# Force exerted by rails on a high speed train

I need to find the ratio between horizontal and vertical accelerations exerted on a train by the rails it is running on , given that the train is traveling East at a speed of $$v_r=250$$ km/h, along the 30 degree North parallel ($$\gamma = 30^o = \pi / 6 \hspace{4px} rad$$). ($$\gamma$$ is the angle generated by the Earth's nucleus - equator line and the Earth's nucleus - train line). I need to account for all the fictitious forces that arise from the fact that the Earth is rotating and the train is going at a high speed, but my solution doesn't match my professor's. This is what I've done:

• The train will be subjected to an apparent acceleration $$\vec{g_{ap}}$$, which will be a composite of accelerations: the "natural" gravitational acceleration ($$\vec{g_o}$$), the centrifugal force ($$\omega^2R_T \hat{\rho}$$) and the Coriolis force ($$2\omega v \hat{\rho}$$, being $$R_T$$ the radius of the planet Earth and $$\omega$$ its angular speed): $$\vec{g_{ap}}=-g_o\hat{r} + \omega^2R_T \hat{\rho} + 2\omega v \hat{\rho}$$

• Therefore, since we can convert the radial unit vector to cylindrical coordinates ($$\hat{r} = \hat{z}\sin\gamma + \hat{\rho}\cos\gamma$$), after substituting I get: $$\vec{g_{ap}} = (-g_o\cos\gamma + \omega^2R_T + 2\omega v)\hat{\rho} - g_o\sin\gamma\hat{z}$$

But I don't know go from here to the final solution, which is related to horizontal and vertical accelations. My professor gets:

$$\frac{R_{H}}{R_{V}}=\frac{\omega^{2} R_{T} \sin \lambda \cos \lambda+2 \omega v_{r} \sin \lambda}{g-\omega^{2} R_{T} \cos ^{2} \lambda-2 \omega v_{r} \cos \lambda}$$

What am I missing? Thanks!

• This is a delightful question, but could you please read through it carefully and edit it, giving definitions of all the symbols? The meaning of $λ$ isn’t obvious to me - though that may be my ignorance - and $R_T$ and $R_r$ similarly. – Martin Kochanski May 24 '19 at 21:17
• Hi, @MartinKochanski. I'm glad you think this is interesting :) I tried to clarify everything. Thanks for you attention! – mar May 24 '19 at 21:37

General remark:
A widely used convention is to used the upper case $$\Omega$$ for the Earth's angular velocity, and the lower case $$\omega$$ for the angular velocity of some small object.

In this problem the component of the effect vertical to the local level surface is called the Eötvös effect, after the geophysicist Loránd Eötvös.

For completeness let me derive the magnitude of the Eötvos effect for motion along the equator.

$$\omega_{\ r}$$ angular velocity of the train relative to the Earth
$$\Omega$$ Angular velocity of the Earth
$$R$$ Radius of the Earth
$$a_s$$ required centripetal acceleration when stationary with respect to the Earth
$$a_{\omega}$$ required centripetal acceleation when moving at angular velocity $$\omega_{\ r}$$

Again, this is the simplified case of motion along the Equator.

As we know, the fact that the Earth rotates has the following effect: at the equator the effective gravitational acceleration is the resultant of two components: the true gravity of the Earth, and the fact that a force is required to sustain circumnavigating motion.

The required centripetal force to remain co-moving with the Earth is small, but not negligable.

When the train is stationary (with respect to the Earth) then the required centripetal force is proportional to $$\Omega^2R$$

When the train is moving (in this example along the equator) then its total angular velocity is ($$\Omega$$ + $$\omega_{\ r}$$), therefore the total required centripetal force is proportional to $$(\Omega + \omega_{\ r})^2R$$

To find the difference in required centripetal force you subtract the one from the other. $$a_\omega - a_s$$ $$(\Omega + \omega_{\ r})^2R - \Omega^2R$$ $$\Omega^2R + 2\Omega\omega_{\ r}R + {\omega_{\ r}}^2R - \Omega^2R$$ $$2\Omega\omega_{\ r}R + {\omega_{\ r}}^2R$$

Notice the difference between the expression for the Eötvös effect and the expression for the fictitious forces that are taken into account in the case of motion relative to rotating frame of reference.

