Derivations of the Formulas for Lift in a Centrifugal Drive

Robert S. Wilson PhD September 28, 2017

Abstract This is the derivation of the formula for the mean lift in a centrifugal drive This is to go with a patent application for which I have applied. It provides explanatory material for the invention.

1 Deﬁnitions

If you have a weight on a string and swing it around, the centrifugal force will act from the center of the spin, along the string, out to the weight. A centrifugal drive is one, which will use centrifugal force. The trouble with a centrifugal drive is that, while the weight is on one side of the center, the force will be going in one direction, but when the weight is on the other side, the force will act in the opposite direction, which will result in a net force of 0. The problem is to come up with a scheme where the force on one side will not be cancelled out by the force on the other side. The basic unit in this invention consists of four major parts. The ﬁrst is motor. There are many possible types of motors available such as: a reciprocating piston internal combustion motor, either gasoline or diesel, an electric motor, which can be energized by such methods as a battery or a generator powered by a gasoline or diesel engine, i.e. a diesel electric or gasoline electric hybrid, or solar among others. If you plug an electric motor into the wall in the US, it will turn at 3600 rpm. For an automobile engine in a car going down a freeway 3000rpm is a common angular velocity, so a motor capable of turning out several thousand rpms would be feasible. The the motor will be in what I will call the center of the ﬁgure where there will be a hub. The hub will be a hollow rectangular prism, and, in the hub, a rod will go back and forth as the motor causes the hub to rotate about the center. There will be a closed track, and the ends of the rod will follow the track with the rod stretching from one side of the track to the other. It would be

1 good if there were roller bearings on the ends of the rod as they follow the track around. See the following ﬁgure.

track rod

hub

The motor is not shown. It causes the hub to rotate about the center, and the rod moves back and forth through the hub as it follows the track. While I am calling the place, where the motor connects with the hub, the center of the figure, it is not the center of the track, because, if we want there to be net lift, the track will have to move the rod around off center. The next seven illustrations show the rod in the first 6 multiples of 30o, as it moves around the track, and the seventh figure shows the rod returning to its original position. At

2 this point the cycle repeats with the rod being oriented in the opposite direction.

3 30o

4 60o

5 90o

6 120o

7 150o

8 180o

And when the rod gets back to 180o, the cycle starts over. A cycle is half of a revolution. Note that, when the rod is at 90o, it does not quite reach from the upper part of the track to the lower. This is because the rod has non zero thickness, so, in order for the rod to avoid contacting the track, the track can not come up to the center. The hub can be oﬀset so that it avoids contacting the track, but since the rod has to follow the track, it can not be oﬀset, so we will need to leave a little space for when the rod is vertical. Also, when the rod

9 is horizontal, the track can not come up to the center, because we need to leave some space for the thickness of the rod. The rod is rectangular in shape. While a cylindrical rod might be possible, a rectangular rod would be stronger and easier to make. The two dimensions which are shown in the ﬁgure are the length and the thickness. The third dimension, the width, goes into the page perpendicularly, so does not appear. If the rod has greater thickness, it will contact the track more often, which we want to minimize. Note that we can increase the width of the rod without having to worry about increasing the chances of the rod intersecting the track, and that would result in a stronger rod. A wider rod would be heavier, but that will not be a problem because if the other factors remain constant, the lift is proportional to the mass of the rod. More mass in the rod will result in greater lift. The hub, through which the rod goes back and forth, would be a hollow rectangular prism. There should probably be bearings where the rod goes through the track. Since the rod is rectangular, the roller bearings at the ends of the rod will be cylindrical. Again, a rectangular rod with cylindrical bearings on the ends would be stronger and easier to make than a cylindrical rod with spherical bearings. If we look at a picture which shows the rod in various positions (without the hub),

10 we see that there will be generally more mass above the center than below. Since the rod stays within the track, the net acceleration will be 0, but force is mass times acceleration, so the fact that there is more mass above the center than below will result in positive lift. Actually, when the rod is close to horizontal, there will be slightly more mass below the center than above, but when the rod is close to vertical, practically all of the rod will be above the center. When the rod is close to horizontal, the diﬀerence between the amount of mass below the center and the amount of mass above is quite small, and so the centrifugal forces will be close to being balanced; whereas when the rod is close to vertical,

