Hovercraft

Hovercraft is an event where teams must design and build a hovercraft. It is a Division B and Division C event for the 2017 and 2018 seasons. Teams will build self-propelled air-levitated vehicles, and also take a test on classical mechanics and other similar topics. The trial rules used at the 2016 National Tournament can be found at the bottom of the page in the "Links" section.

Description
Hovercraft is a dual lab, consisting of a test and a build portion. The test portion should take at least 20 minutes and competitors are allowed only 8 minutes to run their device. A 3-ring binder of any size is allowed, provided that all materials are hole punched and secure. Calculators of any type are also allowed and need not be impounded.

Device
The hovercraft itself can weigh no more than 2000 grams. Since the mass score counts for half of the entire device score, it is optimal that the hovercraft weigh as close to the 2000 g limit as possible, to achieve as high a score as is possible. However, teams should not abandon functionality for the sake of a perfect score - if a hovercraft is too heavy, it may not complete a successful run, meaning an overall build score of 0. Also, heavier hovercraft will often experience more friction and be harder to control, so there is a balance between run score and mass score.

There are no prohibited materials for the construction of the vehicle itself, but there are numerous other specifications and the vehicle may not modify the track provided by the event supervisor. Brushless motors and integrated circuits are only allowed if they are an essential part of a commercial fan. Also, only 2 fans are allowed, typically one lift and one thrust. The 2016 trial event rules are posted below, but there may be some changes in the official rules for next year.

Hovercraft have a few essential parts - the base of the hovercraft itself, the fans, the batteries, the wiring/switch, and the skirt. The best teams will experiment with all these parts to find a successful design. The speed control methods will vary from device to device, ranging from fine-tuned potentiometers to simply covering up the thrust fan with tape to reduce airflow. Unlike some events, there is no single "best" method for all teams; instead, the best method is the one which has been tested most.

Base
The base is likely the simplest thing to construct on the hovercraft. However, it still merits consideration in designing the most efficient device possible. A good base will make it easy to test and swap out components without destroying any structural support, among other considerations. It should also be strong enough to support all components without pushing the weight above 2000 g.

There are many choices of materials when constructing a base. Common selections include foam core and basswood sheets, though others are certainly possible. Materials should be light enough to allow weight adjustment to change while maximizing the mass score.

Also important are fan coverings. The rules require shielding over the fans with holes of 1/4" in diameter or smaller, so the base needs to include some form of covering (eg. mesh) so that no fan blades are exposed.

Besides the basics, any other information will depend entirely on the chosen components.

Fans
Fans are some of the most important components to a functional, efficient hovercraft. Fans have entirely different requirements for lift and thrust, and should be selected accordingly. It can be very helpful to enable swapping of fans, which facilitates testing.

Lift
Centrifugal blower fans are considered the most efficient type of lift fan, both in real-world designs and Science Olympiad devices. To fully support a 2 kg hovercraft, a fairly high pressure is required, making blower fans more efficient than traditional axial fans.

Thrust
Thrust fans should instead prioritize CFM to provide maximum acceleration. Although it is possible to successfully use a powerful computer fan, hobby motors have the potential to be more effective and more adjustable. Computer fans may struggle to provide enough thrust on a high-friction track like plywood, even if they work well on smoother surfaces like lab tables. Devices need to be able to work on all surfaces, not just those they have been tested on.

Batteries
There are a few possible choices for batteries.

Traditional commercial batteries, like AAs or 9Vs could work, but are unlikely to provide the amperage required to lift and propel a heavy device.

Many teams use lithium-polymer batteries, or LiPos. However, most LiPos are only 7.2V. This is significantly lower than the allowed voltage of 9V, and may not be enough to power 12V commercial fans.

The most ideal solution seems to be nickel-metal hydride batteries, or NiMHs. NiMH batteries can be found up to 8.4V, but often overcharge above 9 volts, meaning they can very effectively be used even with 12V components.

