Effect of Earth's rotation on a human?

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aphh

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<p>We fly in space on a giant spacecraft, that keeps spinning around it's axis. If I'm not mistaken, each human orbits the combined center of mass of both the individual and earth. This suggests a constant one-sided centripetal acceleration, the direction of acceleration determined by the hemisphere you're on.</p><p>So how come we don't notice the effect of constant acceleration? I can turn my head to all directions without noticing any effect caused by standing on a rotating platform.</p><p>Is the effect negligible on a individual's scale or are we just so well adjusted to the constant rotation in space?&nbsp;</p>
 
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aphh

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<p><font size="2">I did some research, and it turns out a = v*v/r, where v is the velocity, r is the radius and a is the centripetal acceleration.</font></p><p><font size="2">Earth's radius is about 6378 000 m and the rotational velocity or uniform circular motion at N 28.5 degrees would be 1471 000 m/s. We can assume that the combined center of mass for the individual and the earth resides very very near the center of the earth.</font></p><p><font size="2">Solving this would give us constant centripetal acceleration of 339 m/s2 on latitude equivalent to Florida.</font></p><p><font size="2">Maybe it would better to calculate the force imparted on us by the centripetal acceleration. That would be centripetal force.</font></p><p><font size="2">However, according to wikipedia this should not be confused with centrifugal force, which is a true kinematic force. So now I can't be quite sure what is the net force that is constantly affecting us by Earth's rotation and should be calculated?</font> </p>
 
