way. That is fine, just don't mix the numbers to suit your theory (which is that a turn downwind must cause a big performance problem). Where you get an F in physics is that you ignore the enormous energy pouring through your system from the wind.
If we take a 100 foot square of surface, and pour a 30 knot wind through it, we have a constant energy source of about 90 Horsepower, enough to entirely power a small airplane. But you simply ignore this energy/momentum contribution, and have done so for about 10 of your posts! Where does the aircraft get the ""extra" energy/momentum as it turns downwind? From the Wind!
This thing about reference frames has you boggled, but let me try this:
In this case, there is no absolute energy or momentum. It is all relative to the reference frame you chose. Since you calculate the energy/momentum in one frame, you get confused and wonder where the extra energy/momentum came from when downwind vs upwind. You forget that the wind is a big contributer to the energy/momentum of the situation, perhaps because the air is so clear it seems to not be there.
If you are in a balloon drifting in a 50 knot wind, and you bump into a traffic sign, the energy/momentum is enormous, and you might be killed. If you bump into another balloon, the energy/momentum is small, and you say "Pardon me!". Why? Because these quantities are relative, not absolute, and are dependant on the reference frame you decide to use.
You have decided to stay in the earth reference frame, and calculate all energy/momentum that way. That is fine, just don't mix the numbers to suit your theory (which is that a turn downwind must cause a big performance problem). Where you get an F in physics is that you ignore the enormous energy pouring through your system from the wind.
If we take a 100 foot square of surface, and pour a 30 knot wind through it, we have a constant energy source of about 90 Horsepower, enough to entirely power a small airplane. But you simply ignore this energy/momentum contribution, and have done so for about 10 of your posts! Where does the aircraft get the ""extra" energy/momentum as it turns downwind? From the Wind!
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Sorry folks, don't mind me, this is all to technical for my fragile little mind, however way way back Bertie Thruster asked
My take on that is, compare the movement of air to the movement of the sea, if you ever sail into wind you are going to hit the wave head on and ramp over, but if you are sailing downwind its seems very smooth beacuse the wave is catching up with you, and you are basically surfing down wave,
Won't butt in on the curent debate, I am of the opinion of just be sensible be aware and just fly the damn thing.
Why is it always bumpier flying (at the same IAS) into wind than it is going downwind?
Won't butt in on the curent debate, I am of the opinion of just be sensible be aware and just fly the damn thing.
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I'm sorry, I just can't resist sticking my oar in here.... 
Overpitched, here's another thought experiment for you... imagine you're flying in calm air (ie no wind), circling round and round at constant IAS, bank, etc etc. On the ground below you, as far as the eye can see in all directions, is the world's biggest conveyor belt, which is currently switched off. An observer is standing on the conveyor belt directly underneath the centre of your circles.
Now, what happens if, just as you're turning through North, we turn on the conveyor belt and it starts moving North at, say 25kt? The ground observer and everything else you can see now appears to be moving North, or (more likely) you will believe yourself to be drifting South. Don't touch the controls! Would you expect to feel anything unusual through the seat of your pants, or see anything unusual on the instruments, as you continue circling?
I hope you'll say "no". Now, why not?
cbl.
Overpitched, here's another thought experiment for you... imagine you're flying in calm air (ie no wind), circling round and round at constant IAS, bank, etc etc. On the ground below you, as far as the eye can see in all directions, is the world's biggest conveyor belt, which is currently switched off. An observer is standing on the conveyor belt directly underneath the centre of your circles.
Now, what happens if, just as you're turning through North, we turn on the conveyor belt and it starts moving North at, say 25kt? The ground observer and everything else you can see now appears to be moving North, or (more likely) you will believe yourself to be drifting South. Don't touch the controls! Would you expect to feel anything unusual through the seat of your pants, or see anything unusual on the instruments, as you continue circling?
I hope you'll say "no". Now, why not?
cbl.
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From a scientific point of view it doesn't matter what reference system you use, you should always be able to satisfy the laws of physics and get the same answer.
The simplest and most useful reference system in aviation is that parcel of air that an aircraft flies in. You could also use a point on the ground or even infinity as the origin of your reference system to analyse aircraft aerodynamics, but the maths is more complicated.
Take for example the police radar trap. The radar measures the velocity at which you are approaching the gun ie gives you a direct measurement of your speed. However, if you could measure your velocity relative to the centre of the earth and compare it with the velocity of the policeman relative to the same point you will get the same result - a £60 fine and three points!
But Overpitched has decided to use a point on the earth for his (can I assume that you are male?) reference point. Nothing wrong with that, except that you will get different answers because you are looking at momentums and energies associated with the aircraft and ground interacting - something we all try to avoid. I think what is puzzling Overpitched is that there is a change in the magnitude of aircraft momentum between up and downwind without a force to generate that change. That got me puzzled too, but I think I have the answer:
Lets assume an aircraft speed of 100kts, a wind speed of 30 knots and an aircraft weight of 1000kg. The reference point is an observer on the ground.