Although voltage is an important specification for any batteries used in hovercraft, so is capacity. Many teams use 3000 mAh batteries, which are more than enough for the amount of testing done in the Hovercraft event. 5000 mAh batteries are also available, and allow for longer testing periods but are more expensive. Some companies offer 1600 mAh batteries, but higher capacity is preferable.

Caution is needed when using either LiPo or NiMH batteries, which can be extremely dangerous if used improperly. NiMH batteries are typically less likely to catch fire than LiPOs, which can go up in flames should they be improperly charged. However, should a NiMH battery be damaged, they explode rather than catch fire. However, this is only the case that the battery is very compromised.

A general guideline is to charge NiMH batteries at 1 C. For a typical 3000 mAh battery, this means charging at 3.0 A.

Wiring
Wiring should be fairly cut-and-dry. Wires should be selected based on the power specifications of fans and batteries. There are many options available for switches, and very simple designs will work. Many household switches have two sides, perfect for the two circuits needed to power two fans. JST connectors or similar can be used to enable easy swapping of components, especially batteries and fans.

Skirt
The skirt is likely the most difficult part of a hovercraft to perfect. Beyond hardware like fans and batteries, the skirt is the largest determining factor of a device's success. Skirts need to have as low of friction as possible to have accurate results and to enable low speed runs. Many hovercraft failed at the MI and PA state competitions when the track was created out of plywood, as many designs were unable to compensate for such high friction.

There are practically infinite possible variations of the hovercraft skirt. However, these can be divided into a few smaller subsets.

The bag skirt is exactly what it sounds like - a completely contained cushion of air, possibly with small holes poked in it to decrease friction. This is one of the most common designs in Science Olympiad, seeing as it is easy to construct and fairly stable. However, its shortcoming lies in friction, as bag skirts are generally one of the most high friction skirt designs.

The wall skirt is the other variation likely to be found in Science Olympiad competitions. The skirt itself is only a wall around the pocket of air, but all surfaces must be flat for the purpose of Science Olympiad, so a wall skirt can still contain a pocket of air. Unfortunately, the wall skirt is more prone to air leakage and difficult to construct, which makes it difficult to effectively use this design.

Other innovations can be added to these types of skirts. One design used in many real-world hovercraft is the momentum curtain, which directs airflow to the edges of the hovercraft before bending down to propel the hovercraft up. This has the benefit of placing the highest pressure region near to the ground as opposed to in the center of a cushion of air. Also, many Science Olympiad hovercraft use a flat piece of wood or foam core at the bottom of a bag skirt to create a flat surface and decrease friction.

The finger skirt is the third common category of hovercraft skirts, but is very unlikely to be used as almost all high school students lack the tools to construct such a complicated design.

Testing, testing, and more testing is the only way to guarantee a good skirt design and a good overall score. Many teams are buying the same fans and batteries, so what truly sets teams apart is their skirts.

Written Exam
Part 1 of the event consists of a written test which draws from the AP Physics 1, or algebra-based mechanics, curriculum. This test must contain at least five questions from each of the following topics: Newton's laws of motion, kinematics, kinetic energy, air cushioned vehicles and applications, and fluid mechanics for Div. C only. 20 – 30 minutes is suggested to complete this test. All answers must be provided in metric units along with the appropriate significant figures.

Newton's First Law
An object will remain at rest or in uniform motion unless acted on by an external force.

Newton's first law describes inertia, the tendency of an object to maintain a constant velocity in the absence of an external force.

Newton's Second Law
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object.

[math]F = ma[/math]

Newton's second law describes the relationship between force, mass, and acceleration as represented above. Common problems for this section will give 2 of the values, and ask for the third. This idea is also likely to be used within larger, more difficult problems as a simple conversion step.

Newton's Third Law
Every action has an equal and opposite reaction

Forces always occur in equal action-reaction pairs, no matter what the masses of the objects being acted upon are.

For example, if a 5 (metric) ton collides with a 5 gram mosquito, both experience the same force - the mosquito just experiences a greater acceleration due to its lesser mass. Similarly, a rifle recoils, as the force that propels a bullet forward must have an equal and opposite reaction, which propels the rifle backwards.

