# Hemisphere question **URL:** \ **Category:** Factual Questions\ **Created:** [March 14, 2004, 9:51pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295 "2004-03-14T21:51:20Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![dtilque](https://avatars.discourse-cdn.com/v4/letter/d/d6d6ee/32.png) [@dtilque](https://boards.straightdope.com/u/dtilque)\ **Post date:** [March 15, 2004, 6:55am UTC](https://boards.straightdope.com/t/hemisphere-question/234295/21 "2004-03-15T06:55:58Z")
Well, I just checked my bathroom and kitchen sinks. The bathroom always drains clockwise and the kitchen counterclockwise. Does that mean the equator runs between the two? ---
**Author:** ![Sergio](https://avatars.discourse-cdn.com/v4/letter/s/b77776/32.png) [@Sergio](https://boards.straightdope.com/u/Sergio)\ **Post date:** [March 15, 2004, 12:18pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/22 "2004-03-15T12:18:41Z")
> [@z1221z](#): > > Yes they are right; the coriolis force is much to small to have an effect on the vortex in your drain. But the snopes explanation is very simplistic, and it suggests only objects moving north or south will be affected by the coriolis force, but that´s not true. A cannonball fired due east or west will also be deflected, just like _any_ moving object will be deflected (to the left or to the right, dependig on the hemisphere). And that´s the hard part to understand. The Coriolis effect needs that you move in a way to alter your distance relative to the center of rotation. When you move northwards or southwords you alter your distance relative to the center of the Earth. If you move due east or west you remain over the same paralel and there is no effect. Of course, since the cannonball goes up and then down, somehow it’s distance to the center of the Earth varies and there is a Coriolis effect. If thaer was no air resistance the times up and down would be the same and the net effect would be null. Since there is an atmosphere you have a small Coriolis effect that alters the maximum range, but does not deflect the ball. For a good debunking of the flush effect see the Bad Meteorology page: [http://www.ems.psu.edu/~fraser/Bad/BadCoriolis.html](http://www.ems.psu.edu/~fraser/Bad/BadCoriolis.html) ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [March 15, 2004, 2:59pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/23 "2004-03-15T14:59:35Z")
> [@Sérgio](#): > > The Coriolis effect needs that you move in a way to alter your distance relative to the center of rotation. When you move northwards or southwords you alter your distance relative to the center of the Earth. If you move due east or west you remain over the same paralel and there is no effect. Doesn’t work out that way. Check out the [Bad Coriolis FAQ, The teacher was right](http://www.ems.psu.edu/~fraser/Bad/BadFAQ/BadCoriolisFAQ.html), at the Bad Meteorology page you linked. > [@](#): > > For a good debunking of the flush effect see the Bad Meteorology page: > [http://www.ems.psu.edu/~fraser/Bad/BadCoriolis.html](http://www.ems.psu.edu/~fraser/Bad/BadCoriolis.html) ---
**Author:** ![Sergio](https://avatars.discourse-cdn.com/v4/letter/s/b77776/32.png) [@Sergio](https://boards.straightdope.com/u/Sergio)\ **Post date:** [March 15, 2004, 3:16pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/24 "2004-03-15T15:16:40Z")
> [@RM Mentock](#): > > Doesn’t work out that way. Check out the [Bad Coriolis FAQ, The teacher was right](http://www.ems.psu.edu/~fraser/Bad/BadFAQ/BadCoriolisFAQ.html), at the Bad Meteorology page you linked. In the example of the rotating platform, you don’t have to go straight radially to have the Coriolis effect, but you must have a radial component of the movemnt. If you move keeping a constant distance from the center, there is no Coriolis force. The same applies to the Earth. If you move in any direction but due east or west, there will be a northward or southward component that will arise a Coriolis force. If you move strictly over a paralel there is no Coriolis effect. ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [March 15, 2004, 3:25pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/25 "2004-03-15T15:25:24Z")
> [@r\_k](#): > > Of course, it doesn’t stop a guy at the equator in Kenya doing demonstrations for folks, using a basin of water and a twig. I believe he just gives the water a slight swirl while they’re not looking 😉 But seriously, if you lived on the equator, and what people say were true, then the water would not swirl at all on the equator. When flushed, it would just drop straight down. To switch directions, from clockwise and counterclockwise (or vice versa), there would have to be a point where the angular velocity is zero. Similar to how, with linear motion, you cannot go from a forward motion to reverse (or vice versa) without passing through zero. That place would be along the equator - applying the general public’s understanding, anyhow. - Jinx ---
