This post will discuss the physics of the railroad.

In modern railway systems track geometry is such that all railroads are composed of a series of straight lines and curves. The straight lines are called tangents and the curves are of various radii. In order to allow the trains to travel at higher speeds the curves are canted in order to balance the centrifugal force of the train with the x-component of the gravitational force. The measure of this cant is called super-elevation. Depending on the train speed and the radius, each curve will be canted in order to achieve an equilibrium state where these forces are balanced. The transition section of track from the tangent to the curve body of a specific super-elevation, is called a spiral. The higher the super-elevation the longer the spiral.

The mass of the train need not be known, as it will be cancelled out of the equation. Therefore the only two factors that determine the required super-elevation for a particular curve are it’s radius and the speed of the train. The radius for the curve can be accurately determined given the length of a chord and the distance from it to the arc. The following image is exaggerated for clarity, but the actual chord length would be much smaller compared to the curve’s radius in a real world situation.

The following will demonstrate how the radius of this curve can be calculated given the chord length (DE) and the distance to the arc (BC).

let AB = a, BC = b, BE = c, DE = d, AC = r and AE = r.

We begin with the following equations:

(1)

(2)

(3)

Substitute equations 1 and 2 into equation 3 and solve for the r:

The distance between the two inside heads of the rail where the wheel flanges sit is called the track gauge (G). The height of the high rail also known, as the super-elevation, is h and the angle of the cant is θ. The following images illustrate the gravitational forces acting on the train when is sits on a super-elevated curve.

Using the diagram above we can derive the following equation for the sine of the angle θ.

A break down of the gravitational force components is as follows:

The gravitation force acting on an object is equal to the product of the mass and the acceleration due to gravity, approximately 9.8 m/s^{2} on earth.

The x-component of this gravitational force can be determined using trigonometry as follows and by substituting in the equation above.

The centrifugal force of the train traveling around the curve is calculated using the following formula:

At the equilibrium velocity these two forces should be equal in order to ensure safe travel for the train. In order to determine the appropriate super-elevation we equate these two forces and solve for h. It should be immediately obvious how the mass cancels in this equation.

We can now substitute the value for the sine of θ to solve for height of the high rail or super-elevation.

Depending on the situation, the frequencies of the various types of trains (passenger or freight trains), and their different speeds, the actual super-elevation that is implemented in the real world may be either slightly higher or lower than that of the equilibrium velocity. The term used to describe this is called cant excess and cant deficit.

That concludes this post on railway curve super-elevation. The next article on this topic will discuss the role of the Harsco Mark IV tamper and how it implements these formulas to perform its operations.

Dan Roy, Founder and Creator of Universal Physics.

]]>We previously discussed the concepts of matrices and how they are important in representing complex mathematical abstractions. There are many operations that can be applied to matrices. The first most basic is scalar matrix multiplication where a matrix is multiplied by a constant.

Consider the following matrix A.

The matrix can be multiplied by a constant *k* in the following manner.

Furthermore, two matrices can be added or subtracted if they contain the same number of elements in each row and column. They can also be multiplied if the number of columns in the first matrix is equal to the number of rows in the second matrix.

Example of adding two matrices.

For an example of multiplication consider the two matrices.

The two matrices can be multiplied two ways the first is the inner product.

The second method is by determining the outer product.

If we add these matrices its obvious that both methods produce the same answer in this example.

Another operation that can be performed is the transpose of a matrix. The transpose of a matrix can be found by swapping rows and columns in the following manner.

Moreover, the inverse of a function is another important operation that can be performed on a matrix. The quotient of two matrices can be found my multiplying by the inverse of the second matrix.

If two matrices A and B are inverses of one another the product of the two is equal to the identity matrix.

I will now demonstrate how to find the inverse of a matrix.

Consider the following 3×3 matrix.

Create an augmented matrix with the matrix A on the left and the identity matrix on the left.

Now we perform the necessary elementary row operations to rearrange the identity matrix onto the left similar to the Gauss-Jordan elimination method discussed in the previous post.

Add row 3 to row 1.

Divide row 1 by 3.

Multiply row 1 by 7 and subtract it from row 2.

Subtract row 1 from row 3.

Divide row 2 by -2.

Multiply row 2 by 9 and add it to row 3.

Divide row 3 by 56.

Multiply row 3 by 6 and subtract it from row 2.

Finally we subtract row 3 from row 1 to achieve the reduced form.

Therefore the inverse of matrix A is as follows.