Kinematic Equations
[math]1.\quad v = v(0) + at[/math]

Acceleration times time equals change in velocity, or acceleration equals change in velocity over time - the definition of acceleration. This works well when no distance is given.

[math]2.\quad \Delta x = v(0)t + 0.5at^2[/math]

Though this should not be part of an algebra based test, change in distance is equal to the integral of velocity, which is exactly what this equation is - the integral of the above velocity equation. Another way to think of it is the average velocity times time, where final velocity is calculated using the equation 1 and average velocity is found from equation 4.

This works well when no final velocity is given.

[math]3.\quad \Delta x = vt - 0.5at^2[/math]

The exact same thing as the previous equation, but from the opposite end. Instead of using final velocity, initial velocity is found from equation 1 and plugged into equation 4, again to get average velocity. This can be used when no initial velocity is given, but works almost the same way as equation 2.

[math]4.\quad \Delta x = 0.5(v + v(0))t[/math]

Average velocity times time is equal to distance, which pretty easy to remember. This equation is great when no acceleration is given, but it only works for a constant acceleration.

[math]5.\quad v^2 = v(0)^2 + 2a\Delta x[/math]

How this equation works is based on kinetic energy. Multiply the equation by 0.5m and this is the result: [math]0.5~mv^2 = 0.5~mv(0)^2 + ma\Delta x[/math]. The last term can be converted into [math]Fd[/math], or work, while the first two represent final and initial kinetic energies. This is just a change in kinetic energy calculation!

There are 5 variables in kinematics - initial velocity, final velocity, acceleration, time, and distance; these equations make it possible to find any two quantities given the other three.

Projectile Motion
Projectile motion can be treated as 2D kinematics - 1 equation for x, and 1 equation for y. However, it adds another variable - the angle the projectile is launched at.

99% of the time, x can be treated with a simple equation: [math]x = v(0)~cos(\theta )~t[/math], and y uses one of the kinematics equations.

These three, already derived equations can be used to run calculations faster or to double check.

[math]t = \frac {2v(0)~sin(\theta )}{g}[/math]

[math]h = \frac {v(0)^2~sin(\theta )^2}{2g}[/math], where [math]h[/math] is maximum height

[math]R = \frac {v(0)^~2~sin(2\theta )}{g}[/math], where [math]R[/math] is range of the projectile

When in doubt, the basic kinematic equations will always work.

Example Problems
Here are some examples of working through kinematics with the five equations.

''1. A car moving with constant acceleration covers the distance between two points 50.0 m apart in 5.00 s. Its speed as it passes the second point is 15.0 m/s. What is its speed at the first point?''

We use equation 4 to find

[math]50.0 = 0.5(15.0 + v(0))\cdot 5.00[/math]

[math]50.0 = 37.5 + 2.50\cdot v(0)[/math]

[math]v(0) = 5.00~m/s[/math].

2. What is the car's acceleration from problem 1?

As the initial velocity has already been calulated, we use equation 1.

[math]15.0 = 5.00 + a\cdot 5.00[/math]

[math]a = 2.00~m/s^2[/math].

Let's suppose we did not have the initial velocity already caluculated. Equation 3 would be more suitable.

[math]50.0 = 15.0\cdot 5.00 - 0.5\cdot a\cdot 5.00^2[/math]

[math]50.0 = 75.0 - 12.5a[/math]

[math]a = 2.00~m/s^2[/math].

As you can see, both equations yield the same answer. With 3 out of the 5 values, there's always 1 direct kinematic equation to use, and 1 more indirect path using 2 kinematic equations, which helps if you happen to forget one.

''3. A airplane pilot wishes to accelerate from rest (a stationary position) with a constant acceleration of 4g to reach the speed of sound, 331 m/s. How long would the period of acceleration be?''

We can again use equation 1 to quickly solve the problem.