**Author:** ![slumtrimpet](https://avatars.discourse-cdn.com/v4/letter/s/b5e925/32.png) [@slumtrimpet](https://boards.straightdope.com/u/slumtrimpet)\ **Post date:** [March 15, 2004, 3:41pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/26 "2004-03-15T15:41:24Z")
Hey, thanks for all the info guys! I’m off in about an hour and I’ll keep my eyes out for that red line. 😃 (and you know I’m going to have to drain a sink as soon as we get to Oz just to see 😉 ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [March 15, 2004, 3:45pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/27 "2004-03-15T15:45:30Z")
> [@](#): > > _coop wrote:_ > I lived in Butterworth, Malaysia from December, 1977 to December, 1979. Water drained anti-clockwise all the time I was there. I am now living in a suburb near Sydney, NSW, Australia. The water drains clockwise in my sink. Is that enough of an experiment? It’s not enough - we would need much more data, but your preliminary study seems to indicate that the Coriolis explanation has it backwards. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [March 15, 2004, 4:14pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/28 "2004-03-15T16:14:46Z")
> [@CurtC](#): > > It’s not enough - we would need much more data, but your preliminary study seems to indicate that the Coriolis explanation has it backwards. As the water drains, in the north, it moves to the right–which causes a counter-clockwise swirl. Same reason that hurricanes rotate counterclockwise in the North–but [their tracks turn to the right](http://scienceworld.wolfram.com/physics/CoriolisForce.html). ---
**Author:** ![Sergio](https://avatars.discourse-cdn.com/v4/letter/s/b77776/32.png) [@Sergio](https://boards.straightdope.com/u/Sergio)\ **Post date:** [March 15, 2004, 5:29pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/29 "2004-03-15T17:29:49Z")
> [@CurtC](#): > > It’s not enough - we would need much more data, but your preliminary study seems to indicate that the Coriolis explanation has it backwards. oriolis force is too weak to turn water in any sense. It acts in long term phenomena, like hurricanes, but not in water draining in a sink. As sveral people have mentioned and as is well explained in the bad meteorology site, water turns on a sink because of it’s initial velocity and not because of Earths rotation. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [March 15, 2004, 7:54pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/30 "2004-03-15T19:54:19Z")
> [@Sérgio](#): > > In the example of the rotating platform, you don’t have to go straight radially to have the Coriolis effect, but you must have a radial component of the movemnt. If you move keeping a constant distance from the center, there is no Coriolis force. If you stay at the same distance from the center, you must allow for a centripetal force, as you keep the same radius. Straight lines on the Earth’s surface are not along parallels, either. ---
**Author:** ![z1221z](https://avatars.discourse-cdn.com/v4/letter/z/ee7513/32.png) [@z1221z](https://boards.straightdope.com/u/z1221z)\ **Post date:** [March 15, 2004, 8:14pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/31 "2004-03-15T20:14:27Z")
> [@Sérgio](#): > > In the example of the rotating platform, you don’t have to go straight radially to have the Coriolis effect, but you must have a radial component of the movemnt. If you move keeping a constant distance from the center, there is no Coriolis force. The same applies to the Earth. If you move in any direction but due east or west, there will be a northward or southward component that will arise a Coriolis force. If you move strictly over a paralel there is no Coriolis effect. Sorry, but that´s wrong. _ANY_ moving (relative to the frame of reference) object is affected by the coriolis force, regardless of direction. The coriolis force does vary with latitude, and is zero at the equator and peaks at the poles, but it is totally independent of the direction of the moving object. This is why I originally pointed out that the snopes explanaition was simplistic, because it makes you believe the coriolis force is something it isn´t. Consider your rotating platform example: imagine you´re sitting on the edge of the platform and you´re shooting a ball in a tangential direction. From an outside observer´s point of view, the balls goes straigt. From your point of view, it´s a curveball. And it will always be a curveball, regardless of the direction in which you shoot your ball. ---