11 there will be much more mass above the center than below, resulting in positive net lift. I am not claiming any particular shape in this patent application. I am merely presenting an example of a shape for the track which will provide lift. The significant new contribution in this application is the track. The track is very important. It is the load bearing element of the device. The lift will come from the top of the rod pushing up against the top part of the track, so the rod pushing against the track has to bear the load. As a result , the track and the rod need to be very strong. The rod could be made from cast aluminum or titanium. The rod could be steel, or any other materials which would have the necessary strength. The track is symmetrical about the vertical line through the center. The rod, which goes through the center of the hub, stretches from one side of the track to the other. Let us describe points on the track using the Cartesian coordinate system. There are many shapes possible for the track. One feature which all of the shapes, which we will be considering, is that when the rod is horizontal, it will stretch from (−L/2, 0) to (L/2, 0), where L is the length of the rod. One important feature of the track is that as the rod stretches from one side to the other, contacting the track at both ends as much as much as possible. In this example, the top part of the track is part of the upper half of an ellipse. In the examples, which we will consider, the top of the track is at (0,L). Stretching the top of the track vertically, so that the top is at a distance greater than L from the center, is a possibility. This will result in increased lift, but bottom of the rod would lose contact with the track over a greater interval. One of the purposes of the bottom part of the track is to keep the top end of the rod in contact with the upper part of the track. This is very important because the top end of the rod contacting the track is the load bearing element of the device. If the top of the rod loses contact with the track, there will be no positive lift, which could be serious if the device is in a vehicle traveling several thousand feet above the ground. If we want to have the interval where bottom part of the rod loses contact with the track to be as small as possible, we will want the thickness of the rod to be as small as possible, while still providing enough strength to lift a payload. Over that interval, the top of the rod will stay in contact with the track because of centrifugal force. Actually, since the negative lift, and we will see how there is some negative lift during a cycle, comes from the bottom of the rod contacting the lower part of the track, when the bottom of the rod loses contact with the lower part of the track, there will be no negative lift. That is not very significant, because the interval when the bottom of the rod loses contact with the lower part of the track is when the negative lift from the bottom of the rod contacting the lower part of the track is when the negative lift is smallest., Once we have the part of the track which is above the horizontal line through the center, the bottom part of the track is determined as the curve which the bottom end of the rod will trace as the top end follows the upper part of the track. In order for the hub to avoid contacting the track, the hub could be offset

12 from the track, but the rod has to follow the track, so it will not be oﬀset. The bottom part of the track will have to be machined to close tolerances so that the ends of the rod will stay in contact with both sides of the track, as much as possible. If the track were to be cast, then once we had the mold, it would be fairly inexpensive to mass produce the units. The track which follows these guidelines will look like

As the rod stretches from the top, which is at (O,L), down a distance of

13 L, it will reach the center at (0, 0), and that will put a spike in the middle of the bottom part of the track at (0, 0), which, as we noted earlier, will not work, because of the thickness of the rod. If you look at the illustrations on pp 3 - 9, you will see that if the spike in the bottom part of the track reaches (0, 0), the bottom part of the rod would make contact with the track when the rod goes past the vertical. It looks like the hub will also make contact with the track, but we can avoid that by having the hub offset. However, the rod needs to follow the track. Solving the problem of the rod getting in contact with the spike will remain as a topic for a future revision. The result is that, in the illustrations on pp 3 - 9, the spike has been cut off. The if the interval where the bottom of the rod loses contact with the track is small enough, then centrifugal force will keep the top of the rod in contact with the track. As we have mentioned above, this is important because the point where the top of the rod makes contact with the upper part of the track is the load bearing element of the device. If that contact is lost, the positive lift will vanish. The next problem to solve is to eliminate the horizontal component of the force. The vertical component of the force provides lift; the horizontal component contributes nothing of value, and it would be good to eliminate it. This can be accomplished by having a second unit attached to the first unit, where the hubs and rods are rotating in opposite direction. We will want the rods of both units to be horizontal at the same time and vertical at the same time but rotating in opposite directions. This would cancel out the horizontal component of the force and double the vertical component. It would also cancel out the gyroscopic forces. Later, after we have computed the lift as a function of the angle, we will see that the lift is quite variable, being greater when the rods are vertical and, even slightly negative when the rods are horizontal. As a result, there will be quite a bit of vibration as the rod goes around the track, especially at a high rate of rotation. It would be good to even out this vibration by having another pair of units attached, whose rods would be offset by 90o. This would not only even out the vibrations, it would also further double the lift. The lift could be further homogenized and doubled again by adding another two pairs of units, which would be offset by 45o and 135o, or even another four pairs of units offset by 30o, 60o, 120o, and 150o. All of the units would have to be synchronized, which could be accomplished by using a single motor to power all of the units, which would be linked with gears. Since it would not be good to have all of the units seize up if one unit should break, it would be good to have the units able to disconnect in that event. Having eight or twelve units would not only make it work more smoothly, it would also multiply the lift by a factor of eight or twelve, and spread out the load. While there would still be vibrations even with all of these units, it should not be any more than the vibration from a single cylinder reciprocating piston internal combustion engine. Besides the motor, in each unit, there would be only two main moving parts:

14 the rod and the hub. There may be roller bearings where the rod follows the track around, as well as bearings where the rod moves through the hub, and these could be considered to be additional moving parts. There are many types of motors, which would be suitable. An important feature would be that the motor was capable of a continuous range of angular velocities. If one were to simply plug an electric motor into a wall socket supplying 60cycle/sec alternating current, the unit might take oﬀ through the roof, not only resulting in the device becoming unplugged, but also causing unwanted damage. One would want to start the device at 0rpm and increase the angular velocity until it started to lift the device. Some advantages of a centrifugal drive would be that it could hover and be capable of vertical take oﬀ and landing like a helicopter, but without the prop wash. The centrifugal drive units would replace the rotor on a helicopter. Also, unlike an airplane or helicopter, the device is not dependent upon atmosphere for lift. As such, it could be used as a vehicle in outer space. With a centrifugal drive, the lift is proportional to the length if the rod, the mass of the rod, and the square of the angular velocity. The mean lift would be given by the formula

F = nKmLω2 Where n is the number of units, K is a constant, dimesionless factor which depends only upon the shape of the track. Diﬀerent shaped tracks will have a diﬀerent value of K. In the next section we will calculate the value of K for a particular shape of track. m is the mass of the rod, L is the length of the rod, and ω is the angular velocity. In the next section we will discuss how to compute K. While it may not be the simplest calculus problem, it is perfectly straightforward. Since the mass of the rod is

m = ρV = ρT W L where ρ is the density, V is the volume of the rod, T is the thickness of the rod, W is the width of the rod, and L is the length of the rod, this could also be expressed as

F = nKρT W L2ω2 Which shows how all of these factors contribute to the mean lift. This last factor is what makes a centrifugal drive really interesting. If one uses a motor supplying a few thousand rpm, 3000rpm is 50rev/sec, giving us an ω of approximately 314radians/sec. This will result in an ω2 of close to 100, 000(radians/sec)2. A factor, which will move the decimal point ﬁve places, is rather signiﬁcant.

15 2 Calculations

Acceleration is the change in velocity. Velocity is a vector, which has magnitude and direction. If you change either the magnitude or the direction of the velocity, you get acceleration, and when you multiply that by mass, you get force. Centrifugal force is a reactive force as in ”For every action there is an equal but opposite reaction.” When you swing a weight on a string in a circle, even if the magnitude of the velocity does not change, when the direction changes, there will be force. When you swing a weight on a string around in a circle, the change in velocity is toward the center, and the resulting reactive centrifugal force is directed in the opposite direction. Thus if we are looking to lift an object, the acceleration needs to be downward or negative. It is important to remember that negative acceleration results in positive lift. With the units paired so that the rods are revolving in opposite directions in the units, the horizontal components of the centrifugal force will cancel out, and the vertical components will double. Another advantage of doubling the units with rods revolving in opposite directions is that the gyroscopic forces will also cancel out. So we only need to consider the vertical component of the lift. With the deﬁnition of force

F = ma force equals mass times acceleration, the vertical component of the force is

d2y F = m dt2 This is the centrifugal force for each point on the rod. To get the total lift, we integrate this over the entire rod. If we use polar coordinates.

x = r cos θ and y = r sin θ then the vertical component of the velocity dy v = dt dy dθ = dθ dt Let dθ = ω dt