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BrianSlee

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I did some research, and it turns out a = v*v/r, where v is the velocity, r is the radius and a is the centripetal acceleration.Earth's radius is about 6378 000 m and the rotational velocity or uniform circular motion at N 28.5 degrees would be 1471 000 m/s. We can assume that the combined center of mass for the individual and the earth resides very very near the center of the earth.Solving this would give us constant centripetal acceleration of 339 m/s2 on latitude equivalent to Florida.Maybe it would better to calculate the force imparted on us by the centripetal acceleration. That would be centripetal force.However, according to wikipedia this should not be confused with centrifugal force, which is a true kinematic force. So now I can't be quite sure what is the net force that is constantly affecting us by Earth's rotation and should be calculated? <br />Posted by aphh</DIV><br /><br /><span class="mw-headline">aphh,</span></p><p><span class="mw-headline">&nbsp; I think what you need to do in this case is ignore the coriolis force as it is counteracted by&nbsp;friction for somone standing or sitting on the surface of the earth. &nbsp;then solve the equation for centripdal force at the specific latitude in question and then solve for the reactive centrifigal force and subtract&nbsp;the gravitational constant to give a force vector.&nbsp;</span></p><p>**<font size="3">From Wikpedia</font>**</p><p><span class="mw-headline">Rotating sphere</span></p><div class="thumb tright"><div class="thumbinner" style="width:252px"><img class="thumbimage" src="http://upload.wikimedia.org/wikipedia/commons/thumb/8/8a/Earth_coordinates.PNG/250px-Earth_coordinates.PNG" border="0" alt="Figure 2: Coordinate system at latitude &phi; with x-axis east, y-axis north and z-axis upward (that is, radially outward from center of sphere)." width="250" height="272" /> <div class="thumbcaption"><div class="magnify"><img src="http://en.wikipedia.org/skins-1.5/common/images/magnify-clip.png" alt="" width="15" height="11" /></div>Figure 2: Coordinate system at latitude &phi; with <em>x</em>-axis east, <em>y</em>-axis north and <em>z</em>-axis upward (that is, radially outward from center of sphere).</div></div></div><p>Consider a location with latitude <img class="tex" src="http://upload.wikimedia.org/math/3/5/3/3538eb9c84efdcbd130c4c953781cfdb.png" alt="varphi" /> on a sphere that is rotating around the north-south axis.<sup class="reference">[1]</sup> A local coordinate system is set up with the <span class="texhtml"><em>x</em></span> axis horizontally due east, the <span class="texhtml"><em>y</em></span> axis horizontally due north and the <span class="texhtml"><em>z</em></span> axis vertically upwards.The rotation vector, velocity of movement and Coriolis acceleration expressed in this local coordinate system (listing components in the order East (<em>e</em>), North (<em>n</em>) and Upward (<em>u</em>)) are:</p><dl><dd><img class="tex" src="http://upload.wikimedia.org/math/b/a/3/ba30cd928bf77f8bb4bedbd9e516d5d4.png" alt="oldsymbol{ Omega} = omega egin{pmatrix} 0 cos varphi sin varphi end{pmatrix} ," /> &nbsp; &nbsp; <img class="tex" src="http://upload.wikimedia.org/math/d/7/0/d706b8d70692ab9cd38985785e0be3f7.png" alt="oldsymbol{ v} = egin{pmatrix} v_e v_n v_u end{pmatrix} ," /> </dd><dd><img class="tex" src="http://upload.wikimedia.org/math/c/6/e/c6e62c86103293ddc86811cb26593c3a.png" alt="oldsymbol{ a}_C =-2oldsymbol{Omega imes v}= 2,omega, egin{pmatrix} v_n sin varphi-v_u cos varphi -v_e sin varphi v_e cosvarphiend{pmatrix} ." /> </dd></dl><p>When considering atmospheric or oceanic dynamics, the vertical velocity is small and the vertical component of the Coriolis acceleration is small compared to gravity. For such cases, only the horizontal (East and North) components matter. The restriction of the above to the horizontal plane is (setting <font style="font-weight:normal;font-size:100%;font-style:italic;font-family:BookAntiqua">v<sub>u</sub></font>=0):</p><dl><dd><img class="tex" src="http://upload.wikimedia.org/math/6/e/c/6ece91a538058dc212db6d0ddaf4ab8d.png" alt=" oldsymbol{ v} = egin{pmatrix} v_e v_nend{pmatrix} ," /> &nbsp; &nbsp; <img class="tex" src="http://upload.wikimedia.org/math/a/5/b/a5b1fe2817cccf01853fe9cb89607b42.png" alt="oldsymbol{ a}_c = egin{pmatrix} v_n -v_eend{pmatrix} f , " /> </dd></dl><p>where <img class="tex" src="http://upload.wikimedia.org/math/5/9/2/592b71cdee7f6e4782afdacc7310be29.png" alt="f = 2 omega sin varphi ," /> is called the <em>Coriolis parameter</em>.</p><p>By setting <font style="font-weight:normal;font-size:100%;font-style:italic;font-family:BookAntiqua">v<sub>n</sub></font> = 0, it can be seen immediately that (for positive <img class="tex" src="http://upload.wikimedia.org/math/3/5/3/3538eb9c84efdcbd130c4c953781cfdb.png" alt="varphi" /> and <img class="tex" src="http://upload.wikimedia.org/math/a/7/0/a70d0c9b2e529c999ec05569e1638668.png" alt="omega," />) a movement due east results in an acceleration due south. Similarly, setting <font style="font-weight:normal;font-size:100%;font-style:italic;font-family:BookAntiqua">v<sub>e</sub></font> = 0, it is seen that a movement due north results in an acceleration due east &mdash; that is, standing on the horizontal plane, looking along the direction of the movement causing the acceleration, the acceleration always is turned 90&deg; to the right. That is:<sup class="reference">[2]</sup> <sup class="reference">[3]</sup></p><p>&nbsp;</p><p><span class="mw-headline">Uniformly rotating reference frames</span></p><dl><dd><span class="boilerplate seealso"><em>See also: Circular motion&nbsp;and Uniform circular motion</em></span> </dd></dl><p>Rotating reference frames are used in physics, mechanics, or meteorology whenever they are the most convenient frame to use.</p><p>The laws of physics are the same in all inertial frames. But a rotating reference frame is not an inertial frame, so the laws of physics are transformed from the inertial frame to the rotating frame. For example, assuming a <em>constant</em> rotation speed, transformation is achieved by adding to every object two <em>coordinate accelerations</em> that correct for the constant rotation of the coordinate axes. The vector equations describing these accelerations are:<sup class="reference">[27]</sup><sup class="reference">[42]</sup><sup class="reference">[10]</sup></p><dl><dd><table border="0"><tbody><tr><td><img class="tex" src="http://upload.wikimedia.org/math/b/6/4/b6481b07ca0ef019136bc1a8ab164579.png" alt="mathbf{a}_mathrm{rot}," />
 