That concludes this post on matrix operations. I will introduce the concepts of vectors in the next post and then begin discussing eigenvalues and eigenvectors then move on to introduce the use of complex numbers.

Dan Roy, Founder and Creator of Universal Physics.

]]>I will begin now by introducing a basic example of the intersection of two lines.

Recall in elementary school the following equation for a line.

For this example we have two lines represented by the following two equations.

In order to find the point of intersection of these two lines we can perform a simple substitution and solve for x.

Now we can substitute the value of x back into one of the equations in order to solve for y.

Therefore the point of intersection of the two lines is (-2, 1).

As mentioned in linear algebra we often use matrices for a better mathematical representation of systems. A typical example of a matrix is as follows.

For this example we can represent this system of equations as a matrix. To begin we must rearrange each equation to the following form.

Now we can represent the system of equations as the following set of matrices and create an augmented matrix.

Now that we’ve created the augmented matrix we need to get the matrix into a reduced row echelon form using elementary row operations.

First we subtract row 2 from row 1.

Now we divide row 1 by -5.

Finally to achieve reduced row echelon form we multiply row 1 by 2 and subtract it from row 2.

From looking at this matrix we now have the two equations x = -2 and y = 1 representing the point of intersection of the two lines that we found previously.

For this example using the Guass-Jordan elimination method doesn’t seem practical. But for more complicated systems of equations especially those involving higher dimensions we find that matrices offer a much more practical abstraction.

In the next posts I will discuss matrix operations and their importance into quantum mechanics as well as introduce the concept of eigenvalues and eigenvectors.

Dan Roy, Founder and Creator of Universal Physics.

]]>The concept of the paradox of Gabriel’s horn or Torricelli’s trumpet is very popular and interesting to many mathematicians. The horn object is created using the function of the inverse of *x* in the domain *1 ≤ x ≤ ∞* spun about the *x* axis. The domain is chosen to avoid the asymptote at *x = 0*.

Whats interesting about this shape, as we will see, is that it has an infinite surface area yet has a finite volume.

If we consider the horn to be a container full of paint we can easily visualize why this concept is considered a paradox. It is impossible to imagine a container that can hold a finite amount of paint yet require an infinite amount of paint to coat it’s inside wall.

For this example we will use calculus to prove that this peculiar object has an infinite surface area yet has a finite volume.

First we’ll find the volume using simple integration.

We have just proven that the volume of the horn is π and therefore is finite.

In the last post I introduced the concept of finding the surface area of 3-dimensional objects using a combination of integration and partial differentiation. The previous post introduced a general equation that can be used. However for this example we’re going to use an easier equation to determine the arc length of a function and simply spin it about the x-axis.

Fortunately we don’t actually need to solve this integral since the following is true.

Therefore it is obvious that the following is true as well.

Using this trick we can predict how the integral converges and we can now simply solve the following.

So it can be shown that Gabriel’s horn has an infinite surface area yet has a finite volume.

The mystery behind the paradox can easily be explained however. The volume is finite because of the way the 3-dimensional units of volume converge as x approaches infinity.

The posts so far have focused primarily on integral calculus and differentiation. In the next few posts I will begin to discuss linear algebra and introduce the use of matrices as well as discuss the importance of eigenvalues and eigenvectors.

Dan Roy, Founder and Creator of Universal Physics.

]]>In the partial differentiation of functions involving multiple variables much of the same rules apply as in regular differentiation. We choose the one variable in which we differentiate with respect to and the other variable is considered as a constant. As such there are multiple differential equations each in respect to different variables.

For a function involving two variables we can define the partial derivatives as follows.

For partial differentiation we use the following notation to distinguish it from regular differentiation.

One important application of partial derivatives is in finding the area of a curved surface. Before the discovery of calculus this was done by approximation. Archimedes is credited for finding the formulas for the surface areas of many 3-dimensional curved objects.

The following is the general equation for the surface integral of a 3-dimensional object in terms of *z*.

This equation, and partial differentiation in general, is incredibly useful in solving many problems in both mathematics and physics itself. In the next post, which will be on the subject of the paradox of Gabriel’s horn, I will make use of the above equation. Also in the future, once I’ve establish a strong mathematical foundation, I will begin to demonstrate the importance of calculus in quantum theory. The particular example of the use of partial differentiation that im refering to is in its importance in regards to the Heisenberg’s Uncertainty principal.

Dan Roy, Founder and Creator of Universal Physics.

