Understanding Newton's Laws in Aviation

I'm sure you're all familiar with Newton's laws. The first law states that the body at rest remains at rest, and the body in motion continues to move at a constant velocity along a straight line unless acted upon by an external force. Now this law only applies in an inertial frame of reference, or, in other words, a frame of reference that does not accelerate. Now imagine being in a train with all curtains closed, so you cannot look outside. From your own perspective, everything inside the train is at rest. However, if there would be a ball on the ground, and the train would suddenly have to brake, then the ball will start to move forward. But, there is no external force acting upon it. In fact, the ball just wants to continue moving forwards, whereas the train is stopping. Similarly, if the train starts to turn, the ball will start moving either to the left or to the right. And again, no external force is acting upon the ball. Since the train was experiencing an acceleration in both examples, the frame of reference was not an inertial frame of reference, and thus Newton's first law does not apply. This is very important to realize when we derive the equations of motion for an actual aircraft. We should clearly define the different frames of reference. A logical point to start with is the Earth's surface. So we can start defining a reference frame fixed to the Earth. This reference frame is called the Earth axis system. Of course we are interested in the motion of the aircraft relative to this frame. So is this frame of reference an inertial frame of reference or not? Actually it isn't. If you are flying at a constant altitude above the Earth's surface, you are actually moving in a circle. Thus an acceleration to the center of the circle is required in order to create that motion. So the frame of reference is accelerating and it is not an inertial frame of reference. As a result there appears to be a centrifugal force which works in the outward direction perpendicular to the circular motion. Realize that this is not a real force but an apparent force due to the force of the centrifugal force. just like in the train example. The question now is, how large is this acceleration? From high school physics we already know that the acceleration equals V squared over r. Most aircraft have a cruise altitude of about 10 kilometers, so the radius in the equation equals the radius of the Earth plus 10 kilometers. The radius is about 6000. kilometers and that results in a total radius of 6381 kilometers. A typical velocity for a commercial aircraft at this altitude is about 250 meters per second or Mach 0.8. So the acceleration actually equals 250 squared divided by the radius squared and that yields 0.0097 meters per second squared. Now realize that this is only a single plane 0.1% of the gravitational acceleration. So even though the acceleration is present, it is so small that we can neglect it and make what we call the flat earth assumption. Now there is another effect due to the earth, it is in fact rotating. So the velocity of a point on the earth's surface depends on the latitude, because the radius to the axis of the earth's rotation is in fact changing. So, if a person is moving in a north-south direction, the velocity actually changes as well. Hence there is an acceleration and it is not an inertial frame of reference. This can also actually be visualized from the perspective of somebody on a roundabout. The ball moves in a curved motion. It should be noted, however, that the accelerations related to the fact that the earth is rotating are very small. They can be neglected for most airplane motions. They can, however, become important when high-altitude, high-speed vehicles are of interest. So, if we assume that the earth is flat and not rotating, we can safely state that the earth-fixed reference frame is an inertial frame of reference. Now let's define some more useful reference frames. The forces acting on the aircraft depend on its orientation and its motion relative to the air. So it would be useful to define reference frames attached to the aircraft. So we can start by defining a reference frame with the same orientation as the original earth-fixed frame, but now attached to the aircraft itself. This is called the moving earth axis system. Now the nose of the aircraft is actually pointing in a different direction. We can therefore also attach an access system. fixed to the aircraft body, which we call the body axis system. Now when the pilot is looking outside, he or she is looking along the x-axis of the body axis system and can directly observe the pitch attitude of the aircraft, which we call theta, with respect to the horizon. So, summarizing, we have a reference frame from the point of view of the aircraft attitude and we have two frames relative to the Earth. However, Since we are interested in the motion of the aircraft, there is one final element of interest, and that is the direction of the airspeed vector. You should realize that an aircraft is not always following its nose. In this picture you can see that the smoke is not moving in the opposite direction of the aircraft nose, but that the nose actually has a slight angle with respect to the oncoming air. In most cases, the nose of the aircraft has a slight angle, with respect to the airflow in order to generate the appropriate amount of lift and, unfortunately, aerodynamic drag as well. The angle of the airspeed vector with respect to the horizon is called the flight path angle, and the reference system attached to the airspeed vector is called the air path axis system. So now we have the reference frames defined and we can start working on Newton's second law. We can define all forces acting on the aircraft, Then we can calculate the acceleration a and thereby the motion of the aircraft. So let us identify which forces act on the aircraft. Of course there is the gravitational force W, the weight. This force acts on the aircraft and points to the center of the earth, hence it can be drawn along the z-axis of the moving earth axis system. The gravitational force is a function of the distance to the center of the earth, of course. This variation is measurable, but quite small. For applications of aircraft flying below 80 kilometers, we can neglect this effect and assume that the gravitational acceleration is in fact constant. Let's draw the aerodynamics forces. By definition, the lift force is always perpendicular to the airspeed vector, and the drag is parallel. to the airspeed factor. Thus they are defined in the airpath axis system. The final force acting on the aircraft is the thrust of the propulsion system. The direction of this force depends on how the propulsion system is positioned relative to the airframe. So, if we draw it in a general direction, We can actually see it as an angle with respect to the airspeed vector. We call this the thrust angle of attack. So now we have defined all forces acting on the aircraft and we can have a look at the accelerations. Since