16 the angular velocity. Then dy v = ω dθ dr = ω(sin θ + r cos θ) dθ

To find the acceleration, we differentiate the velocity. At this point, let us assume that ω is constant i.e. dω/dt = 0. This will not always be true. When the device is started, the angular velocity will be 0, and then, the operator will increase it until the device starts to lift off. This will involve an increase in angular velocity. But let us first consider the case where ω = dθ/dt is constant. We will deal with the case of variable angular velocity in a later section. Even then, the case where the angular velocity is constant is an important part of the calculation. Let then us assume that the angular velocity is constant. This will result in simpler calculations, and it will enable us to get an idea about the forces which will be at our disposal. Then the vertical component of the acceleration

d2y a = dt2

dv dθ dv dr d2r = = ω = ω2(−r sin θ + 2 cos θ + sin θ ) dθ dt dθ dθ dθ2

The −r sin θ is the usual centrifugal force. If the point mass were going in a circle, then dr/dθ would be 0, and since it would be constantly 0, d2r/dθ2 would also be zero, and the last two terms would not be there. This is the vertical component of the acceleration. For an inﬁnitesimal element of the rod, the mass is

dm = ρAdr

where ρ is the density of the rod, A is the cross sectional area of the rod, and dr is an inﬁnitesimal increment in the length of the rod. The centrifugal force from this inﬁnitesimal section of the rod is

d2y d2y ρA dr = ρAω2 dr dt2 dθ2

To get the total force from the rod at an angle of θ, integrate

Z R 2 2 d y F = ρAω 2 dr R−L dθ

17 Let R = R(θ) be the distance from the center of the hub to the point on the track which is at an angle of θ. Quite often, we will have R < L, and the rest of the rod will be on the other side of the center of the hub below the horizontal line through the center of the ﬁgure, and the centrifugal force which the points on the other side of the center contribute will have the opposite sign. The trick is to get more mass above the center than below as often as possible. Of course, the force will vary with respect to θ. To get the total force from all the units, multiply the mean force by the number of units. To get the mean force, integrate this result with respect to θ as θ varies from 0 to π and divide by π, the length of the interval.

ρAω2 Z π Z R dr d2r F = (−r sin θ + 2 cos θ + sin θ 2 )drdθ π 0 R−L dθ dθ

When we consider the dr, it will be very helpful to note that the rod is solid, so dr will be the same at all points on the rod for any given value of θ. If we deﬁne R as above, then dr dR = dθ dθ

for every point on the rod as a function of θ. Hopefully, we will have a formula for R as a function of θ, which will enable us to complete the calculations. Indeed, we will choose a shape of the track with this in mind. The general formula for the lift will be

ρAω2 Z πZ R dR d2R = −r sin θ + 2 cos θ + sin θ 2 drdθ π 0 R−L dθ dθ

In general, once we have decided the shape of the upper half of the track, the part of the track, which is above the horizontal line through the center, the lower part will be determined as the set of points, which the lower end of the rod will trace as the upper end of the rod follows the upper part of the track. The ﬁrst term is simply the standard centrifugal force. Since there is more mass above the center than below, it will be negative acceleration resulting in positive lift. The last term is easy enough to recognize as the vertical component of the force from the acceleration of the rod as it moves back and forth through the hub. The middle term is a problem. The whole point is that as θ increases from 0 to π/2, except for possible rare exceptions, the track moves away from the center, so that there will be more mass above the center than below, and then dR/dθ will be positive, and between 0 and π/2, cos θ is positive, so the integrand in the middle term is positive for θ between 0 and π/2. When θ is between π/2 and π, the track is moving back toward the center so dR/dθ will be negative, but cos θ is also negative over this interval, so the integrand will also be positive

18 here. dR/dθ and cos θ will both be antisymmetric functions about θ = π/2, and the product of two antisymmetric functions is a symmetric function. Conclude that the middle term will give us positive acceleration resulting in negative lift over the entire interval between 0 and π. One can see why this would be. As the rod moves up from θ = 0, the upper end will move up away from the center. It will be pushed up by the bottom end of the rod following the lower part of the track. This will result in positive acceleration or negative lift. As the rod gets closer to vertical, dR/dθ becomes close to 0 as it prepares to have its sign changed, so we will see that the negative lift is much more pronounced when the rod is close to horizontal. The situation is aggravated by the fact that this middle term is doubled. This negative lift is unfortunate, and we will need to take steps to deal with it.

