Science quick question. Need help! [More stuff to know!]

[Error]

The middle of an average person should work quite well. The velocity wouldn't matter as long as it is constant; humans can't "feel" velocity in any way unless they have an external reference point (such as the trees beside a road when you're moving in a car, or the stars in space) that can tell you that you are moving. The 1 g I mentioned is just the Earth's gravitational acceleration of 9.81 m/s^2 again; the constant g is equal to 9.81. m/s^2.

OK that makes sense... but what you are hinting at is that lets say the base rotates where the person is at 5 m/s.
They will not feel that then unless for example they look out the window? If that is the cast then as long as the wheel doesn't accelerate or decelerate then it should be good then. right?

I'm just making sure that no one gets sick if the thing turns too fast hehe. Would windows cause people looking outside to feel sick or as I am thinking the same as you go on a plane, you wont feel it but you will feel the acceleration if it slows down or speeds up.

Edit: posted on wrong account but w.e
~nomnomnom

Yup, exactly that! You're fine as long as the ring doesn't accelerate or decelerate - like you said in your plane example, you can't feel velocity directly but you can feel changes in it - but they may indeed get a bit motion-sick if they look out a window. That'll vary from person to person and depend on the speed of your wheel is rotating at, though.

[Lord Caedus]

Well make sure you have a second, counter-rotating ring, to keep the station stable.

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[nOmnomnOm]

Well make sure you have a second, counter-rotating ring, to keep the station stable.

exactly that yeah. on my other thread : http://discoverygc.com/forums/showthread.php?tid=128579

I have videos posted on 1 wheel turning and what happens.
Its interesting and yeah. Everything has to balance everything else to that it is all stable. hehe

As this problem is almost fully solved, once i get overall dimensions then i can get moving again hehe. Atm right now I am working on graphs to see what is the best radius to choose from.

[Lord Caedus]

Well, if you want, you could have concentric rings, all spinning at the same speed, which would allow for different levels of gravity for different uses.

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[nOmnomnOm]

Well, if you want, you could have concentric rings, all spinning at the same speed, which would allow for different levels of gravity for different uses.

yep OR I can have one ring inside another ring spinning at different speeds but maintaining the same gravity.
it all depends as you said as to what is it used for and if humans are allowed access to that place...

hehe fun stuff to think about

[nOmnomnOm]

You know just for kicks.... I did a experiment with maths to see how fast the earth spins. I compared to google and got it completely wrong....

what I did was:

F = (mv^2) / r
F=9.81
m= 62 kg (average person's weight)
v = unknown
r= 6371000 meters

so it came out that v was equal to....1004 m/s or 278 km/h...... which is wrong! D:

It is weird i was doing the same process for the calculations for the base. With a radius of 100 meters you only need to apparently spin the wheel at 1.1 km/h to match 9.81 for the eatch's G.
So now i am wondering if I am doing it wrong or is it right? If it is right then wow! No problem with motion sickness at all!

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[Error]

The 9.81 m/s^2 of acceleration that we experience as gravity on the Earth's surface is not a centripetal acceleration, but, as the name implies, a gravitational acceleration, which is given by a = (k_g)*m/r^2, where k_g is the gravitational constant, m is the total mass of the attracting object (here the Earth) and r is the distance from the centre of the object (i.e. the Earth's radius). The formula for centripetal acceleration is not usable for that case, so it not giving out (something similar to) the equatorial rotational velocity is not surprising.

And I seem to get a velocity of about ~31.2 m/s = ~112.2 km/h for a ring with a radius of 100 m; could you post your calculations? I'm using v = sqrt(a*r) as given from the equation a=(v^2)/r where a = 9.81 m/s and r = 100 m.

EDIT: Important note: The force in your calculation above would /not/ be 9.81 Newton; it's given by F = m*a, where m is the mass of your human and a is the gravitational acceleration of 9.81 m/s^2.

[nOmnomnOm]

The 9.81 m/s^2 of acceleration that we experience as gravity on the Earth's surface is not a centripetal acceleration, but, as the name implies, a gravitational acceleration, which is given by a = (k_g)*m/r^2, where k_g is the gravitational constant, m is the total mass of the attracting object (here the Earth) and r is the distance from the centre of the object (i.e. the Earth's radius). The formula for centripetal acceleration is not usable for that case, so it not giving out (something similar to) the equatorial rotational velocity is not surprising.

And I seem to get a velocity of about ~31.2 m/s = ~112.2 km/h for a ring with a radius of 100 m; could you post your calculations? I'm using v = sqrt(a*r) as given from the equation a=(v^2)/r where a = 9.81 m/s and r = 100 m.

EDIT: Important note: The force in your calculation above would /not/ be 9.81 Newton; it's given by F = m*a, where m is the mass of your human and a is the gravitational acceleration of 9.81 m/s^2.

Hmm ok that makes sence then that the earth cant do that... xD WOOOOPS

ok so my calculations are the following:

F=9.81 m/s^2
r=100 m
v= unknown
m=62 kg (average weight. I am 65kg or 145 pounds)

So from:

F = ((m x v) ^2) / r

9.81 = ((62 x v)^2) / 100


which is

v = Square-root((100 x 9.81) / 62)

I got 3.97 m/s or 1.104 km/h



OK so the F then is wrong? because the goal is to get 1 g

[Error]

There's no need to use a mass in there, and the force F isn't 9.81 m/s^2 as explained above; the force F would be given by F = m*a here, not just F = a = 9.81 m/s^2. If you set it up with F = m*a, where m is 62 kg and a = 9.81 m/s^2 you get:

F = m*a = (m*v^2)/r (not (m*v)^2)

From here we can see that we've got the constant mass m on both sides, so we divide by m on both sides to remove it from the equation:

a = (v^2)/r

We can solve this for v to get the equation I mentioned above by multiplying by r on both sides and taking the square root of both sides:

v = sqrt(a*r)

It's again important to note here that 1 g is not a force (given in Newton with unit kgm/s^2) but rather an acceleration (given in g where g = 9.81 m/s^2), if that makes any sense.

[nOmnomnOm]

Alright xD I'll recalculate now hehe.
Yeah I was doing my solution and was like wtf why so low!

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