# Frictionless Banked Curve on Mars Simulates Earth Gravity

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1. Aug 2, 2011

### TheClockmaker

1. The problem statement, all variables and given/known data

A train on the planet surface of Mars is traveling along a banked track shaped as large circle with a radius of 500m.

Q. Assuming Mars gravity is 3.8m/s$^{2}$ what Velocity v must the Martian train travel to simulate a total artificial gravity equal to Earth's 9.81m/s$^{2}$?

Q. And what is the correct angle $\theta$ for this to be a frictionless banked turn?

2. Relevant equations

v=Velocity
$\theta$=Angle of track bank
g=Mars Gravity of 3.8m/s$^{2}$
e=desired artificial gravity of 9.81m/s$^{2}$

v=$\sqrt{rg \tan \theta}$

$\theta$=arctan $\frac{v^{2}}{rg}$

3. The attempt at a solution

I'm not in school, I'm just doing this for fun out of curiosity, but I've been working on this for over a week now and I am stuck.

Using Excel I'm able to calculate the bank angle with my formula: =DEGREES(ATAN(v^2/(r*g)))

The best that have come up with is to calculate v so that it will equal my desired earth gravity of 9.81 so I came up with this: =SQRT(r*9.81)

I derived at that using a=$\frac{v^{2}}{r}$ and substituting 9.81 for a.

So this calculates to: v=70.0357051795725 and $\theta$ angle of 68.8256331451085 degrees.

But my formula 9.81=$\frac{v^{2}}{r}$ does not factor in the gravity existing on Mars at all and I don't know how to do this; i assume that i need to be sutbracting the downward acceleration of 3.8 in relationship to the cosine of the angle; but then I won't know the angle until I know the aceleration, so I am just stuck. Any help is appreciated, I've spent so long on this problem I can't put it down now, it's driving me crazy. If you would, please help me to know what I need to do to my formulas. Thanks.

Last edited: Aug 2, 2011
2. Aug 2, 2011

### PeterO

On banked tracks, the Normal force [there is no friction in this case] has to have a vertical component to balance the force of gravity, and a horizontal component to provide the required centripetal force. The Normal force is 9.81.m, the gravity force is 3.8.m

From that you should be able to work it all out - helped by a good diagram to keep you focused.

3. Aug 2, 2011

### TheClockmaker

I don't think I'm explaining why this is so challenging to me (other than the fact that I've been out of school for 30 years).

I want the person to "feel" gravity as being perfectly downward to the floor of the room, even though the floor is truly angled due to the bank of the curve.

So when you see me writing about "frictionless" this is why--it's not due to the rails, it's because I don't want the occupants to feel thrown to the side of their train car while sitting in it--this requires the bank angle to be that of a frictionless design.

With this in mind, my challenge is how do I correctly calculate the velocity needed to reach a preceived 9.81m/s/s because I would think that I need to know angle to correctly know the amount of velocity needed--and yet I won't know angle until I know velocity... so it's a chicken and egg thing. I can't figure it out, I am stuck because of this.

Last edited: Aug 3, 2011
4. Aug 3, 2011

### PeterO

The chicken and egg is solved by the radius of the track. Currently you have not stated it. Until you know the radius you will not know the velocity.
My answer was going to offer no lateral force to the occupants.
The centrtipetal acceleration required is Square root of[(9.81)^2 - (3.8)^2]

Once you know that value: [just a little bit over 9] you know (v^2) / r = 9

Once you know the radius, you will know the velocity.

The angle is already defined - arcsin(3.8/9.81) EDIT: it may be arc-cos(3.8/9.81) depending which angle you are defining - angle from horizontal or angle from vertical.

5. Aug 3, 2011

### TheClockmaker

Thank you. That is where I went wrong. I knew I needed to subtract and every time I tried to get my math to double check i was not working for me. I was not squaring the numbers before subtracting them. It makes perfect sense now.

My radius is 500m; I wrote it in the original problem but I did not carry them down to step #2 to make it more obvious. Thank you very much.

It occurs to me that a person could also use this to determine how much extra g-force they wanted to add to passengers on a rollercoaster here on Earth.

I can now sleep; it's almost 1am and I'm still up trying to solve this. Thanks.

So:

Extra Gravity Needed (X) =SQRT((9.81)^2 - (3.8)^2) = 9.044119637
V = SQRT(X * 500) = 67.2462624854992m/s = 242.0865449kph = 150.4256051mph
Angle =DEGREES(ATAN(V^2/(500*3.8))) = 67.2096798262262 degrees

I see that my Angle formula here checks out with your ACOS(3.8/9.81) so that made me fell good; however, your mentioning arcsin(3.8/9.81) threw me because then it made me doubt my understanding of which is which. When I calculate 67.21 degrees I'm assuming a very steep bank; however, am I wrong, and really it's going to be the less agressive 22.79 degree angle sloping upward from the center?

One thing that caught me by surprise is that the angle of the bank remains consistent throughout the entire range of track diameters; the required speed increases as the radius increases, but the angle remains constant--I was expecting the bank angle to increase with radius due to the increased speed. I was wrong with my expectation.

Last edited: Aug 3, 2011
6. Aug 3, 2011

### PeterO

This track will be very steep!! The 67 degrees you calculate is the angle up from the horizontal I hope. 90 degree banking would be a vertical track, not a horizontal track.

The vertical component of the "artificial gravity" created balances the real gravity on the planet. If the banking was only 45 degrees the normal reaction force would clearly outweigh the gravity on Mars. Even 60 degree banking would balance a gravity of 4.9. We only have to balance only a gravity of 3.8, so the angle is bigger that 60 degrees !!