If the aircraft is flying directly into the wind, the ground speed is 70kts, giving momentum RELATIVE TO the Earth of 70,000kt.kg. In the same way down wind the momentum is 130,000kt.kg. But why the difference - how has the momentum changed by 60,000kt.kg by turning down wind without any extra forces acting on the system? We have to remember that we are using the earth as a reference system so when we apply aerodynamic theory to the aircraft we have to do it relative to the ground.
If we maintain a constant total rotor thrust vector by maintaining power and disk attitude, the only other horizontal force acting on a helicopter is air resistance, which is depends on velocity of the air flowing over and around the airframe. When using the air as a reference system, TAS is used to calculate skin friction and form drag. But we are using the earth as a reference system so we have got to account for the motion of the aircraft reative to the air and the air relative to the ground.
Into wind, the skin friction and form drag (relative to the earth) are calculated from the speed of the aircraft through the air PLUS the speed of the air over the ground (wind). In this case 100 + 30 = 130 kts.
Down wind, the skin friction and form drag (relative to the earth) are calculated from the speed of the aircraft through the air MINUS the speed of the air over the ground (wind). In this case 100 - 30 = 70 kts.
Hence the skin friction RELATIVE TO THE EARTH changes and that accounts for the change in the magnitude of momentum when using the ground as a reference system.
It is interesting in the above example that regardless of the reference system you use, the change in momentum for a 180 degree turn is the same - 200,000kt.kg. That's because in the turn the momentum is changed by the centripetal force acting towards the centre of the turn. This centripetal force is generated by the horizontal component of total rotor thrust and is therefore independent of wind or ground.
It makes sense to me, but I am still going to duck for incoming!
The simplest and most useful reference system in aviation is that parcel of air that an aircraft flies in. You could also use a point on the ground or even infinity as the origin of your reference system to analyse aircraft aerodynamics, but the maths is more complicated.
Take for example the police radar trap. The radar measures the velocity at which you are approaching the gun ie gives you a direct measurement of your speed. However, if you could measure your velocity relative to the centre of the earth and compare it with the velocity of the policeman relative to the same point you will get the same result - a £60 fine and three points!
But Overpitched has decided to use a point on the earth for his (can I assume that you are male?) reference point. Nothing wrong with that, except that you will get different answers because you are looking at momentums and energies associated with the aircraft and ground interacting - something we all try to avoid. I think what is puzzling Overpitched is that there is a change in the magnitude of aircraft momentum between up and downwind without a force to generate that change. That got me puzzled too, but I think I have the answer:
Lets assume an aircraft speed of 100kts, a wind speed of 30 knots and an aircraft weight of 1000kg. The reference point is an observer on the ground.
If the aircraft is flying directly into the wind, the ground speed is 70kts, giving momentum RELATIVE TO the Earth of 70,000kt.kg. In the same way down wind the momentum is 130,000kt.kg. But why the difference - how has the momentum changed by 60,000kt.kg by turning down wind without any extra forces acting on the system? We have to remember that we are using the earth as a reference system so when we apply aerodynamic theory to the aircraft we have to do it relative to the ground.
If we maintain a constant total rotor thrust vector by maintaining power and disk attitude, the only other horizontal force acting on a helicopter is air resistance, which is depends on velocity of the air flowing over and around the airframe. When using the air as a reference system, TAS is used to calculate skin friction and form drag. But we are using the earth as a reference system so we have got to account for the motion of the aircraft reative to the air and the air relative to the ground.
Into wind, the skin friction and form drag (relative to the earth) are calculated from the speed of the aircraft through the air PLUS the speed of the air over the ground (wind). In this case 100 + 30 = 130 kts.
Down wind, the skin friction and form drag (relative to the earth) are calculated from the speed of the aircraft through the air MINUS the speed of the air over the ground (wind). In this case 100 - 30 = 70 kts.
Hence the skin friction RELATIVE TO THE EARTH changes and that accounts for the change in the magnitude of momentum when using the ground as a reference system.
It is interesting in the above example that regardless of the reference system you use, the change in momentum for a 180 degree turn is the same - 200,000kt.kg. That's because in the turn the momentum is changed by the centripetal force acting towards the centre of the turn. This centripetal force is generated by the horizontal component of total rotor thrust and is therefore independent of wind or ground.
It makes sense to me, but I am still going to duck for incoming!