3 The Track

There are many shapes for a track where the mean net force will be 0. Since the force from the doubled middle term will be generally greater than the force from the first term, if the net force is 0, we conclude that the last term will be negative resulting in positive lift. We need to use a track which will collect the least negative force from the middle term and the greatest positive force from the first and last terms. The calculations from the previous section will hold no matter what the shape of the track. Let us now consider an example of a shape for the track. The example used here is not chosen to optimize the lift, which could be done using something like Calculus of Variations. It is chosen to provide an example where the calculations will not be too complicated. The calculations will be complicate enough even with this simpler example, that we will not ask the reader to verify them for an optimal case. You would find that this example will provide better lift than if we just used the upper half of an ellipse which would have the cardinal points on the horizontal line through the center, and vertical tangents when the track went through those points. If we can find an example which provides adequate positive lift, an optimal example, or even a better example would be better. Given that the factor of ω2 could move the decimal point five places, and if the mass of the rod were 10kg, that would move the decimal point another place and if we are using eight or twelve units, that would move it another place for total of an amazing seven places, a value of K of +0.1 would be sufficient, and we would want K to be at least this large. Less than that would put K too close to 0 The upper part of the track, which we will use in this example comes from taking the middle part of the upper half of an ellipse. An ellipse, centered at the center, is given parametrically by

x = a cos θ and y = b sin θ

19 We will take the part of the upper half of the ellipse where π/3 ≤ θ ≤ 2π/3. We stretch it vertically so that the distance from the horizontal line going through the two endpoints to the top of the ellipse is L. Then stretch it horizontally so that the distance between the two endpoints is L, and then translate it vertically so that the endpoints are on the horizontal line through the center. This curve will then be given parametrically by θ + π x = L cos( ) 3 and √ θ + π √ y = L((4 + 2 3) cos( ) − (3 + 3)) 3 There are two reasons for taking this part of the ellipse. First, this is the part of the ellipse where the usual centrifugal force from the ﬁrst term is greatest. The other reason is that the negative lift from the second term, which will contain two cosine factors, is smaller here since cos(π/2) = 0. So the distance from the center to the point on the track at an angle of θ would be r θ + π √ θ + π √ R = L (cos( ))2 + ((4 + 2 3) sin( ) − (3 + 2 3))2 3 3 Note that when θ + π π θ = 0, = 3 3 and when θ + π 2π θ = π, = 3 3 Thus, when

θ = 0, π, R = ±L/2 and when π θ + π π θ = , = 2 2 2 and then

R = L so this curve goes through the points we want. Let us look at the track thus generated

20 Note that this curve contains a spike at the center of the ﬁgure, with which we will deal presently. Let us look at our three terms in light of this example,

ρAω2 Z πZ R dR d2R = (−r sin θ + 2 cos θ + sin θ 2 )drdθ π 0 R−L dθ dθ

21 Remembering that R, dR/dθ, and d2R/dθ2 are constants with respect to r, depending only on θ, the ﬁrst integration, with respect to r is simple enough that we may as well complete it forthwith.

ρAω2 Z π R2 (R − L)2 dR d2R = −( − ) sin θ + 2(R − (R − L)) cos θ + (R − (R − L)) sin θ 2 dθ π 0 2 2 dθ dθ

ρAω2 Z π L2 dR d2R = − RL sin θ + 2L cos θ + L sin θ 2 dθ π 0 2 dθ dθ

All of these terms have a factor of L in them, and when we factor it out, recall that

ρAL = m

the mass of the rod so we can simplify it further to get

mω2 Z π L dR d2R = − R sin θ + 2 cos θ + sin θ 2 dθ π 0 2 dθ dθ In order to understand the ﬁrst two terms, consider the following illustration.

22 It shows the upper part of the track together with a semicircle centered at the center of radius L/2. When the ellipse is inside the semicircle, R < L/2, and when it is outside the semicircle, R > L/2. We can see that R > L/2 much more than R < L/2. As a result, the first term in the integrand, L/2 − R, is generally negative, and negative acceleration results in positive lift. So the first term in the integrand contributes positive lift. The problem term is the middle term, cos θdR/dθ. Since the track goes from lying L/2 from the center when the rod is horizontal, to L when the rod is vertical, even if there may be some places where dR/dθ < 0, as there is in this example, generally dR/dθ will be positive when 0 ≤ θ ≤ π/2 which is when cos θ is positive, and dR/dθ will be negative when π/2 ≤ θ ≤ π which is when cos θ is negative. To see how this works, consider the following figure

