Rationale: of nowhere, especially here in the Netherlands,

Rationale: In the beginning of my mathematical journey weather was a topic that stood out to me because the whole job of predicting weather revolves around the use of very complex mathematical equations dedicated to try and model possible future wind shifts and weather patterns. Of course, doing a IA on this would prove way too difficult as even pure mathematicians have trouble modelling patterns in weather, but since the topic of meteorology still interested me I decided to instead divert my attention to modelling something related to weather but being a little more manageable.

After scouring previous IA’s I realized that while modelling rainfall was a very interesting subject it was overdone, and as such I decided to focus my IA on modelling hailfall, specifically on the falling motion. I am quite interested by hail, as it exists in a weird limbo between rain and snow, and sometimes when it hails it appears out of nowhere, especially here in the Netherlands, while for rain it is much easier to predict when and where it will fall. To continue, hailfall is different from rainfall since the density, average mass, shape, size, weather patterns associated with it, and the state of matter it is in are different for the two, which makes this IA similar in question but different in application as with the rainfall modelling IA’s. Introduction: This exploration aims to correctly model hailfall, which is important since hail is known to damage homes, cars, and aircrafts, all while being potentially deadly to people and livestock alike, so by modelling the fall of hail I can predict the force, and as such damage, behind an average hailstorm. To model hailfall I need to consider several variables that could affect the model, with the biggest variable being the size of hail. According to the TORRO Hailstorm Intensity Scale hail can have a diameter of 5 mm at minimum and be considered a H0 (causes no damage) up to a diameter of 100 mm and greater still, being considered a H10 hail-piece, capable of extensive structural damage and a risk of severe or even fatal injuries to persons caught in the open. I thus decided to focus the second part of the exploration on falling motion in three different hail diameters: the hail size at 25 and 45 mm, the extremities of the average hail size that can cause damage, and the larger hail-pieces that have a diameter of d = 10.

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0cm. However, the motion of falling objects also depends on the air resistance on top of the gravity and mass, which causes the hail-pieces to experience an upward drag, and must be accounted when modelling hailfall. This then allows me to determine the velocity upon impact of the hail (terminal velocity), and the time it takes the hail to reach the ground.Lastly, the model was applied to another example, that of a falcon diving in the air, falling in much the same way as the hail. However, the diving falcon has a different shape and changes it shape mid-dive by tucking its wings in different way to be as aerodynamic as possible, a difference that must be kept in mind when creating a model for the falcon.

First Model, no air resistance: To develop the first model, I have to use the Newtonian forces on falling objects. One of Newton’s laws is the Second Law of Motion, which states that mass times acceleration equals force. This can be written as: F=ma, where F is the net force, m the mass of the object and a is its acceleration.?Newton’s Law of Motion can be used to model a piece of hail falling with an acceleration (ignoring the air resistance). Since acceleration (a) is the first derivative of velocity (v), and the second derivative of position (s), this can be written as the equation: For a falling object the only acceleration acting upon it is vertical acceleration, which is equal to gravity (g). This gives me: As such a differential equation for a piece of hail with an acceleration g is obtained. The value for gravitational acceleration at or around the earth’s surface (in the atmosphere) is:, and since its value does not change much in the upper atmosphere (at 6000 meters of altitude the acceleration of gravity decreases by 2 per thousand), we will consider g a constant.

However, in order to solve the differential equation, I first need to set some initial conditions: The hail starts at rest from 6,100 meters in the air, which is the average height where hail is brought up to by upwards air currents in a thunderstorm cloud, making this the origin for descent. It can also be stated that, at origin, the velocity (v), position (s), and the time (t), are all equal to zero. This differential equation can now be solved by use of basic integration: To find the constant of integration k, the initial conditions explained above must be applied: As such v = gt (since k=0), and from this I can use another position equation, to continue, while again using basic integration to solve for the upcoming differential equation. To determine the constant of integration, the initial conditions have to be used again, with t(0)=0. ?This gives the equation: As previously mentioned, and . These values can be substituted in the equation above to solve for time (t): Time has three significant values since so does the constant of gravity.Now, to determine the velocity of the piece of hail upon arriving on the ground I have to use the first equation, . The falling motion of a piece of hail falling from a given distance of 6100 meters which experiences a constant acceleration of has now been modelled.

This model works for any size of piece of hail, since neither acceleration nor velocity is dependent on mass or size. As I aimed, the results obtained show how long it takes the piece of hail to reach the ground, the time it takes, and its velocity when it reaches the ground. As demonstrated in this previous section it would take the piece of hail approximately 35.3 seconds to reach the ground and it would hit at a velocity of 346 meters per second (or 1246 km/hr). This value seems to be rather large, as the hail would fall slightly faster than a jet and be able to punch through thin steel, but this is because this model is unrealistic, as only the basic Newtonian gravitational force has been taken into account for the fall, however in reality hail, as all other falling objects, is subject to air resistance which would affect it’s falling motion, and hail would at one point reach a terminal asymptotic velocity.Second Model, has air resistance: Having found this first model for hail is good to lay the ground-work, but I now need to build upon it with a more realistic model. To do this, I have to take into account the air resistance, a force that acts against anything that moves through the atmosphere (since the atmosphere is not a vacuum and as such consists of particles which slows down anything they come into contact with).