The fictitious centrifugal force is proportional to the angular velocity of the rotating system. That is, the expression for the fictitious centrifugal force is proportional to $$\Omega^2R$$. On the other hand, the expression for magnitude of the Eötvös effect has a term $${\omega_{\ r}}^2R$$ ; that is not the fictitious centrifugal force.

Moving the problem to 30 degrees North

The task of the problem is to decompose the local effect into two components: perpendicular to the local level surface, and parallel to the local level surface.

Note that both these components are still the Eötvös effect, just decomposed.
The amount weight that the train exerts on the rails will be diminished.
Since the train is moving East the train will tend to swing wide, that is: the rail closest to the Equator will have a bit more weight on it than the rail furthest from the Equator.

General discussion

I hope it is clear that this problem is unrelated to the case of taking fictitious forces into account. The Eötvös effect is a true physical effect.

My favorite illustration is the case of an airship that makes a U-turn, from eastward velocity to westward velocity. With eastward velocity the airship needs less buoyancy force, as the Eötvös effect reduces its weight. After the U-turn, moving westward the airship needs more buoyancy force, since moving westward the airship has more weight than when stationary (with respect to the Earth). That means that after the U-turn the airship needs to be re-trimmed, back to neutral buoyancy.

[LATER EDIT]
Additional discussion, triggered by the information that the question is from a practice exam.

You mention that the answer that was intended by the creator of the practice exam uses a horizontal component ($$R_H$$) and a vertical component ($$R_V$$).

The vertical component is presented as follows:
$$R_V = g - \omega^2 R_T \ cos^2 \lambda - 2\omega v_{r} \cos \lambda$$

We assume that here the $$\omega$$ is used for the angular velocity of the Earth.

I reconstruct that here 'g' is intended as the true gravity, not the measured gravity, as the expression subtracts a term with a factor $$\omega^2 R_T$$ to arrive at the answer.

(It's very unusual to have 'g' stand for true gravity. By universal convention when the letter 'g' is used it stands for the measured gravitational acceleration. It's not clear why the creator of that practice exam has not followed that convention.)

In the derivation that I gave for the magnitude of the Eötvös effect you see that a term $$2\Omega \omega_{\ r}R$$ arises. That term is identical to the Coriolis term of course.

[LATER EDIT]
Here is my attempt at reconstructing the thought process of the Practice Exam Creator.

First start with a spherical non-rotating Earth, and define 'g' as the gravitational acceleration of that spherical Earth. Next, retain that spherical shape, but introduce rotation of the Earth.

Now, first consider the following 2-dimensional case: a rotating disk, with angular velocity $$\Omega$$. An object that moves in a straight line in inertial space will move along a curvilinear trajectory with respect to the rotating disk. As we know, the amount of centrifugal acceleration (wrt the disk) will be $$\Omega^2r$$. You can turn that around and recognize that in order to prevent that centrifugal acceleration a centripetal force proportional to $$\Omega^2r$$ must be exerted.

In this case the train has a velocity relative to the rotating disk, in tangential direction. If there would be zero friction then the acceleration wrt to the disk is given by the Coriolis term $$2\Omega v$$ (Where $$v$$ is the velocity wrt the disk). You can turn that around and figure that in order to prevent that acceleration a force proportional to $$2\Omega v$$ must be exerted.

The above is for the case of spherical, rotating Earth.
As we know, our Earth is somewhat flattened; the diameter along the equator is about 40 kilometers more than the distance from pole to pole. This means a line from the Equator to a pole is a downhill slope.

At the poles and along the Equator the direction of local measured gravity is towards the Earth's geometrical center. At all latitudes away from the poles and the Equator the direction of local measured gravity is not exactly towards the geometrical center. For instance, at 45 degrees latitude the angle between the direction of measured gravity and the line towards the geometrical center is about 0.1 degree. That's a 0.1 degree downhill slope. That slope provides the required centripetal force that keeps the water of the oceans at their own latitude. Conversely: if the Earth would not be flattened all water would flow towards the Equator.

Repeating what I wrote in a comment:
At the Equator a gravimeter measures a value of 9.78 $$m/s^2$$ Given the Earth's angular velocity, co-rotating with the Earth (along the Equator) requires 0.034 $$m/s^2$$ So we can infer that the true gravity at the Equator is 9.82 $$m/s^2$$ Of course, since gravitational mass and inertial mass are equivalent we cannot measure the true gravity directly. But by taking the required centripetal acceleration into account we can infer the value of the true gravity.