23 In addition to the upper part of the track and the semicircle of radius L/2, we also have the semicircle centered at the center which is tangent to the part of the ellipse that forms the upper part of the track. Between the horizontal on the right, and the point of tangency, dR/dθ will be negative, resulting in negative acceleration and positive lift, but the positive lift will be very small. However, any positive lift, no matter how small, is better than negative lift, even if there is very little of it. The point of tangency is where the track comes closest to the center, and after that, we will have dR/dθ > 0 until the rod becomes vertical, during which interval, cos θ is positive. Then after the rod passes the vertical point, the track moves back toward the center, and dR/dθ will be negative until the point of tangency on the left. During that interval cos θ ≤ 0, and a negative times a negative is a positive. Conclude

cos θdR/dθ ≥ 0 between the points of tangency. Outside of the points of tangency,

24 cos θdR/dθ ≤ 0 resulting in positive lift, but there is very little of that. So the middle term in the integrand will contribute negative lift. This is the reason why we are only using a part of the ellipse for the upper part of the track. This part of the ellipse is the part which contributes the greatest positive centrifugal force, but where the negative lift from the middle term is lowest.

4 Calculation if the Lift

In order to simplify the displays of the calculations let us let √ q = 4 + 2 3 ≈ 7.464 and √ h = 3 + 2 3 = q − 1 ≈ 6.464 Then r θ + π θ + π R = L (cos( ))2 + (q sin( ) − h)2 3 3 Dividing by 3 inside parentheses will stretch the graph horizontally by a factor of 3. This will cut dR/dθ down by a factor of 3. However, when we integrate to ﬁnd the mean lift, this eﬀect will be reversed for no net change. However, it will cut d2R/dθ2 down by a factor of 9, so this will result in a net decrease of a factor of 3 after we integrate.

2 θ+π θ+π θ+π dR L (q − 1) sin 3 cos 3 − hq cos 3 = q dθ 3 θ+π 2 θ+π 2 cos 3 + (q sin 3 − h) so

d2R (q2 − 1)[cos2 θ+π − sin2 θ+π ] + hq sin θ+π = L 3 3 3 2 q dθ θ+π 2 θ+π 2 9 cos 3 + (q sin 3 − h) 2 ((q2 − 1) cos θ+π sin θ+π − hq cos θ+π −L 3 3 3 q 3 θ+π 2 θ+π 2 9 cos 3 + (q sin 3 − h)

The ﬁrst term in the lift is then

r ! mω2 Z π θ + π θ + π F = sin θ L/2 − L (cos( ))2 + (q sin( ) − h)2 dθ π 0 3 3

We can factor out another factor of L from these two terms

25 r ! mLω2 Z π θ + π θ + π F = sin θ 1/2 − (cos( ))2 + (q sin( ) − h)2 dθ π 0 3 3

There is a factor of L in the other terms as well, so they become

2 Z π 2 θ+π θ+π θ+π 2mLω (q − 1) sin 3 cos 3 − hq cos 3 + cos θ q dθ π 0 θ+π 2 θ+π 2 3 cos 3 + (q sin 3 − h)

2 Z π 2 2 θ+π 2 θ+π θ+π mLω (q − 1)[cos 3 − sin 3 ] + hq sin 3 + sin θ q dθ π 0 θ+π 2 θ+π 2 9 cos 3 + (q sin 3 − h)

2 mLω2 Z π ((q2 − 1) cos θ+π sin θ+π − hq cos θ+π − sin θ 3 3 3 dθ π q 3 0 θ+π 2 θ+π 2 9 cos 3 + (q sin 3 − h)

Since the middle term involves the derivative of a square root, the square root 1 is the 2 power, so when you lower the power by 1 when you take its derivative, 1 you get a − 2 power, or a square root in the denominator. The chain rule will provide a trigonometric function for the numerator, giving us a fraction, and we conclude that we will need to use the quotient rule to get the second derivative. This will provide a difference of two terms, so the last term will split up into a difference of two terms, as we have seen in the formulas above. Let us call the first of these terms, the first part of the last term and the other one the second part of the last term. To get an idea of the sign on all of these terms, let us look at all the graphs. The first graph is

26 0.5 1.0 1.5 2.0 2.5 3.0

- 0.1

- 0.2

- 0.3

- 0.4

- 0.5

s θ + π √ θ + π √ sin θ(1/2 − (cos )2 + ((4 + 2 3) sin − (3 + 2 3))2 3 3

As we expected, this function takes on negative values except for a slight bit of positive at the ends. This will give us negative acceleration and positive lift. The value of this integral, after dividing by π to get the mean is

≈ −0.2139 This makes sense. If the rod were to rotate about one of its ends, K would 1 be 2 . But it is not being rotated about one of its ends, and so we would expect K to be less than that; a little less that half of that is not surprising. Now we look at the graph of the function for the second term.