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aphh

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<p>Thanks Brian, looks helpful. I will study the math next.</p><p>Meanwhile I did some layman thinking and realized that not even astronauts have reported feeling the effect of being in a merry-go-round with the diameter of some 13 000 000 metres and period of 90 minutes or 5400 seconds.</p><p>That gives the ratio for period to diameter of 5.4/13 000 or 0.054/130.&nbsp;</p><p>If we think about a merry-go-round on a amusement park with the diameter of 130 metres, using that ratio would give rotational period of only 0.054 seconds.</p><p>I would think that if on the ground the merry-go-round rotated once in 0.054 seconds, one would feel the effect of centripetal acceleration?</p><p>Something can not be quite right in that comparison.</p><p>I'm still baffled, I can see the coriolis effect each time I empty the kitchen sink, yet I can not feel it. Obviously the effect of gravity is vastly more powerful than the centripetal or centrifugal acceleration. </p><p>Like I said, I'll study the math and physics next. &nbsp; </p>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>We fly in space on a giant spacecraft, that keeps spinning around it's axis. If I'm not mistaken, each human orbits the combined center of mass of both the individual and earth. This suggests a constant one-sided centripetal acceleration, the direction of acceleration determined by the hemisphere you're on.So how come we don't notice the effect of constant acceleration? I can turn my head to all directions without noticing any effect caused by standing on a rotating platform.Is the effect negligible on a individual's scale or are we just so well adjusted to the constant rotation in space?&nbsp; <br /> Posted by aphh</DIV></p><p>I don't think it has anything to do with the individual's scale or being well adjusted to it.&nbsp; I think it has more to do with the overall scale of the system.</p><p>Centripital force is a what keeps your velocity from going in a straight line.&nbsp; Your velocity becomes circular.&nbsp; Acceleration is the rate of change of that velocity.&nbsp; Acceleration isn't always dependent on a change of speed in a linear direction.&nbsp; Acceleration can also be felt by changing direction while your speed remains constant.&nbsp; Speed is linear while velocity is both speed and direction.</p><p>While in orbit or a person standing on earth, that speed remains constant, but the change in direction is not.&nbsp; However, that change in direction is really small and negligible and will not be 'felt'.&nbsp; </p><p>On a merry-go-round, your change in direction is quite pronounced.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> <br /> Posted by BrianSlee</DIV></p><p>Brian... If you are going to cut n paste from another website, it is policy to attribute the source.&nbsp; </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p>[<br /><br />QUOTE]We fly in space on a giant spacecraft, that keeps spinning around it's axis. If I'm not mistaken, each human orbits the combined center of mass of both the individual and earth. This suggests a constant one-sided centripetal acceleration, the direction of acceleration determined by the hemisphere you're on.So how come we don't notice the effect of constant acceleration? I can turn my head to all directions without noticing any effect caused by standing on a rotating platform.Is the effect negligible on a individual's scale or are we just so well adjusted to the constant rotation in space?&nbsp; <br />Posted by aphh[/QUOTE]</p><p>Hopefully you can enlarge and read these pages, which contain the work to address your question.&nbsp; They are not polished, but they are correct and fairly complete, starting from first principles. What is in them is the analysis of the acceleration due to the rotation of the earth in the directions of a radial vector from the center of the earth, a tangential vector in the direction of the rotational velocity, and a tangential vector normal to both the velocity and the radial vector.&nbsp; It is this last direction in which you might feel the acceleration, as in the radial direction you would feel a force dominated by gravity and in the direction of the velocity there is no acceleration.&nbsp; But the reason that you feel essentially nothing is because the omega-squared R term is very small, only 0.00105 g (which is the centripetal acceleration at the equator), and the reason that this is so small is because the angular velocity of the earth is very small, 7.23*10^-5 radians/sec.&nbsp;&nbsp; The maximum tangential acceleration takes place at 45 degrees north or south latitude and is&nbsp;just 1/2 the maximum radial acceleration (which occurs at the equator).&nbsp;</p><p>&nbsp;</p><p><br /><img src="http://sitelife.space.com/ver1.0/Content/images/store/0/11/50f44a72-5791-4171-abcc-c7343f49d092.Medium.gif" alt="" /><img src="http://sitelife.space.com/ver1.0/Content/images/store/4/12/f442518e-4649-4f02-bcea-06dc2fc68f6c.Medium.gif" alt="" /><img src="http://sitelife.space.com/ver1.0/Content/images/store/0/1/5057dbd3-c702-49bc-a13f-c77103f8763d.Medium.gif" alt="" /><img src="http://sitelife.space.com/ver1.0/Content/images/store/6/3/d64fc16d-b63f-40de-93b0-48ca3953e957.Medium.gif" alt="" /></p><p><br /><br /><br /><br /><br /><br />&nbsp;</p> <div class="Discussion_UserSignature"> </div>
 
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BrianSlee

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Brian... If you are going to cut n paste from another website, it is policy to attribute the source.&nbsp; <br />Posted by derekmcd</DIV><br /><br />Sorry thought I grabbed the header when I did it.&nbsp; Reference has been added :) <div class="Discussion_UserSignature"> <p> </p><p>"I am therefore I think" </p><p>"The only thing "I HAVE TO DO!!" is die, in everything else I have freewill" Brian P. Slee</p> </div>
 
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aphh

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Hopefully you can enlarge and read these pages,<br /> Posted by DrRocket</DIV></p><p>Thanks for the material. It looks like I am able to get the information from the scans and will study the papers next.</p><p>I'm on Astrophysics 101 in the Open University, and it has sparked my brain to start thinking about things. Once I have learned to think, math and physics will follow (I'm actually almost getting to the point of starting to refresh calculus). </p>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Something can not be quite right in that comparison.I'm still baffled, I can see the coriolis effect each time I empty the kitchen sink, yet I can not feel it. Obviously the effect of gravity is vastly more powerful than the centripetal or centrifugal acceleration. Like I said, I'll study the math and physics next. &nbsp; <br /> Posted by aphh</DIV></p><p>Your water going down the drain has nothing to do with the coriolis effect.&nbsp; It is a myth that it swirls opposite in the southern hemisphere versus the northern hemisphere.&nbsp; At least, it is a myth under real world circumstances.&nbsp; I've been in both hemispheres... it doesn't happen.&nbsp; Maybe in tightly controlled lab experiments, but not in your kitchen sink.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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aphh