we are looking at two-dimensional motion of the aircraft, its acceleration vector can be defined by two components. In principle we are free to choose in which axis system we define the accelerations. However, experience has shown us that defining them in the airpath axis system is the most convenient way. If the aircraft has a forward acceleration along the x-axis of the airpath axis system, it will change the magnitude of the airspeed vector. Hence this acceleration can be described as dv dt, so the change of velocity with time. If there is a perpendicular acceleration, then the path of the airplane will be curved. In fact, a perpendicular acceleration can be interpreted as a centripetal acceleration of a circle with a radius r. And as you know, probably from high school physics, a centripetal acceleration has the value v squared over r. However, r is a bit meaningless in the context of our story. It is in fact the radius of an imaginary circle which you see here on the screen. Now, from this circle we know that the angular velocity multiplied with the radius yields the speed v. And you can also observe from the geometry of the circle that the angular velocity is in fact the same as the change of flight path angle with time. equation 3 in our equation. So, from the visualization you can observe that V can also be written as the change of flight path angle with time multiplied with the radius r. Therefore, v squared over r from the first equation can be rewritten as v times d gamma d t times r divided by r. So I'm basically replacing one of the v's in the equation with the combination of d gamma d t and r. So the final result gives us then that the acceleration is also equal to v times the change of flight path angle with time. And that is a nice form, because the equation of motion of the aircraft is purely defined by gamma and V. And these are the only variables in the acceleration vectors. Based on the forces and accelerations, the point mass equations of motion can be written down now. It is most convenient to express them parallel and perpendicular to the airspeed vector. And these equations will form the basis for our aircraft performance calculations. Let us derive the equations for motion based on Newton's law. Now Newton's law states that F equals m times a, whereas the forces and the accelerations are in fact vectors, so they have a direction. So we started with a free body diagram in which we can draw all the forces on the aircraft. So let us draw a simple representation of the aircraft. Let us include the airspeed vector v. which has an angle of gamma with respect to the horizon. Now in this diagram we can draw all the relevant forces acting on the aircraft. So let's do that in a slightly different color and let's start with the wave. weight, w. It points straight down to the center of the earth. Now as you know, the aerodynamic forces are defined relative to the airspeed vector, whereas lift is perpendicular to the airspeed vector and drag is parallel but in the opposite direction of the airspeed vector. Now we have one more force which is the propulsive force or the thrust, t, which can point in an arbitrary direction and which therefore has a small angle with respect to the airspeed vector v. Now these are all the forces in the equation f is m times a. Now let's have a look at the kinetic diagram in which we represent all accelerations. So we take the same aircraft, again it is flying in a direction slightly upwards. with an angle gamma with respect to the horizon. And in the kinetic diagram we can define two accelerations. Now let's first define an acceleration parallel to v which is the change of the airspeed vector with time and an acceleration perpendicular to v, which is v times the change of flight path angle with time. So with these two pictures we can start deriving the equations of motion. So we start doing that by summing all the forces parallel to the airspeed vector. The acceleration of forces must be equal to mass times the acceleration parallel to the airspeed vector, which is this one over here. So this is m times dv. and that must equal all the forces parallel to the airspeed vector. So we have a thrust vector, and this thrust vector has a slight inclination with respect to the speed vector. So the component parallel to the airspeed is the cosine of alpha t. Now parallel to the airspeed vector is defined the aerodynamic drag, and this drag is in the opposite direction. So we get a minus d in the equation. And finally we have the aircraft weight which is pointing straight down, and it's defined in the moving earth axis system. So the angle with respect to the airspeed vector is gamma, and the component of the weight parallel to the airspeed vector is then the sine of the flight path angle. So this is basically our first equation of motion. But let's simplify it. simplify it a little bit, because in normal conditions the thrust vector will be fairly parallel to the airspeed vector if you consider a conventional aircraft, and therefore we can assume that this cosine alpha t is actually quite close to one. Furthermore, this mass we have over here, we know that in fact gravitational acceleration can vary as a function of altitude. So it's better to express it as the weight of the aircraft. So let's now rewrite the equation slightly. We have weight divided by gravitational acceleration, g, times dv dt equals t thrust. minus the drag minus weight times the sine of the flight path angle, and that is our first equation of motion. Now since we're actually looking at symmetric flight in two dimensions, we have two equations of motion. Now the second one we have is defined perpendicular to the airspeed vector. Again, we have a mass. And we can multiply that with the acceleration, so mv dgamma dt. And that should equal all the forces perpendicular to the airspeed vector. Now, by definition, lift is perpendicular to the airspeed vector, so in this equation it simply is lift. Now, again, our thrust vector has a component which points slightly upwards. So, in this case we will have plus t times the sine of alpha t. And finally we have a weight component and instead of the sine gamma we see here now we have minus weight times the cosine of the flight path. Now again, just like in the first equation, this αt can be assumed quite small, and therefore we can say, well, this is almost equal to zero. This of course is relevant for normal aircraft operations. but you must be careful not to do this step if you consider, for example, helicopters or other more exotic aircraft. So, now we're almost at the end of our derivation. Let us write down the mass in terms of weight and gravitational acceleration times V times d gamma dt equals lift minus weight. cosine of gamma. So now we have two equations of motion and these are used to describe the performance of an aircraft in a two-dimensional plane. Summarizing, we have obtained two equations of motion. And these two equations of motion we will use in the next lectures for our aircraft performance calculations.