27 0.4

0.2

0.5 1.0 1.5 2.0 2.5 3.0

- 0.2

2 θ+π θ+π θ+π (q − 1) sin 3 cos 3 − hq cos 3 2 cos θ q θ+π 2 θ+π 2 3 cos 3 + (q sin 3 − h) When we double this integral and divide by π to get the mean lift from the middle term, we get

≈ 0.4472 We next take the ﬁrst part of the last term

√ √ √ 2 Z π 2 2 θ+π 2 θ+π θ+π mLω ((4 + 2 3) − 1)[cos 3 − sin 3 ] + (4 + 2 3)(3 + 2 3) sin 3 + sin θ q √ √ dθ π 0 θ+π 2 θ+π 2 9 cos 3 + ((4 + 2 3) sin 3 − (3 + 2 3)) The graph of the ﬁrst part of the last term looks like

28 0.4

0.2

0.5 1.0 1.5 2.0 2.5 3.0 - 0.2

- 0.4

- 0.6

- 0.8

We see that the acceleration from this term will be positive when the rod moves a little way oﬀ of the horizontal, but when it is closer to vertical, the acceleration is negative, an there is more negative acceleration than positive. The value is of this integral is

≈ −0.4159 giving us positive lift. The graph for the second part of the last term looks like

0.5 1.0 1.5 2.0 2.5 3.0

- 0.02

- 0.04

- 0.06

- 0.08

- 0.10

29 √ √ √ 2 ((4 + 2 3)2 − 1)2 cos θ+π sin θ+π − (4 + 2 3)(3 + 2 3) cos θ+π sin θ 3 3 3 q √ √ 3 θ+π 2 θ+π 2 9 cos 3 + ((4 + 2 3) sin 3 − (3 + 2 3)) We can tell that this term will give us positive lift because the numerator is a square, and the denominator is 9 times the cube of a distance, so it will also be positive. The whole fraction is positive, but the quotient rule says that it is being subtracted to get its contribution to the lift. The fact that the graph is always below the x-axis, and the large value of q causes this term to contribute more positive lift than any other term. The value for this integral is

≈ −0.1607 This will give us a value for the K in the nKmLω2 of

K ≈ 0.4472 To compute the lift, which is the reactive force to the acceleration of the rod, I have changed the sign on the accelerations. When one considers that the factor of ω2 can move the decimal point 5 places, this will give us a signiﬁcant amount of lift.

5 Variable Angular Velocity

In the previous section, when we computed the lift, we made the assumption that the angular velocity was constant. It may be that, when operating a vehicle, you will want to keep the angular velocity constant, but there will be times when you will want to vary it, not only when you will need a sudden increase in lift to avoid obstructions, but, when starting the machine, you will start with an angular velocity of 0 and will then need to increase it to get up to operating speed. So we will need to know the eﬀect of varying the angular velocity. The ﬁrst place we used the fact that ω, the angular velocity, was constant was when we would pull factors of ω out of our integrals. The integrals were to compute the mean lift over a cycle, which, in our case, is half a revolution. At operating speeds, the length of a cycle is a very small fraction of a second, and there will not be enough time for ω to vary much over such a small interval. The maximum value of ω will be very close to the minimum value over a cycle, so we can approximate the lift using the assumption that ω is constant over the interval. For the next feature, if we go back to the computation of the vertical component of the acceleration, we started with the y coordinate of the position

y = r sin θ

30 Then

dy/dt = (r cos θ + sin θdr/dθ)ω

where ω = dθ/dt. So the product rule gives us the vertical component of the acceleration to be d dr dr dω d2y/dt2 = (r cos θ + sin θ )ω + (r cos θ + sin θ ) dt dθ dθ dt

