# Inertia and flying

• B
• Mejeko
Mejeko
TL;DR Summary
Landing a plane which flies E to W on a N to S runway at the equator violates Newton's first law.
As someone who only knows elementary physics (so pardon me for maybe getting some things wrong), I have a question which troubles me and I'm having difficulties in finding an answer to:

If a plane takes off at the equator and flies east to west, counter to earth's rotation, how it would be able to land on a north to south runway considering that it has lost its initial angular momentum which it had from the ground? Wouldn't the runway be moving sideways and the atmosphere blow laterally, making a 90 degrees maneuver impossible without the plane flying in the opposite direction in order to regain its initial lost momentum that it had from the ground before takeoff? Doesn't this type of landing violate Newton's first law because the thrust of the engines would act against its initial momentum which the plane can never recover without flying west to east, so that it may be able to align again with earth's rotational speed and its atmosphere?

Newton’s first law says that a body moves in a straight line at constant speed if it is not subject to an external force. A plane is subject to external forces.

Dale said:
Newton’s first law says that a body moves in a straight line at constant speed if it is not subject to an external force. A plane is subject to external forces.
So, the way I understand it is that the plane moves, while on a ground, at a constant speed of 1000 miles/h, west to east, relative to earth's axis. After it takes off, the engines act as an external force, lowering its initial momentum (that would be 1000 miles/h minus the planes minimum speed required to stay in the air eg. 150 miles/h). The plane could never regain the lost 150 miles/h without flying in the opposite direction.

Mejeko said:
The plane could never regain the lost 150 miles/h without flying in the opposite direction.
It doesn't need to fly in the opposite direction, just have some force applied that points in the opposite direction. That it gets as it turns the corner, banking and applying rudder

Ibix said:
It doesn't need to fly in the opposite direction, just have some force applied that points in the opposite direction. That it gets as it turns the corner, banking and applying rudder
Sorry, but I don't understand... How could such a force point in the opposite direction without a 180 degrees turn? In a 90 degrees maneuver the plane would still maintain a west to east inertia of 850 miles/h.

Mejeko said:
How could such a force point in the opposite direction without a 180 degrees turn?
Lift points downwards from the plane. If it tips over the force has a horizontal component inwards - which is the centripetal force pushing the aircraft in a circle. Doing a quarter circle turn takes the aircraft from an east/west heading to a north/south one.

You'd be correct if this were a rocket in vacuum. The rocket would have to point at least slightly backwards in order to make the turn. But the aircraft can also exert forces against the air, and vice versa.

Dale
Mejeko said:
##\dots## how it would be able to land on a north to south runway considering that it has lost its initial angular momentum which it had from the ground?
Can you be a bit more specific about this angular momentum that the plane loses? What is the initial expression for it and with respect to what point?

Ibix said:
Lift points downwards from the plane. If it tips over the force has a horizontal component inwards - which is the centripetal force pushing the aircraft in a circle. Doing a quarter circle turn takes the aircraft from an east/west heading to a north/south one.
I'm sorry, I'm having issues wrapping my head around this. The engines still thrust the plane east to west before attempting the maneuver, so I fail to see how at any moment the plane can regain the lost 150 miles/h west to east inertia. Newton's first law says that the plane would continue its 850 miles/h west to east momentum until a force acts (as a vector?) in the opposite direction in order to achieve again the 1000miles/h it had at the start. A 90 degrees turn, even if possible (for a fighter jet) with a 150 miles/h lateral wind of the atmosphere (which still rotates at the same speed with the earth), would have no effect of its lost inertia, unless this lateral wind pushes on the side of the plane, drifting it from 150 to 0 (which I'm not if it's possible for a commercial plane to sustain such sideways wind speeds).

kuruman said:
Can you be a bit more specific about this angular momentum that the plane loses? What is the initial expression for it and with respect to what point?
I'm referring to the velocity it has relative to earth's axis of 1000 miles/h and how it will lose some of it after takeoff. I'm having troubles figuring out how it can regain it without rolling over due to lateral winds in a 90 degrees turn.

Mejeko said:
due to lateral winds
What lateral winds? If it's a calm day, i.e. no wind is blowing on the planes parked on the tarmac, the only "wind" on the landing plane is generated by air resistance as the plane moves forward.

Mejeko said:
so I fail to see how at any moment the plane can regain the lost 150 miles/h west to east inertia.
The air is moving 1000 miles/hour west to east. If the plane simply turns right to a heading due north, the moving air will carry it eastward at the requisite 1000 miles/hour.