[math]331 = 0 + 4\cdot 9.81\cdot t[/math]

[math]t = 8.44~s[/math].

4. How far will the pilot in problem 3 travel during the period of acceleration?

Using the amount time above, we choose to work with equation 2, getting

[math]\Delta x = 0 + 0.5\cdot 4\cdot 9.82\cdot 8.44^2[/math]

[math]\Delta x = 1398~m[/math].

Of course, there's another equation we could have used if we hadn't already calculated the time, so let's try using equation 5.

[math]331^2 = 0 + 2\cdot 4\cdot 9.81\cdot \Delta x[/math]

[math]109600 = 78.48\Delta x[/math]

[math]\Delta x = 1396~m[/math].

This time the answer is slightly different due to intermediate rounding, but both answers would be accepted and would probably end up the same with proper sig fig usage.

Now let's work through some 2D kinematics problems, since we've covered all 5 basic kinematics equations.

''5. A swimmer dives off a 10.00 m high diving board with a horizontal leap. What must her velocity be just before her leap if she is to clear the edge of the pool, which extends 2.00 m beyond the edge of the diving board?''

Since initial velocity is perfectly horizontal, we can use equation 2 to find the time it takes her to fall without worrying about initial y velocity.

[math]10.00 = 0 + 0.5\cdot 9.81\cdot t^2[/math]

[math]t = 1.428~s[/math].

Now it's just a basic question - what velocity is required to clear 2.00 m in 1.428 seconds? No kinematics needed.

[math]2.00 = 1.428v[/math]

[math]v = 1.401~m/s[/math].

''6. A quarterback throws a football with a initial velocity of 15 m/s and angle of 30 degrees. What is the highest point of the football's trajectory?''

Lets start by running a kinematic equation. We'll use equation 5 to get to the highest point, where velocity must be equal to 0 in the y direction.

[math]0 = (15~sin(30))^2 - 2\cdot 9.81\cdot h[/math]

[math]0 = 56.25 - 19.62h[/math]

[math]h = 2.867~m[/math].

We could also use one of the projectile motion equations as well.

[math]h = (15~sin(30))^2/2\cdot 9.81[/math]

[math]h = 2.867~m[/math].

You can see where we got the maximum height equation from earlier.

7. How long will it take the football from problem 6 to reach a wide receiver of the same height?

We can do this equation 2.

[math]\Delta y = 15~sin(30)\cdot t - 0.5\cdot 9.81\cdot t^2[/math].

Solve where change in y is equal to 0 - this will yield 2 possible points, one where t is equal to 0 and the t we are looking for.

[math]0 = 7.5t - 4.905t^2[/math]

[math]t = 1.529~s[/math].

We can also just use the time equation from the projectile motion equations.

[math]t = 2\cdot 15\cdot sin(30)/9.81[/math]

[math]t = 1.529~s[/math].

I originally made a dumb mistake when working through this problem, but I caught it using the second method. I cannot stress how important it is to check both ways whenever possible - don't just rely on the projectile motion equations.

8. How far away does the wide receiver need to be to catch the ball?

Thankfully, we already have t, which makes this a bit easier. We can now just plug this into a basic x equation.

[math]\Delta x = 15\cdot cos(30)\cdot 1.529[/math]

[math]\Delta x = 19.862~m[/math].

Given how long this calculation has been, it's a good idea to check with the range equation.

[math]R = 15^{2}~sin(30\cdot 2)/9.81[/math]

[math]R = 19.862~m[/math].

As you can see, it's always possible to rely on those 5 kinematic equations, so if you ever get confused, just go back to those 5 equations.

Calculations
Even though the section says "kinetic energy", tests are more likely to cover multiple energy calculations.

Kinetic energy: [math]K = 0.5mv^2[/math]

Gravitational potential energy: [math]U(g) = mgh[/math]

Elastic potential energy: [math]U(s) = 0.5kx^2[/math]

Work: [math]W = Fd[/math]

Just like [math]F = ma[/math], these equations can be used by themselves or in larger problems. Momentum problems will be tough and could require instant application of multiple of these equations.