However, if there would be a ball on the ground, and the train would suddenly have to brake, then the ball will start to move forward. But, there is no external force acting upon it. In fact, the ball just wants to continue moving forwards, whereas the train is stopping.

Similarly, if the train starts to turn, the ball will start moving either to the left or to the right. And again, no external force is acting upon the ball. Since the train was experiencing an acceleration in both examples, the frame of reference was not an inertial frame of reference, and thus Newton's first law does not apply. This is very important to realize when we derive the equations of motion for an actual aircraft. We should clearly define the different frames of reference.

A logical point to start with is the Earth's surface. So we can start defining a reference frame fixed to the Earth. This reference frame is called the Earth axis system. Of course we are interested in the motion of the aircraft relative to this frame. So is this frame of reference an inertial frame of reference or not?

Actually it isn't. If you are flying at a constant altitude above the Earth's surface, you are actually moving in a circle. Thus an acceleration to the center of the circle is required in order to create that motion. So the frame of reference is accelerating and it is not an inertial frame of reference.

As a result there appears to be a centrifugal force which works in the outward direction perpendicular to the circular motion. Realize that this is not a real force but an apparent force due to the force of the centrifugal force. just like in the train example. The question now is, how large is this acceleration? From high school physics we already know that the acceleration equals V squared over r.

Most aircraft have a cruise altitude of about 10 kilometers, so the radius in the equation equals the radius of the Earth plus 10 kilometers. The radius is about 6000. kilometers and that results in a total radius of 6381 kilometers. A typical velocity for a commercial aircraft at this altitude is about 250 meters per second or Mach 0.8. So the acceleration actually equals 250 squared divided by the radius squared and that yields 0.0097 meters per second squared.

Now realize that this is only a single plane 0.1% of the gravitational acceleration. So even though the acceleration is present, it is so small that we can neglect it and make what we call the flat earth assumption. Now there is another effect due to the earth, it is in fact rotating. So the velocity of a point on the earth's surface depends on the latitude, because the radius to the axis of the earth's rotation is in fact changing.