Note that the north/south runway is also slewing to the east at 1000 miles/hour. So a heading due north will result in a course (relative to the non-rotating earth-centered frame) that is mostly eastward.

Those of us who do physics routinely will automatically shift to a more natural frame of reference where the surface of the earth is at rest and everything is easy to understand.

The distinction between course and heading is natural for sailors.

Lnewqban and Mejeko
Mejeko said:
I'm referring to the velocity it has relative to earth's axis of 1000 miles/h and how it will lose some of it after takeoff.
The atmosphere travels with the runway and the Earth's surface that is the reference. An aircraft can fly in a circle.

jbriggs444 said:
The air is moving 1000 miles/hour west to east. If the plane simply turns right to a heading due north, the moving air will carry it eastward at the requisite 1000 miles/hour.
So, I guess my question would be better addressed to an engineer who works with commercial planes, that is, if it's possible for such an airplane to sustain lateral winds of over 150 miles/h without being flipped over, until it gets again to 0 relative to earth's surface moving west to east.

Motore
Mejeko said:
The engines still thrust the plane east to west before attempting the maneuver
The engines are not the only force on the plane. In fact, the engines are not even the largest force. Commercial planes have more lift and more weight than thrust.

The aerodynamic forces are very large. They are completely sufficient to turn the aircraft.

Mejeko said:
if it's possible for such an airplane to sustain lateral winds of over 150 miles/h without being flipped over, until it gets again to 0 relative to earth's surface moving west to east
That simply doesn’t happen in normal commercial flights.

Mejeko said:
So, I guess my question would be better addressed to an engineer who works with commercial planes, that is, if it's possible for such an airplane to sustain lateral winds of over 150 miles/h without being flipped over, until it gets again to 0 relative to earth's surface moving west to east.
Aircraft do not turn instantaneously. They take time to turn.

Here's a picture of an aeroplane flying straight towards you.

I've added the lift vector in red, pointing straight up, and the weight in green pointing straight down. Now the aeroplane wants to make a right turn, so it tips its wings:

The weight vector remains pointing straight down, but the lift is generated by the wings, and remains perpendicular to them. If you add those two force vectors you get a net force to the aeroplane's right (left on the diagram because the plane is coming out of the page).

Now let's draw a sequence of images of the plane making its turn.

This time I haven't drawn the lift and the weight, just the resultant vector pointing inwards in the turn. Notice that, after the very first frame, all of those blue arrows point slightly backwards relative to the initial direction of the plane. That's the force that reduces its groundspeed in the east-west direction to zero. It comes from the aircraft's wings pushing on the air - it does not come from the engines. Their thrust is solely used here to counteract drag.

Lnewqban, Mejeko and Dale
Mejeko said:
if it's possible for such an airplane to sustain lateral winds of over 150 miles/h without being flipped over
The plane does not make a 90 degree right hand turn on a dime. It does so over time. About 30 seconds for a small plane making a 90 degree right turn at the standard rate.

During the turn, the plane is banking so that the turn is coordinated. There is no lateral airflow across the plane's fuselage during the turn. No side slip. The pilot usually needs to use both rudder and ailerons to maintain coordination, though the plane may naturally pinwheel without much rudder. The pilot may also manipulate power and elevator to maintain speed and altitude. A slight increase to power and pitch up from the elevator.

The "ailerons" are the control surfaces on the trailing edge of the outer part of the wing. One on each wing. They control right and left roll. When used, one is shifted up and the other is shifted down to match.

The "rudder" is the control surface on the trailing edge of the vertical part of the tail (that is, on the trailing edge of the vertical stabilizer). It controls right and left yaw.

The "elevator" is the pair of control surfaces on the trailing edge of the horizontal part of the tail (that is, on the trailing edge of the horizontal stabilizer). It controls up and down pitch.

The classic control layout is with a stick for elevator and ailerons. Right on the stick to roll right. Left to roll left. Push to pitch down. Pull to pitch up. Then pedals for right and left rudder.

It is possible to mismatch the roll and yaw inputs and force a loss of coordination. A slip. In a slip, the plane is banked right (for instance) while the rudder is commanding a left turn. The result is that the plane will tend to slip to the downhill side. The airflow will then run across the fuselage at an angle. This can be a useful way to lose energy, for instance, when approaching the runway with a glider.