Momentum
Momentum is an important part of all collisions.

Momentum: [math]p = mv[/math]

Impulse: [math]J = \Delta p = F\Delta t[/math]

In any collision, momentum is conserved. This is called the Law of Conservation of Momentum.

[math]\Sigma p(initial) = \Sigma p(final)[/math]

Inelastic
In a perfectly inelastic collision, two objects combine to form essentially one mass. Kinetic energy is not conserved in such collisions, and instead can be lost to factors such as sound or heat.

[math]m(1)\cdot v(i) = (m(1) + m(2))\cdot v(f)[/math]

Inelastic collision problems are relatively easy as they only require solving 1 problem.

Elastic
In a perfectly inelastic collision, both kinetic energy and momentum are conserved as both objects simply bounce off of each other.

Thus two equations can be used to solve for the final velocities of both objects:

[math]m(1)\cdot v(1,i) + m(2)\cdot v(2,i) = m(1)\cdot v(1,f) + m(2)\cdot v(2,f)[/math]

[math]m(1)\cdot v(1,i)^2 + m(2)\cdot v(2,i)^2 = m(1)\cdot v(1,f)^2 + m(2)\cdot v(2,f)^2[/math]

Note that not all of the 0.5 terms cancel out, but are present in each of the kinetic energies used in the second equation. Both equations should yield a quadratic equation with two solutions - the initial conditions and the more useful final conditions.

Example Problems
''1. A 70 kg astronaut is installing a new sensor onto the outside of a space station. If he were to throw the 10 kg sensor at a speed of 2.8 m/s, how quickly would he begin to move, and in which direction?''

We will use Law of Conservation of Momentum, assuming an initial momentum of 0.

[math]0 = 70\cdot v + 10\cdot 2.8[/math] [math]v = -0.4~m/s[/math].

The negative sign indicates the astronaut is moving in the opposite direction of the sensor.

''2. A 1000 kg car moving at 15.0 m/s collides with a stopped 6000 kg truck. What is the final velocity of the two vehicles, assuming they lock together after colliding?''

Since this is an inelastic collision, we only use conservation of momentum.

[math]1000\cdot 15.0 = (1000+6000)\cdot v[/math]

[math]15000 = 7000v[/math]

[math]v = 2.142~m/s[/math].

Pretty easy, right?

''3. A 1.00 kg electric vehicle veers off course and collides with a 0.20 kg can. If the initial velocity of the vehicle is 5.00 m/s and the can is at rest, what are the final velocities of both objects assuming a perfectly elastic collision?''

Let's start with conservation of momentum:

[math]1.00\cdot 5.00 = 1.00\cdot v(1) + 0.20\cdot v(2)[/math]

[math]v(1) = 5.00 - 0.20v(2)[/math].

And now for conservation of kinetic energy, since this is perfectly elastic:

[math]1.00\cdot 5.00^2 = 1.00\cdot v(1)^2 + 0.20\cdot v(2)^2[/math]

[math]25.0 = (5.00 - 0.20v(2))^2 + 0.20\cdot v(2)^2[/math]

[math]25.0 = 25.0 - 2.00v(2) + 0.24v(2)^2[/math]

[math]0.24v(2) = 2.00[/math]

[math]v(2) = 8.33~m/s[/math]

[math]v(1) = 5.00 - 0.20(8.33)[/math]

[math]v(1) = 3.33~m/s[/math].

And we're done! Feel free to plug the values in for both equations to verify your results. That really is the best way to check these types of problems.

Density
Density is a material's mass per unit volume, applicable to any form of matter, not just fluids.

[math]\rho = \frac {m}{V}[/math]

The specific gravity is the ratio of its density to the density of water at 4.0 degrees C, or [math]1000~kg/m^3[/math]

Pascal's Law
Pressure is the normal force per unit area.

[math]p = \frac {dF}{dA} = \frac {F}{A}[/math]

When the weight of a fluid is ignored, pressure is equal throughout the entire volume of the fluid. However, this is not always the case.