Overpitched - does an aeroplane wing stall at a specific groundspeed? No but it does at a specific airspeed. If an aircraft flys past you (you are on the ground not moving) at an indicated 60 kts airspeed into a 20 kts head wind, it will fly past you at 40 kts GS. If it turns round and flys past with a 20 kt tailwind (still with 60 kts IAS) it will fly past you at 80 kts GS. As far as you are concerned the plane has twice the speed and therefore momentum because you are stationary on the ground. As far as the pilot of the plane is concerned he is flying at 60 kts airspeed both ways and the plane behaves in exactly the same manner (same power setting, attitude etc). His reference is airspeed (which is moving) and yours is a fixed point on the ground, which is not.
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Crashondeck,
Your last paragraph is the correct and complete solution to the mystery - I'd delete the rest if I were you...
In your 30kt wind example, you have missed the fact that momentum is a vector quantity. The change in momentum is from 70,000kt.kg in one direction to 130,000kt.kg in the opposite direction. The change is therefore:
70,000 - (-130,000) = 200,000
not 60,000 as you stated. If you change the example to a 30kt tailwind, you get a change from 130,000 to -70,000:
130,000 - (-70,000) = 200,000
As you stated in your last paragraph, the change in momentum is completely independent of the reference frame - you could be moving with the wind, fixed on the ground, or watching from a passing train - the change in the aircraft's momentum will always be 200,000kt.kg, and is the result of the same forces (tilted lift vector) acting in the same directions over the same periods of time, regardless of what the air is doing relative to the ground.
A bit of algebra may help further: stick with a 1000kg aircraft flying at 100kt IAS in a headwind of W knots:
Starting momentum = 1000 x (100 - W) = 100,000 - 1000W
After the 180 turn = 1000 x (-100 - W) = -100,000 - 1000W
The change is therefore:
(100,000 - 1000W) - (-100,000 - 1000W)
The two 1000Ws cancel out, leaving just
100,000 - (-100,000) or 200,000!
There is no "W" left in the answer, which is the key - the wind has no effect.
cbl.
Your last paragraph is the correct and complete solution to the mystery - I'd delete the rest if I were you...
In your 30kt wind example, you have missed the fact that momentum is a vector quantity. The change in momentum is from 70,000kt.kg in one direction to 130,000kt.kg in the opposite direction. The change is therefore:
70,000 - (-130,000) = 200,000
not 60,000 as you stated. If you change the example to a 30kt tailwind, you get a change from 130,000 to -70,000:
130,000 - (-70,000) = 200,000
As you stated in your last paragraph, the change in momentum is completely independent of the reference frame - you could be moving with the wind, fixed on the ground, or watching from a passing train - the change in the aircraft's momentum will always be 200,000kt.kg, and is the result of the same forces (tilted lift vector) acting in the same directions over the same periods of time, regardless of what the air is doing relative to the ground.
A bit of algebra may help further: stick with a 1000kg aircraft flying at 100kt IAS in a headwind of W knots:
Starting momentum = 1000 x (100 - W) = 100,000 - 1000W
After the 180 turn = 1000 x (-100 - W) = -100,000 - 1000W
The change is therefore:
(100,000 - 1000W) - (-100,000 - 1000W)
The two 1000Ws cancel out, leaving just
100,000 - (-100,000) or 200,000!
There is no "W" left in the answer, which is the key - the wind has no effect.
cbl.
Last edited by CBLong; 26th Feb 2004 at 23:52.
CBL, but by the time you've worked all that out, you have hit the ground because you didn't maintain the airspeed as you turned downwind!
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Downwind turns equal bull sh#t
O.K. Enough already.
We went flying yesterday and I forgot to take my calculator along. We were flying into a reasonable head wind and I was too scared to turn as I couldn't figure out if it was going to be safe to do so. Ended up coming back by train.
Can we just get on with something not so scary so those of us that are not mathematicians can get back into the air without being petrified with the apparent new physics laws.
STL
We went flying yesterday and I forgot to take my calculator along. We were flying into a reasonable head wind and I was too scared to turn as I couldn't figure out if it was going to be safe to do so. Ended up coming back by train.
Can we just get on with something not so scary so those of us that are not mathematicians can get back into the air without being petrified with the apparent new physics laws.
STL
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So, let me get this right.
Nearly everyone is saying that if you stay in the same airmass, and it happens to be moving across the earth, that you won't see any acceleration or deceleration as you do a level turn.
So, your inertial navigation system wouldn't register any change then.
Better tell this to all the people using inertial navigation systems...
Nearly everyone is saying that if you stay in the same airmass, and it happens to be moving across the earth, that you won't see any acceleration or deceleration as you do a level turn.
So, your inertial navigation system wouldn't register any change then.
Better tell this to all the people using inertial navigation systems...
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Oh Shawn! Of course your inertial will show the acceleration, it has a little window, and the guy inside peeks out to see.
Last edited by NickLappos; 27th Feb 2004 at 06:40.