I heard a second or third hand story some 50 years ago about a glider pilot whose spoilers were not working. He slipped as much as he could to lose energy and altitude. But once he was almost down and riding on ground effect, he couldn't slip any more without touching a wing tip in the grass. I think he just flew the thing down the last few feet and landed way too fast. I do not remember how it came out.

I am not a pilot. Just someone who has paid attention a few times.

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Dale, Lnewqban and Mejeko
Mejeko said:
I'm sorry, I'm having issues wrapping my head around this. The engines still thrust the plane east to west before attempting the maneuver, so I fail to see how at any moment the plane can regain the lost 150 miles/h west to east inertia.
It's just aerodynamic drag. The atmosphere is moving with the rotation of the Earth, so to a plane flying it is no different from if the earth were stationary.

It might help if you drew diagrams, labeling the motions and forces.

Ibix said:
Now let's draw a sequence of images of the plane making its turn.
View attachment 342633
This time I haven't drawn the lift and the weight, just the resultant vector pointing inwards in the turn. Notice that, after the very first frame, all of those blue arrows point slightly backwards relative to the initial direction of the plane. That's the force that reduces its groundspeed in the east-west direction to zero. It comes from the aircraft's wings pushing on the air - it does not come from the engines. Their thrust is solely used here to counteract drag.
So, the plane starts to regain the lost momentum during this maneuver where the force points back to the eastward direction. Now it makes sense that I visualize it.

Thanks to everyone who has chipped in. My understanding of how force distribution works for an airplane was lacking some critical points :)

WWGD, Ibix, Dale and 1 other person
Mejeko said:
So, the plane starts to regain the lost momentum
This doesn't make sense to me. A plane (any object) will only 'lose' / change momentum when an external force acts on it. Where does your lost momentum come from?

The airfield is batting along very fast as the Earth spins. When a balloon lifts off, it is stationary above a point on the ground and so is the (still) air around it. After a while, the balloon will still be in the same spot above the ground no change in momentum for anyone. The balloon will not be left behind. Now consider a plane, taking off in any direction. It will have momentum, added by the engines and will encounter stationary air as it flies through it - giving momentum to the air (work against the drag force). The engines will make up for this if the plane wants to stay up. All this is the same, whichever direction the plane takes of in. Everything local can be regarded as stationary and the arguments involve just relative forces and velocities.

Mejeko said:
My understanding of how force distribution works for an airplane was lacking some critical points :)
No one finds this easy. You or I would fly a plane into the ground with no difficulty because we don't know 'the rules'. The only planes that use just the engines to change direction are high performance fighters which can use vector thrust by firing the engines in a chosen direction. This is very sexy and wins dogfights but costs a vast amount in fuel, compared with what the normal control surfaces achieve.

Mejeko said:
I'm referring to the velocity it has relative to earth's axis of 1000 miles/h and how it will lose some of it after takeoff.
It's not relevant because the plane is not navigating relative to Earth's axis. It's navigating relative to Earth's surface. So it's velocity relative to Earth's surface is all that's relevant.

Now, when a spacecraft is launched into orbit its velocity relative to Earth's center is relevant. That's why engineers launch them eastward from a low latitude.

Mejeko said:
I'm referring to the velocity it has relative to earth's axis of 1000 miles/h and how it will lose some of it after takeoff.
You need to re-think this. It doesn't work this way.
There is an effect (see Coriolis) when you try to travel in directions to the North or South on a rotating surface. (Like a kids' roundabout or - more complicated- over the Earth's surface. This is because you carry your original momentum with you and need to steer (add another force) to follow a wanted straight line course. Could that be something that's been bothering you?
It is a factor that affects long distance flying (along with all the other navigation calculations). If you can find yourself a kid's playground, you can experience it on simple rides. That's if you can get away without being plagued with protective parents!!. Best to take your own kids with you. (I know that's not funny in the present climate - very sad.)

sophiecentaur said:
There is an effect (see Coriolis) when you try to travel in directions to the North or South on a rotating surface.
Coriolis also has an effect on east-west motion. It can be more difficult to detect since it manifests as a very slight increase or decrease in apparent gravity.

The range corrections (about 107 yards over on an due east shot of 26,000 yards with an 8"/55 at the equator and about 107 yards under on a westward shot with the same gun at the same location) listed on page 449 here arise from this source. Remember, this is also why you launch space craft on an eastward trajectory and use a tropical launch point if you can.

The left/right deflections on north/south shots in the temperate latitudes are comparable to this in magnitude. Those are on page 452.

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