So, if a person is moving in a north-south direction, the velocity actually changes as well. Hence there is an acceleration and it is not an inertial frame of reference. This can also actually be visualized from the perspective of somebody on a roundabout.

The ball moves in a curved motion. It should be noted, however, that the accelerations related to the fact that the earth is rotating are very small. They can be neglected for most airplane motions. They can, however, become important when high-altitude, high-speed vehicles are of interest.

So, if we assume that the earth is flat and not rotating, we can safely state that the earth-fixed reference frame is an inertial frame of reference. Now let's define some more useful reference frames. The forces acting on the aircraft depend on its orientation and its motion relative to the air.

So it would be useful to define reference frames attached to the aircraft. So we can start by defining a reference frame with the same orientation as the original earth-fixed frame, but now attached to the aircraft itself. This is called the moving earth axis system. Now the nose of the aircraft is actually pointing in a different direction.

We can therefore also attach an access system. fixed to the aircraft body, which we call the body axis system. Now when the pilot is looking outside, he or she is looking along the x-axis of the body axis system and can directly observe the pitch attitude of the aircraft, which we call theta, with respect to the horizon.

So, summarizing, we have a reference frame from the point of view of the aircraft attitude and we have two frames relative to the Earth. However, Since we are interested in the motion of the aircraft, there is one final element of interest, and that is the direction of the airspeed vector. You should realize that an aircraft is not always following its nose.

In this picture you can see that the smoke is not moving in the opposite direction of the aircraft nose, but that the nose actually has a slight angle with respect to the oncoming air. In most cases, the nose of the aircraft has a slight angle, with respect to the airflow in order to generate the appropriate amount of lift and, unfortunately, aerodynamic drag as well. The angle of the airspeed vector with respect to the horizon is called the flight path angle, and the reference system attached to the airspeed vector is called the air path axis system.

So now we have the reference frames defined and we can start working on Newton's second law. We can define all forces acting on the aircraft, Then we can calculate the acceleration a and thereby the motion of the aircraft. So let us identify which forces act on the aircraft.

Of course there is the gravitational force W, the weight. This force acts on the aircraft and points to the center of the earth, hence it can be drawn along the z-axis of the moving earth axis system. The gravitational force is a function of the distance to the center of the earth, of course. This variation is measurable, but quite small.

For applications of aircraft flying below 80 kilometers, we can neglect this effect and assume that the gravitational acceleration is in fact constant. Let's draw the aerodynamics forces. By definition, the lift force is always perpendicular to the airspeed vector, and the drag is parallel. to the airspeed factor.

Thus they are defined in the airpath axis system. The final force acting on the aircraft is the thrust of the propulsion system. The direction of this force depends on how the propulsion system is positioned relative to the airframe.

So, if we draw it in a general direction, We can actually see it as an angle with respect to the airspeed vector. We call this the thrust angle of attack. So now we have defined all forces acting on the aircraft and we can have a look at the accelerations.

Since we are looking at two-dimensional motion of the aircraft, its acceleration vector can be defined by two components. In principle we are free to choose in which axis system we define the accelerations. However, experience has shown us that defining them in the airpath axis system is the most convenient way. If the aircraft has a forward acceleration along the x-axis of the airpath axis system, it will change the magnitude of the airspeed vector.

Hence this acceleration can be described as dv dt, so the change of velocity with time. If there is a perpendicular acceleration, then the path of the airplane will be curved. In fact, a perpendicular acceleration can be interpreted as a centripetal acceleration of a circle with a radius r.

And as you know, probably from high school physics, a centripetal acceleration has the value v squared over r. However, r is a bit meaningless in the context of our story. It is in fact the radius of an imaginary circle which you see here on the screen.

Now, from this circle we know that the angular velocity multiplied with the radius yields the speed v. And you can also observe from the geometry of the circle that the angular velocity is in fact the same as the change of flight path angle with time. equation 3 in our equation. So, from the visualization you can observe that V can also be written as the change of flight path angle with time multiplied with the radius r. Therefore, v squared over r from the first equation can be rewritten as v times d gamma d t times r divided by r.