]]>This post will discuss an interesting type of pattern called an involute. Using integration I will demonstrate in detail how to calculate its area and its applications to real world problems.

Consider the following scenario about a grazing cow on a farm.

You have a round silo on your farm with a radius of r. Attached on west side of the silo theres a rope of length πr. Tied to the rope, is a cow. What is the total grazing area of the cow?

The pattern outlined by the maxium distance the rope can reach is called an involute. The following is an image of the involute of a circle.

In this example the involutes occur when the rope is past north and south, in the eastward direction. When the rope is at a tangent to the silo towards the north or south, travelling in the western direction, the area is simply just a half circle with radius πr. Therefore the total grazing area is the sum of the area of the two involutes and the half circle.

In order to solve this problem we need to express the function as a set of parametric equations. The following is a more detailed representation of the involute.

The point *T* can be described using the coordinates* (r cos Θ, r sin Θ ). *It can be shown that the line segment *TP* has a length of rΘ since it was unwound from the arc *TA*. In order to determine the set of parametric equations we need to first determine the angle *PTQ*.

With this, it can be shown that the set of parametric equations representing the involute of a circle are as follows.

If we differentiate the two parametric equations we obtain the following two equations.

So with this information we can construct an integral, perform the necessary substitution and solve.

So finally we have determined the total grazing area of the cow.

In conclusion, as previously discussed, calculus can be incredibly useful to solve a number of problems. In the future we will see just how important calculus is in quantum theory. In my next post I introduce the concept of the paradox behind Gabriel’s horn. This problem will make use of partial differentiation in its application to determine the area of curved surfaces.

Dan Roy, Creator and Founder of Universal Physics.

]]>This post will include a second example, of greater complexity, showing the proof for a volume of a torus. A torus is a “doughnut” shaped object created by spinning a circle about the y-axis. This example is similar to that of a volume of a sphere and I will, in greater detail, explain the steps taking to solve the integral. The following is a 2-dimensional representation of a 3-dimensional torus shaped object.

This example is going to be a bit different cause I’m going to integrate with respect to y as apposed to x. So instead of determining the area under the graph we’re going to be finding the area to the left between the graph and the y-axis.

The image above shows a circle at distance R from the y-axis and of radius r. The circle is divided into two parts. The right half of circle is represented by the function *f(y)* and the left of the circle is represented by the function *g(y).*

The area of this circle can be determined by the difference between the areas to the left of f(y) and to the left of g(y).

To determine the volume of the torus we take the above equations, spin them about the y-axis and integrate.

Apply the rules of symmetry

Expand the polynomials then collect like-terms.

To continue we must convert to polar coordinates by performing some substitution.

Substitute the equations above.

Next we make use of the following trigonometric identity.

Perform the substitution.

Finally to solve the integral we make use of the following half angle formula.

Substitute the above equation and complete the integral.

We have determined the equation for the volume of a torus. This equation was first discovered by Archimedes who reasoned that the volume of the torus is equal to the volume of a cylinder. We can see from the formula that he was indeed correct.

In conclusion, integral calculus is an essential part of mathematics and as we shall see, is a fundamental tool in physics aswell.

In the future I will provide more interesting examples and problems involving calculus. Two interesting and upcomming posts will include the area of a grazing cow problem and the paradox of gabriel’s horn.

Dan Roy, Founder and Creator of Universal Physics.

]]>This post will introduce some of the basic concepts of integration and provide a detailed example for determining the volume of a sphere.

As previously discussed trigonometry and natural logarithmic functions play an important role in Calculus. I’ll begin first with an introduction to logarithmic functions in Calculus. In future posts I’ll discuss in greater detail logarithmic functions and provide some more examples.

In differential calculus the general derivative of a logarithmic function is as follows.

For examples of greater complexity we implement the use of the chain rule but for reasons of simplicity consider the following example of the differentiation of natural logarithmic functions.

I mentioned before that integration and differentiation are inverses of one another. This concept is of great importance as we have seen by the use of anti-derivatives to solve integrals. By this logic the following should be clear.

An interesting example of the inverse relationship between differentiation and integration can be seen in the following special case of the exponentiation of the natural e.

It can be proven using the “Exponentiation Rule” that the differential of the function e to the power of x is simply e to the power of x. By this logic the following should also be apparent.

In a past post regarding the area under the normal curve I demonstrated the use of Exponentiation Rule and anti-derivatives for the function of the natural power of e.