The amount of air resistance is dependent upon a variety of factors, with the most important being the speed of the object, its cross-sectional area, and the shape of the object. An increased speed and cross-sectional area of a moving object in turn leads to an increase in the amount of air resistance. This air resistance, also commonly referred to as drag, can also be called Fair or Fdrag. The drag is proportional to the cross-sectional area (A) of the object and is in turn proportional to the velocity of the object squared, as derived from Bernoulli’s equation. Accordingly, an expression for the drag force is:, the drag equation, where k is a parameter defined as:, where ? is the density of air and Cd is the Reynold’s number.

The Reynold’s number is a dimensionless quantity, which depends upon the shape of the object. In our case, according to tables, I can assume that the drag coefficient (Reynold’s number) is 0.5 since hail is substantially spherical in shape. During the fall of hail two forces act on it, the downwards force of mg (mass*gravity) and the upwards force caused by the drag, For now I’ll assume that ? is constant, even if in reality it changes slightly based on the altitude at which the hail is located. The resultant force equation between mg and can then be written as: , but to make my life easier I’ll substitute parameter for , so that. I’ll also consider this parameter constant.

ie: I will assume that A is constant for a specific piece of hail.According to Newton’s Second Law , which can consequently be written as, in our case: I want to solve this equation in relation to the acceleration, so by isolating the acceleration it becomes: Remember, acceleration is equal to the change in velocity over the change in time, and this can be written as: , where and represents the differentials of v and t, respectively.By comparing equations and together we find that: To keep the following section as clear as possible I’ll substitute parameter for , so that (which in the present case is a constant).Solving with respect to gives the following chain of equations:If we define x so that: We get, by substituting in (1): From the tables of fundamental integrals (Gieck), we get:Where c is the integration constant.Substituting in the previous equation, we have:Where d is another integration constant. Since we have, by substitution:,where , n being a constant of integration.By reintroducing the initial conditions: v=0, t=0, we get:, and since ln(1) = 0 that implies that n is equal to 0 as well.

So we have:And getting rid of the natural logarithm, ln, gives us:Now, to find v, we will have to isolate it:This is the formula for the velocity of a piece of hail, with air resistance, and the previous conditions (g is constant, density of air is constant, Reynold’s number is 0.5, and the cross-sectional area is constant), at time t.To quickly check we can plug 0 for t and we get v = 0, which is correct.

However, we still have to find the time it takes for the hail to fall to the ground. To do that, we will write the velocity as the derivative of s (height) over time, and then solve the differential equation by separation of our variables. ie: In order to integrate the equation, I will use the following substitution:This can then be substituted into the equation as follows:Integrating the differential equation, we get:From integration tables (Gieck):Since , when t = 0 and s = 0, u = 1, then we have:So, by substituting c back into the equation we get:Then, by substituting back into the equation we get:Now we have to find t as a function of s:This is an implicit equation for t, which has to be resolved by trial and error.From the previous equations we can write b as: We will now assume that the hail has a diameter of 2.5 cm, or 0.

025 meters.This makes the volume of the hail become , which is equal to 0.00000818 cubic meters. Since ice is 900kg per cubic meter, the mass of this piece of hail is 0.00736 kg. The cross-sectional area (pi*r2) will be 0.000491 m2.

The density, assuming it is constant, is, at the sea level, 1.22500 kg/m3. By substituting all these values into the equation we get:Which is equal to b = 0.0204 m-1.

By substituting this number into the equation we get:Now to solve for t I will have s be 6100 meters:And by using nSolve on my calculator I find that the time it takes for the 2.5 cm piece of hail under the specified conditions to fall 6100 meters with air resistance is 279.8 s.The velocity of this piece of hail when it hits the ground is going to be:Since the time is 279.

8 s, this equation becomes:So hail at 2.5cm under these specified conditions and with air resistance falls at 78.8 km/hr.Now, we can try different sizes of hail (4.

5 cm and 10 cm).For convenience, we will simplify parameter b. So now we will calculate b for the different radii:At d = 0.045 m the radius is 0.0225 m, so b is:And finally, at d = 0.10 m the radius is 0.05 m, so b is:Now, we will write the equation for each of these two pieces of hail with varying diameters:  But to make my life easier, I will simplify the equation with respect to t.So the time the 4.5 cm piece of hail takes to touch the ground and the velocity it has right before it hits the ground is:Finally, the time it takes for the piece of hail with 10.0 cm in diameter to reach the ground and the velocity at which it reaches the ground is:As such the bigger the hail is, the faster its velocity is when it hits the ground, and as such the less time it spends free-falling in the air.