So I'm basically replacing one of the v's in the equation with the combination of d gamma d t and r. So the final result gives us then that the acceleration is also equal to v times the change of flight path angle with time. And that is a nice form, because the equation of motion of the aircraft is purely defined by gamma and V. And these are the only variables in the acceleration vectors. Based on the forces and accelerations, the point mass equations of motion can be written down now. It is most convenient to express them parallel and perpendicular to the airspeed vector.

And these equations will form the basis for our aircraft performance calculations. Let us derive the equations for motion based on Newton's law. Now Newton's law states that F equals m times a, whereas the forces and the accelerations are in fact vectors, so they have a direction. So we started with a free body diagram in which we can draw all the forces on the aircraft. So let us draw a simple representation of the aircraft.

Let us include the airspeed vector v. which has an angle of gamma with respect to the horizon. Now in this diagram we can draw all the relevant forces acting on the aircraft. So let's do that in a slightly different color and let's start with the wave. weight, w.

It points straight down to the center of the earth. Now as you know, the aerodynamic forces are defined relative to the airspeed vector, whereas lift is perpendicular to the airspeed vector and drag is parallel but in the opposite direction of the airspeed vector. Now we have one more force which is the propulsive force or the thrust, t, which can point in an arbitrary direction and which therefore has a small angle with respect to the airspeed vector v. Now these are all the forces in the equation f is m times a.

Now let's have a look at the kinetic diagram in which we represent all accelerations. So we take the same aircraft, again it is flying in a direction slightly upwards. with an angle gamma with respect to the horizon. And in the kinetic diagram we can define two accelerations.

Now let's first define an acceleration parallel to v which is the change of the airspeed vector with time and an acceleration perpendicular to v, which is v times the change of flight path angle with time. So with these two pictures we can start deriving the equations of motion. So we start doing that by summing all the forces parallel to the airspeed vector.

The acceleration of forces must be equal to mass times the acceleration parallel to the airspeed vector, which is this one over here. So this is m times dv. and that must equal all the forces parallel to the airspeed vector. So we have a thrust vector, and this thrust vector has a slight inclination with respect to the speed vector.

So the component parallel to the airspeed is the cosine of alpha t. Now parallel to the airspeed vector is defined the aerodynamic drag, and this drag is in the opposite direction. So we get a minus d in the equation.

And finally we have the aircraft weight which is pointing straight down, and it's defined in the moving earth axis system. So the angle with respect to the airspeed vector is gamma, and the component of the weight parallel to the airspeed vector is then the sine of the flight path angle. So this is basically our first equation of motion. But let's simplify it. simplify it a little bit, because in normal conditions the thrust vector will be fairly parallel to the airspeed vector if you consider a conventional aircraft, and therefore we can assume that this cosine alpha t is actually quite close to one.

Furthermore, this mass we have over here, we know that in fact gravitational acceleration can vary as a function of altitude. So it's better to express it as the weight of the aircraft. So let's now rewrite the equation slightly.

We have weight divided by gravitational acceleration, g, times dv dt equals t thrust. minus the drag minus weight times the sine of the flight path angle, and that is our first equation of motion. Now since we're actually looking at symmetric flight in two dimensions, we have two equations of motion. Now the second one we have is defined perpendicular to the airspeed vector. Again, we have a mass.

And we can multiply that with the acceleration, so mv dgamma dt. And that should equal all the forces perpendicular to the airspeed vector. Now, by definition, lift is perpendicular to the airspeed vector, so in this equation it simply is lift.

Now, again, our thrust vector has a component which points slightly upwards. So, in this case we will have plus t times the sine of alpha t. And finally we have a weight component and instead of the sine gamma we see here now we have minus weight times the cosine of the flight path.

Now again, just like in the first equation, this αt can be assumed quite small, and therefore we can say, well, this is almost equal to zero. This of course is relevant for normal aircraft operations. but you must be careful not to do this step if you consider, for example, helicopters or other more exotic aircraft. So, now we're almost at the end of our derivation.

Let us write down the mass in terms of weight and gravitational acceleration times V times d gamma dt equals lift minus weight. cosine of gamma. So now we have two equations of motion and these are used to describe the performance of an aircraft in a two-dimensional plane.

Summarizing, we have obtained two equations of motion. And these two equations of motion we will use in the next lectures for our aircraft performance calculations.

Transcript for:Understanding Newton's Laws in Aviation

Transcript for:
Understanding Newton's Laws in Aviation