To adjust for the weight of the liquid, we can use this equation: [math]\Delta p = -\rho g\Delta y[/math]

This can also be expressed as: [math]p = p(0) + \rho gh[/math], where p(0) is the pressure at the surface of the fluid and h is the depth.

Pascal's Law states that pressure applied to an enclosed fluid is transmitted, undiminished, to every portion of the fluid and the walls of the containing vessel.

In equation form: [math]p = \frac {F(1)}{A(1)} = \frac {F(2)}{A(2)}[/math]

This leads to another important equation: [math]F(2) = \frac {F(1)A(2)}{A(1)}[/math]

Pascal's Law is why the hydraulic lift works, which can be used to multiply force using water. However, the hydraulic lift face the same problem as any simple machine - for every increase in force, there is a proportional decrease in distance and work is kept the same.

Buoyancy
Archimedes' principle describes buoyancy: When a body is completely or partially immersed in a fluid, the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body.

This explains why objects seem to weigh less in water than in air, and why some things float.

[math]F(B) = \rho Vg[/math], where V is the volume of the object that is submerged in the water.

Bernoulli's Principle
Volume flow rate, or dV/dt, is equal to fluid velocity times area.

The (fluid) continuity equation states that the volume flow rate is constant for an incompressible fluid.

[math]A(1)v(1) = A(2)v(2)[/math]

Bernoulli's principle or equation combines all of the factors affecting a flowing incompressible fluid, by applying conservation of energy.

[math]p + \rho gy + 0.5\rho v^2 = constant[/math] for any point in the fluid.

Note that when there is no velocity, and the fluid is instead still, the equation simply becomes the pressure equation from above.

Viscosity
Viscosity is the internal friction within a fluid, or the resistance to flow. A viscous liquid tends to cling to the solid surface of its container - this is why there is always a thin boundary layer near the outside of a pipe where fluid velocity is almost 0.

The pressure difference required to maintain a given flow rate is proportional to [math]\frac {L}{R^4}[/math], where L is the length of the pipe and R is the radius. Even a small reduction in radius can drastically increase the required pressure difference.

Previous equations assume a laminar flow with zero viscosity. However, this is actually impossible as a small amount of viscosity is required to ensure laminar flow. Above a certain critical speed, fluid flow will change from even, laminar patterns to irregular and chaotic turbulence.

Air Cushioned Vehicles
While these sections may no longer be tested under the 2018 rules, a grasp on the history of the Hovercraft may yield useful insights into the device building process. Many important ideas for hovercraft construction are historically documented, and are most certainly still applicable.

History

 * 1716: Emanuel Swedenborg first uses the term "hovering" to describe surface effect vehicles.
 * 1870s: John Thornycroft develops a patent for an ACV, but lacks the appropriate engines to power his ideas.
 * 1930s: Charles Fletcher constructs a walled ACV, but is unable to file a patent due to military classification.
 * 1950s: Christopher Cockerell (credited as the inventor of the hovercraft) develops the concept of an annular ring to support a hovercraft.
 * 1958: Cecil Latimer-Needham uses rubber to construct a double-walled skirt to make functional hovercraft possible.
 * 1959: The SR.N1 is first demonstrated to the public, crossing the English Channel with 3 passengers.
 * 1960s: Saunders-Roe develops larger hovercraft for commercial use.

In recent history hovercraft have developed into more specialized niches, but have not experienced similar sweeping innovations.

World Records

 * Largest Civil Hovercraft: SR.N4 Mk. III
 * Largest Military Hovercraft: Zubr class LCAC
 * Largest Developer of Military Hovercraft: Soviet Union
 * Fastest English Channel Crossing: 22 minutes, by the Princess Anne SR.N4 Mk. III
 * World Hovercraft Speed Record: 137.4 km/h
 * Land Hovercraft Speed Record: 90.53 km/h
 * Longest Continuous Use: 20 years, by the prototype SR.N6

Links
Trial Events rules