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Your water going down the drain has nothing to do with the coriolis effect.&nbsp; It is a myth that it swirls opposite in the southern hemisphere versus the northern hemisphere.&nbsp; At least, it is a myth under real world circumstances.&nbsp; I've been in both hemispheres... it doesn't happen.&nbsp; Maybe in tightly controlled lab experiments, but not in your kitchen sink.&nbsp; <br /> Posted by derekmcd</DIV></p><p>You might be right. I tested on 2 separate sinks using soap bubbles, and a very rapid testing seemed to result in vortex of opposite directions, however the most easily detectable vortex in the first sink was clearly running counter-clockwise.</p><p>Do you have a link for this to be a myth for the vortex spiral backwards in Australia? I guess I can dig myself too. </p>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>You might be right. I tested on 2 separate sinks using soap bubbles, and a very rapid testing seemed to result in vortex of opposite directions, however the most easily detectable vortex in the first sink was clearly running counter-clockwise.Do you have a link for this to be a myth for the vortex spiral backwards in Australia? I guess I can dig myself too. <br /> Posted by aphh</DIV></p><p>http://www.badastronomy.com/mad/1996/coriolis.html</p><p>&nbsp;</p><p>edit:&nbsp; here's an even better one:</p><p>http://www.ems.psu.edu/~fraser/Bad/BadCoriolis.html</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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aphh

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>http://www.badastronomy.com/mad/1996/coriolis.htmledit: here's an even better one:http://www.ems.psu.edu/~fraser/Bad/BadCoriolis.html <br /> Posted by derekmcd</DIV></p><p>Thanks. Well, in the latter link, he didn't completely rule out the possibility of seeing the coriolis effect in draining a large sink in lab conditions. Also, in his own test he let the water run while observing the vortex whereas I was sure to let the water calm down before draining the sink, which then occurred with a counter-clockwise vortex.&nbsp;</p><p>Naturally I realize that the dynamics and geometry of the sink itself has an impact on the vortex.&nbsp;</p>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks. Well, in the latter link, he didn't completely rule out the possibility of seeing the coriolis effect in draining a large sink in lab conditions. Also, in his own test he let the water run while observing the vortex whereas I was sure to let the water calm down before draining the sink, which then occurred with a counter-clockwise vortex.&nbsp;Naturally I realize that the dynamics and geometry of the sink itself has an impact on the vortex.&nbsp; <br /> Posted by aphh</DIV></p><p>I'm sure in tightly controlled lab experiments, you can see the coriolis effect, but not in natural real world situations. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks. Well, in the latter link, he didn't completely rule out the possibility of seeing the coriolis effect in draining a large sink in lab conditions. Also, in his own test he let the water run while observing the vortex whereas I was sure to let the water calm down before draining the sink, which then occurred with a counter-clockwise vortex.&nbsp;Naturally I realize that the dynamics and geometry of the sink itself has an impact on the vortex.&nbsp; <br />Posted by aphh</DIV></p><p>The coriolis effect is pretty small at the scale and velocities in your sink.&nbsp; The direction of the swirl is probably dominated by any momentum in the fluid flow (there will be some flow) when you pull the plug than by the coriolis effect.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>
 
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DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks for the material. It looks like I am able to get the information from the scans and will study the papers next.I'm on Astrophysics 101 in the Open University, and it has sparked my brain to start thinking about things. Once I have learned to think, math and physics will follow (I'm actually almost getting to the point of starting to refresh calculus). <br />Posted by aphh</DIV></p><p>You will need a little calculus to follow the calculations.&nbsp; That dot over some of the quantities represents a derivative with respect to time (to get velocity) and two dots the second derivative (for acceleration).&nbsp; You will also need the vector dot product -- multiply vector components pairwise and add them up, and the length of a vector (noted by parallel upright lines like two absolute value signs) which is the square root of the sum of the squares of the components.</p><p>You might want to dig out your old calculus book and look at the material on vectors in 3 dimensions and spherical coordinates.&nbsp; To figure out how I got the tangential component of acceleration that is normal to both the radial vector and to the velocity vector you will need to understand the cross product of vectors, but to simply understand the result you can get by without it.&nbsp; But that tangential direction is important, since it is the direction in which you would feel something if there were enough of a force to feel (there isn't).</p> <div class="Discussion_UserSignature"> </div>
 
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