The ﬁrst term is the acceleration we computed in the last section under the assumption that the angular velocity was constant, and the other term has a factor of dω/dt. When the angular velocity is constant, dω/dt = 0, and this term vanishes leaving us with our previous result. So to consider the case of variable angular velocity, we need only consider the eﬀect of the dω/dt term. As we saw in the last section, the fact that the rod is solid allows us to use the fact that, at each point in the rod,

dr/dt = dR/dt Where R is the distance from the center to a point on the track. To get the acceleration,

d2y/dθ2 To get the mean lift from This term ,

mω Z πZ R dR dω (r cos θ + sin θ ) drdθ π 0 R−L dθ dt

Again, we do the ﬁrst integration with respect to R

mω Z π L dR dω = (L( − R) cos θ + sin θ ) dθ π 0 2 dθ dt

Since dω/dt is under the control of the operator, we will not be able to tell what dω/dt will be. When computing the mean lift, we are integrating over a cycle, which is half of a rotation, and at operating speeds, a cycle takes such a small amount of time, we could also assume that dω/dt is constant over each interval, but let us just leave the factor in the integrand. As we have seen previously, L/2 − R is symmetric about θ = π/2, but cos θ is antisymmetric, so this term will balance out. When we consider the other term, sin θ(dR/dθ), as we have seen previously, dR/dθ will be generally positive when 0 ≤ θ ≤ π/2, and negative when π/2 ≤ θ ≤ π, but now, instead of having a factor of cos θ with it, we have a factor of sin θ. The diﬀerence is that instead of the product

31 always being nonnegative, the negative lift when 0 ≤ θ ≤ π/2 will be balanced out by positive lift when π/2 ≤ θ ≤ π and the contribution from this term will balance out to be practically 0. Conclude that this term will contribute no net lift. The result is that when ω is variable, the dω/dt term will not contribute signiﬁcantly to the net lift. Since dω/dt varies over the interval, the dω/dt term will add some more vibration, but with the lift spread out over eight or twelve units, it should be smoothed out .

6 Adjustments

These calculations are theoretical. One of the nice things about using the vari- ables q and h in our formulas is that it makes it easier to see what kind of an eﬀect it will have to change them. One of the problems with this development is that it does not take into account that the rod will have thickness. This is important because the rod needs to be strong in order to supply the lift which the device will require to lift a signiﬁcant payload. The entire vehicle will be hanging on the top ends of the rods. The spike can not come any higher than the point where the leading edge of the lower roller bearing crosses the vertical line through the center. In the illustration of the track

32 we see a spike where the track comes back to the center. Since the rod always goes through the center, and will have thickness, it will overlap this part of the track. Moreover, when the rod goes past the vertical, it will hang up on this spike, either causing the device to come to a halt or snapping oﬀ this spike. The bottom part of the track is deﬁned to be the locus of points which the bottom of the track will follow as the top part follows the upper part of the track. In this illustration, we are assuming that the bottom end of the rod is a simple point, and that is not the case. Not only does the rod have thickness, but I am

33 suggesting that there be a roller bearing on its ends to facilitate its following the track. The bottom part of the track will need to be the set of points which the roller bearing will trace as the top end of the rod follows the top part of the track. It will look something like

If we look at this track with just the rod in the vertical position

34 We see that the bottom of the rod no longer touches the lower part of the track when the rod is in the vertical position. In fact, there will be an interval, when the rod is near the vertical, where the bottom of the rod does not make contact with the track. This could be a problem, because the purpose of the lower part of the track is to push the rod up until it makes contact with the upper part of the track. This is important, because the point where the top of the rod makes contact with the upper part of the track is the load bearing element of the device. When the top of the rod loses contact with the upper part of the track, there will be no lift. The solution to this problem would be to

35 depend upon the centrifugal force will keep the top of the rod in contact with the top part of the track. If that happens the calculations from the previous section will hold. For this reason it is important to keep the interval where the bottom of the rod loses contact with the track to be a small as possible. If the bottom of the rod loses contact with the bottom part of the track, one good thing would be that there would be no negative lift. The negative lift would come from the bottom of the rod pushing against the bottom part of the track. That negative lift has to come from the middle term. If you look at the graph of the function for the middle term, you will see that the bottom part of the track exerts very little force when the rod is close to vertical, which would be when the bottom of the rod could lose contact with the bottom part of the track. So we can expect the centrifugal force from the other terms to keep the top of the rod in contact with the upper part of the track. This is important because the points where the tops of the rods contact the upper parts of the tacks are the source of all positive lift.