Furthermore, I will now discuss the importance of trigonometry in Calculus. The use of trigometry in calculus was also demonstrated in the example where I found the area under the normal curve. In this example it was fundamental in the conversion of cartesian to polar coordinates.

I wont discuss the basics of trigonmetric differentiation or integration in this post but I will however provide an extensive example of how its used. The example I will show you is how to calculate the volume of a sphere by spinning a circle about the y axis and integrating.

We begin by determining that the area of a cirle is pi times its radius squared. To accomplish this we make use of the general equation for a circle.

Now we can calculate the volume of the sphere by determining the sum of the infintesimal disks along the x-axis. Therefore we can use the following simple integral.

So as you can see integral calculus can be a powerful tool in both science and mathematics.

I will show a few more examples of interesting problems such as those done by Archimedes in my future posts.

Dan Roy, Founder and Creator of Universal Physics.

]]>Differentiation can be regarded as being the measure of the rate of change of a function. In two dimensions it can be used to find the slope of a tangent to the curve of a continuous function. As such, we can define the derivative of a function as the following.

The differential of a function determined by operations acting on two or more functions can easily be determined using the equation above. These operations include multiplication, division and exponentiation.

Consider the following example of the product of two functions.

The function *b(x)* is the result of the product of the two functions *f(x)* and *g(x)*. According to what is known as the Product Rule, the differential equation can be determined by the following.

The following proves that the equation stated above is correct.

Continuing on to division now, the quotient of two functions, according to the Quotient Rule, can be determinded by the following.

Using a similar method as before the following proves this is true.

Furthermore, for exponentiation we make use of the Power Rule, which states the following.

In order to prove that this is true we make use of the following two rules of combinatorics theory.

Again it depends on the use of the fundamental definition of a derivative, as stated at the beginning of this post.

Finally, the last type of derivative I’ll discuss is that of a composite function. A composite function is created by inputing the results of one function into another. For two functions *f(x)* and *g(x)* the resulting composite function is *F(x) = f(g(x))*. To determine the differential of such a function we make use of the Chain Rule. If the functions *f* and *g* are both differentiable then the following is true.

Another way to describe this uses Leibniz notation where *y = f(u)* and *u = g(x)* are both differentiable functions. We can also describe the Chain Rule as the following.

A partial proof to the above can be described using the following.

Up to now I’ve only described a few example of how differentiation is used in mathematics. This last example was useful in the example demonstrated in the post entitled “Introduction to Integral Calculus in Physics” where I made use of its anti-deritive to solve the integral. In the next post I will discuss the differentiation of trigonometric and logarithmic functions as well as exponentiation. An example of an anti-deritive of this was used in the example to find the area under the normal curve. I will also discuss the special case in the function *e* to the power *x* in my next post as well as begin to further explain partial differentiation and its importance in quantum mechanics.

Dan Roy, Founder and Creator or Universal Physics.

]]>I will now provide a detailed proof that the area under the normal curve, represented by the gaussian function, is in fact one.

Recall that a normal curve is represented by the following probability density function and produces the graph thats shown.

Again for this example we’re gunna allow the variance to be one and the mean to be zero. In order to solve for the area under this curve we can’t use traditional methods. So for this example I’m going to introduce the use of multi-variable calculus and perform double integrals in polar coordinates. To do this we make use of the following general rule.

Also we’re gunna make the assumption that the image is symmetrical and by doing so we can multiply by two and we’ll only need to calculate the area from 0 ≤ x ≤ ∞. We begin by stating the following.

First we apply the rules of symmetry.

Now we create a double integral by squaring both sides then taking the square root of both sides. We can also remove the constant from the integral.

Next expand the double integral into two independants. Since they are independant we can change the variable x in the second integral to y. Then we combine these two integrals together.

Recall from the definition above the following.

Now we must convert from cartesian to polar coordinates using the rule we defined above.

We will now substitute the following two equations to simplify the exponent component of our function.

Finally, apply the substitution and solve the integrals.

In this case we know the number is supposed to be positive so we’ll disregard the negative root.

In conclusion, it can be shown using multivariable calculus, that the area under the normal curve represented by the Gaussian probability density function, is equal to one. As mentioned, this is significant in quantum mechanics where the sum of all probabilities of a quantum system is equal to one. In the posts to come I will introduce the other branch of calculus and provide some more detailed proofs of the concepts in differentiation.

Dan Roy, Founder and Creator of Universal Physics.

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