So it does matter what 'g' is said to stand for. Whether it is thought of as standing for true gravity, (which cannot be measured directly), or for the effective gravity, which is what is actually measured, is something that must be stated explicitly.

• I don't understand what "required centripetal acceleration" means in this context. Also, if I understood your answer right, does this mean I need to account for the angular speed of the train due to its motion along a parallel and therefore my calculations are incorrect so far? – mar May 25 '19 at 8:12
• @Manuel I used the expression 'required centripetal force' as short for 'required centripetal force to sustain circular motion'. At the Equator a gravimeter measures a value of 9.78 $m/s^2$ Given the Earth's angular velocity, co-rotating with the Earth (along the Equator) requires 0.034 $m/s^2$ So we can infer that the true gravity at the Equator is 9.82 $m/s^s$ Of course, since gravitational mass and inertial mass are equivalent we cannot measure the true gravity directly. But by taking the required centripetal acceleration into account we can infer the value of the true gravity. – Cleonis May 25 '19 at 8:25
• @Manuel The motion is very simple: cilindrically symmetrical with respect to the Earth's axis. My recommendation: express everything in terms of angular velocity with respect to the Earth's axis. For the entire calculation, use motion relative to the inertial frame of reference. The final answer is force exerted by the weight of the train on the rails. All measurable force. Hence my recommentation: don't use fictitious, use Eötvös effect. – Cleonis May 25 '19 at 8:59
• I understand what you meant now, but sadly this comes from a practice exam about fictitious forces and the Eötvös effect isn't even mentioned in our curriculum: I would get negative points because the exam is intended to test out ability dealing with fictitious forces. – mar May 25 '19 at 11:39
• @Manuel Wow, that practice exam is badly broken. I guess that means you have to set out two tracks. One to understand the problem, the other to try and reconstruct just how the problem is misunderstood by the person who created that practice exam. – Cleonis May 25 '19 at 12:19  If you seat in the train that travel with the velocity $$v$$ , you will feel this acceleration:

$$\vec{a}=\begin{bmatrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \\ \end{bmatrix}=-\vec{g}_v-\vec{\omega}\times \left(\vec{\omega}\times\vec{R}\right)+2\left(\vec{\omega}\times \dot{\vec{R}}\right)\tag 1$$

with:

$$\vec{g}_v=\begin{bmatrix} 0\\ 0\\ g\\ \end{bmatrix}$$, Gravitation vector

$$\vec{\omega}=\begin{bmatrix} 0\\ 0\\ \Omega\\ \end{bmatrix}$$,Earth rotation about the $$z$$ axis

$$\vec{R}=\left[ \begin {array}{c} {\it R_s}\,\cos \left( \varphi \left( s \right) \right) \cos \left( \lambda \left( s \right) \right) \\ {\it R_s}\,\cos \left( \varphi \left( s \right) \right) \sin \left( \lambda \left( s \right) \right) \\ {\it R_s}\,\sin \left( \varphi \left( s \right) \right) \end {array} \right]$$

Position vector discribed the path of the rail on a sphere surface.

$$s$$ is the path line parameter

and

$$\dot{\vec{R}}=\frac{d}{dt}\vec{R}=\frac{d}{ds}\vec{R}\frac{ds}{dt}= \frac{d}{ds}\vec{R}\,v$$

the acceleration components ($$\vec{a}\quad$$equation (1)) are given in inertial coordinate system, we transform the components to a trail fixed coordinate system $$x'\,,y'\,,z'$$

where:

$$\vec{x}'$$ is tangential vector on the line $$\varphi$$

$$\vec{x}'=\frac{d}{d\varphi}\vec{R}\quad, ||\vec{x}'||=1$$

$$\vec{y}'$$ is tangential vector on the line $$\lambda$$

$$\vec{y}'=\frac{d}{d\lambda}\vec{R}\quad, ||\vec{y}'||=1$$

$$\vec{z}'=\vec{R}\quad,||\vec{z}'||=1$$

thus the transformation matrix (Rotation matrix) is:

$$S=[\vec{x}'\,,\vec{y}',\vec{z}']\quad S^T\,S=I_3$$

$$S=\left[ \begin {array}{ccc} -\cos \left( \lambda \left( s \right) \right) \sin \left( \varphi \left( s \right) \right) &-\sin \left( \lambda \left( s \right) \right) &\cos \left( \varphi \left( s \right) \right) \cos \left( \lambda \left( s \right) \right) \\ -\sin \left( \lambda \left( s \right) \right) \sin \left( \varphi \left( s \right) \right) &\cos \left( \lambda \left( s \right) \right) &\cos \left( \varphi \left( s \right) \right) \sin \left( \lambda \left( s \right) \right) \\ \cos \left( \varphi \left( s \right) \right) &0 &\sin \left( \varphi \left( s \right) \right) \end {array} \right]$$

$$\vec{a}_T=S^T\,\vec{a}_o$$

thus:

$$\vec{a}_T=\begin{bmatrix} \ddot{x}'\\ \ddot{y}'\\ \ddot{z}'\\ \end{bmatrix}=\left[ \begin {array}{c} -\cos \left( \varphi \left( s \right) \right) \left( \sin \left( \varphi \left( s \right) \right) { \Omega}^{2}+2\,\sin \left( \varphi \left( s \right) \right) \Omega\, v{\frac {d}{ds}}\lambda \left( s \right) -g \right) \\ 2\,\Omega\,v\sin \left( \varphi \left( s \right) \right) {\frac {d}{ds}}\varphi \left( s \right) \\ \left( \cos \left( \varphi \left( s \right) \right) \right) ^{2}{\Omega}^{2}+2\, \left( \cos \left( \varphi \left( s \right) \right) \right) ^{2}\Omega\,v{\frac {d}{ds}} \lambda \left( s \right) +\sin \left( \varphi \left( s \right) \right) g\end {array} \right]$$

• In eq. (1) you subtract a vector with magnitude $\omega^2R$ from $\vec{g}_v$. However, by universal convention when the letter 'g' is used for the gravitation vector it stands for the measured gravitational acceleration. I recommend that you state explicitly what vector quantity you have in mind for $\vec{g}_v$ – Cleonis May 26 '19 at 8:46
• As we know, the Earth is not a sphere. At every latitude the measured local gravity is perpendicular to the local level surface. It may be that you are treating the problem as if the Earth is a sphere. However, the effect that problem asks about (force exerted by train on rails as it moves eastward at 250 km/h), is small. Treating the Earth as a sphere introduces an error that is significantly larger than the effect that the problem is asking about. – Cleonis May 26 '19 at 8:46
• eq. (1) give you the components of the acceleration vector $\vec{a}$ written in inertial system $x\,,y\,,z$, so the gravitational vector has just component toward the $z$ axis $[0\,,0\,,g]$. correct ? Of course the Earth is not a sphere but ellipsoid. To get the equations for ellipsoid you can use the same methode , write the acceleration components in Initial Frame then transformed the components to a local frame. I don't any problem? – Eli May 26 '19 at 10:30
• I don't understand the part where you "transform the componets to a trail fixed coordinate system". I know those are rotation matrices, but what is the purpose of each one? – mar May 26 '19 at 13:39
• @Manuel, please look at the new description of the local coordinate system. This is a Rotation Matrix (Transformation Matrix) between two coordinate frames initial and train. – Eli May 26 '19 at 16:58 I generally set up these coriolis and centripetal forces on the Earth as shown in the figure.

$$x$$ is south

$$y$$ is east

$$z$$ is vertically upward.

The train is moving east so the velocity vector is

$$\text{vvec}=(0,v,0)$$

where v is the velocity of the train.

The position vector $$R0$$ from the center of the Earth:

$$R0 = (0,0,Re)$$

where Re is the radius of the Earth.

The angular velocity vector of the Earth's rotation.

$$\omega0=(-\omega \cos (\lambda ),0,\omega \sin (\lambda ))$$

where $$\omega$$ is the angular velocity of the Earth's rotation.

We can now compute all the accelerations.

The coriolis acceleration

$$acor=-2\ \omega0\times vvec$$

Gravitational acceleration

$$agrav = (0,0,-g)$$

Centripetal Acceleration

$$acent=-\omega0\times (\omega0\times R0)$$

The total acceleration vector is then:

$$avec = agrav+acor+acent$$