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April 21, 2014

April 21, 2014

Posted by **Mark Main** on Sunday, July 31, 2011 at 10:15pm.

Q: Using 3.8m/s^2 for Martian gravity what is the train car speed and track angle required to simulate Earth’s 9.81m/s^2 gravity? And what are the formulas to derive this answer?

Mars Gravity Source: “An Introduction to Physical Science” by James Shipman, Jerry D. Wilson, and Aaron W Todd

This is not a homework assignment, I'm just doing it for fun in writing a sci-fi screenplay

- Physics -
**drwls**, Monday, August 1, 2011 at 5:46amA centripetal acceleration (horizontal) of 9.04 m/s^2, combined with the existing (vertical) Mars gravity acceleration of 3.8 m/s^2, will reasult in a resultant acceleration of 9.81 m/s^2.

Require that V^2/R = 9.04 m/s^2

V = 134 m/s = 300 mph

It does not seem practical. You'd need a bullet train. Why bother? Enjoy the low g.

- Physics -
**Mark Main**, Monday, August 1, 2011 at 2:45pmGood sci-fi strives to limit itself to the laws of physic within the boundaries of future science and technology; within the relm of my future world of science, abundant clean energy is reality and so powering a railed community is possible.

"Why bother?" Because our muscles and bone denisity will become dramtically weak over extended time in low-g; our Martian colonists would deteriorate to a point where they cannot return to Earth and at that point they are no longer able to "Enjoy the low g." because it no longer seems "low" to them, it's now become normal.

I will have a low-g track used for senior citizen assisted living; the hospital will be on Mars gravity to make recovery easier and also moving patients around; also seniors in their final years may opt to live 100% on Mars gravity.

Regarding the math, can you help me understand how you arrived 9.04?

I can determine the proper angle for a frictionless bank if I know Velocity:

R = 152.336 meters

Bank Angle (A) = DEGREES(ATAN(V^2/(R*9.81))) = 85.24243368

And I can determine Velocity if I know the bank Angle:

Velocity = SQRT(R*9.81*TAN(RADIANS(A))) = 134

But I cannot figure out how to correctly determine them without knowing what the other value is.

Thanks for your help.

- Physics -
**Mark Main**, Monday, August 1, 2011 at 3:24pmI forgot to mention that can reduce the speed by making the radius smaller, but studies show that at a certain point it becomes uncomfortable for long-term living; it seems that nausia effects mildly increase below -.5 mile diameters and the nausia increases as the diameter decreases from there; so I intend to keep my minimum diameter at 1km and build my track rings outward from there up to 2km diameters. If the city grows larger than this initial set of community rings (perhaps called communirings, commrings or c-rings) then a new ringset would be built with it's center point located slighly more than 2km from the first center point so that it can also grow to a diameter of 2km.

- Physics -
**Mark Main**, Monday, August 1, 2011 at 4:51pmI searched around to find some links regarding the bone loss, but I'm not allowed to link it here. Just google this long sentence here:

Tests indicate bone loss rates of up to 1.5% a month johnson space center

and you'll find it right away.

In just a few years the colonist could not return back to Earth.

- Physics -
**Mark Main**, Wednesday, August 3, 2011 at 3:05amI see that 9.04 that you provided comes from SQRT(9.81^2 - 3.8^2).

But I think that your velocity calculation is wrong because velocity should be the SQRT(R * 9.04) and that would be 67.25m/s, 242.09kph, 150.43mph.

Please let me know if you concur.

The angle will always be the same; it's =DEGREES(ACOS(3.8/9.81))

Thanks again for your help.

Mark

- Physics -
**Mark Main**, Wednesday, August 3, 2011 at 11:58amI see that 9.04 that you provided comes from SQRT(9.81^2 - 3.8^2).

But I think that your velocity calculation is wrong because velocity should be the SQRT(R * 9.04) and that would be 67.25m/s, 242.09kph, 150.43mph.

Please let me know if you concur.

The angle will always be the same; it's =DEGREES(ACOS(3.8/9.81))

Thanks again for your help.

Mark

- Physics -
**Mark Main**, Wednesday, August 3, 2011 at 12:26pmI am positive now; the correct answer to this problem is this:

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

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