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Is plannet's surface in game in non-inertial frames?

Cesrate

By Cesrate
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Cesrate

Cesrate

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I mean... Is the sideral rotational acceleration counted? So there's a tiny difference between the gravity on pole and on equator... And foucault pendulum if the game's accurate enough...

Edit: So, it's confirmed at least within the error range(bigger than the value to be measured, LoL). Any idea from Modders to make a Foucault pendulum?:D

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Cesrate

Cesrate

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I know the sidereal rotational acceleration is counted, an easy test is launch east and the launch west. East is about 800Dv easier.

I didn't mean velocity; I mean acceleration effect. So, both on Kerbin sea level the gravity on equator should be about 0.05m•s^(-2) lesser than on pole.

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Cesrate

Cesrate

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It basically means that because the equator is moving faster than the poles in a circular motion, the gravity that you feel is reduced due to centrifugal force from your rotation pulling you up slightly. Does this help :P

Usually we don't use inertial force to describe motion to avoid the possible misunderstanding, or you'd better declare it's in non-inertial reference frame and "centrifugal force" is an inertial force.

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Cesrate

Cesrate

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I have no idea what any of this means :(

As for the description in inertial frames, if an object is on or near the surface and following planet's sidereal rotation, part of gravitational force from the planet will perform as centripetal force to help the object staying at the surface, and the gravity effect, causing objects to fall, should be vector G - vector v^2/R.

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Westo454

Westo454

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While I don't think the centripetal force is taken into account (a good way to test this would be to take a marker with a horizontal stip and have a kerbal jump at both the pole and the equator) but it does take into account the Coriolos Effect, which is why it's cheaper D-V whose to go east.

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Splode

Splode

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While I don't think the centripetal force is taken into account (a good way to test this would be to take a marker with a horizontal stip and have a kerbal jump at both the pole and the equator) but it does take into account the Coriolos Effect, which is why it's cheaper D-V whose to go east.

We don't need jumping kerbals, we have sensors. And they work. Use the accelerometer.

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AustralianFries

AustralianFries

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I love how there is so much mathematics and science in KSP as well as the basic: "Add more boosters and struts until your rockets reach space and don't blow up" aspect. :)

There are two kinds of people in this world community. The ones that math. And the ones that booster.

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AmpsterMan

AmpsterMan

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It basically means that because the equator is moving faster than the poles in a circular motion, the gravity that you feel is reduced due to centrifugal force from your rotation pulling you up slightly. Does this help :P
Usually we don't use inertial force to describe motion to avoid the possible misunderstanding, or you'd better declare it's in non-inertial reference frame and "centrifugal force" is an inertial force.

So If I'm understanding correctly, it's not that the force of gravity is less at the equator but rather the fact that you are being accelerated faster at the equator than at the poles means, in essence, there is less gravitational acceleration toward the center of the planet acting on you.

If this is the case, and I understand correctly, the acceleration due to gravity should be greatest at exactly the North and South Poles and should be the least at the Equator.

Assuming all these things, I would think KSP simulates these things just because it is a natural affect of stuff; kind of like launching east requires less delta-v and accelerating in a lower orbit reduces delta v costs.

As for the description in inertial frames, if an object is on or near the surface and following planet's sidereal rotation, part of gravitational force from the planet will perform as centripetal force to help the object staying at the surface, and the gravity effect, causing objects to fall, should be vector G - vector v^2/R.

I think, for whatever reason you have just explained to me something I already knew intuitively. Why it is I have to accelerate to a certain velocity to attain a certain altitude.

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hubbazoot

hubbazoot

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Follow-up: There's too much atmospheric drag on kerbin for me to feel confident with these results. If someone wants to head to Tylo, though, they may get MUCH better data landing at the poles there.

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Cesrate

Cesrate

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There are two kinds of people in this world community. The ones that math. And the ones that booster.

Maybe... Exactly...

Follow-up: There's too much atmospheric drag on kerbin for me to feel confident with these results. If someone wants to head to Tylo, though, they may get MUCH better data landing at the poles there.

Could you post your whole structure?:) So it's using KAS?

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