**Author:** ![Sergio](https://avatars.discourse-cdn.com/v4/letter/s/b77776/32.png) [@Sergio](https://boards.straightdope.com/u/Sergio)\ **Post date:** [March 15, 2004, 8:33pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/32 "2004-03-15T20:33:52Z")
> [@z1221z](#): > > Sorry, but that´s wrong. _ANY_ moving (relative to the frame of reference) object is affected by the coriolis force, regardless of direction. The coriolis force does vary with latitude, and is zero at the equator and peaks at the poles, but it is totally independent of the direction of the moving object. > This is why I originally pointed out that the snopes explanaition was simplistic, because it makes you believe the coriolis force is something it isn´t. > > Consider your rotating platform example: imagine you´re sitting on the edge of the platform and you´re shooting a ball in a tangential direction. From an outside observer´s point of view, the balls goes straigt. From your point of view, it´s a curveball. And it will always be a curveball, regardless of the direction in which you shoot your ball. Let’s keep it simple to see if you can understand the principle of Coriolis acceleration. Imagine a platform rotating at an angular velocity Omega. You stand in a point P1 at a distance R1 from the axis of rotation. You have a tangential velocity V1 = R1_Omega relative to an inertial reference. Now you move to a point P2, distant R2 from the axis. Your tangential velocity is V2 = R2_Omega. In order to change your velocity you need an acceleration. Coriolis established that this acceleration is 2_Omega_Vr, where Vr is the radial component of the velocity in wich you moved from P1 to P2. Now move from P2 to P3, distant R2 from the axis. Your tangential velocity will still be R2\*Omega, so no acceleration is needed. If you use the formula you see that Vr = 0, so the formula is consistent with the cncept I used. Note that the Coriolis acceleration does not depend of where is P1. So your affirmation that it is dependent of the latitude does not hold. ---
**Author:** ![z1221z](https://avatars.discourse-cdn.com/v4/letter/z/ee7513/32.png) [@z1221z](https://boards.straightdope.com/u/z1221z)\ **Post date:** [March 15, 2004, 8:35pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/33 "2004-03-15T20:35:58Z")
> [@z1221z](#): > > _ANY_ moving (relative to the frame of reference) object is affected by the coriolis force … This should be: _ANY_ object moving relative to a rotating frame of reference is affected by the coriolis force. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [March 15, 2004, 9:08pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/34 "2004-03-15T21:08:22Z")
> [@Sérgio](#): > > Note that the Coriolis acceleration does not depend of where is P1. So your affirmation that it is dependent of the latitude does not hold. The coriolis effect is zero at the equator, right? So, it does have a latitude dependence. ---
**Author:** ![z1221z](https://avatars.discourse-cdn.com/v4/letter/z/ee7513/32.png) [@z1221z](https://boards.straightdope.com/u/z1221z)\ **Post date:** [March 15, 2004, 9:19pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/35 "2004-03-15T21:19:05Z")
> [@Sérgio](#): > > Let’s keep it simple to see if you can understand the principle of Coriolis acceleration. Imagine a platform rotating at an angular velocity Omega. You stand in a point P1 at a distance R1 from the axis of rotation. You have a tangential velocity V1 = R1_Omega relative to an inertial reference. > Now you move to a point P2, distant R2 from the axis. Your tangential velocity is V2 = R2_Omega. In order to change your velocity you need an acceleration. Coriolis established that this acceleration is 2_Omega_Vr, where Vr is the radial component of the velocity in wich you moved from P1 to P2. > Now move from P2 to P3, distant R2 from the axis. Your tangential velocity will still be R2\*Omega, so no acceleration is needed. If you use the formula you see that Vr = 0, so the formula is consistent with the cncept I used. > Note that the Coriolis acceleration does not depend of where is P1. So your affirmation that it is dependent of the latitude does not hold. Yes I do understand the coriolis force. You don´t have to keep it simple for me. And yes I also understand what you´re trying to say. But your expamle also holds true if you´re stepping in a tangential direction. Imagine your rotating turntable is in space (no atmosphere rotating with you etc.), and your “steps” are mini jumps. If you jump off in a tangential direction, you will _NOT_ land at the same distance from the axis as you took off. And it is dependent on the latitude (or distance from the center of rotation in the turntable example). If you jump straight up at the edge of the turntable, your “takeoff” point will be farther away from your “landing” point than if you jumped staight up nearer to the axis of rotation. ---
**Author:** ![z1221z](https://avatars.discourse-cdn.com/v4/letter/z/ee7513/32.png) [@z1221z](https://boards.straightdope.com/u/z1221z)\ **Post date:** [March 15, 2004, 9:28pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/36 "2004-03-15T21:28:37Z")
> [@z1221z](#): > > f. > And it is dependent on the latitude (or distance from the center of rotation in the turntable example). If you jump straight up at the edge of the turntable, your “takeoff” point will be farther away from your “landing” point than if you jumped staight up nearer to the axis of rotation. Oh, sorry, that example was _REALLY_ bad. I apologize. That was a really, really, bad and very wrong example. Shame on me. Sorry. Actually, it is more like the other way around. ---
**Author:** ![Sergio](https://avatars.discourse-cdn.com/v4/letter/s/b77776/32.png) [@Sergio](https://boards.straightdope.com/u/Sergio)\ **Post date:** [March 15, 2004, 10:51pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/37 "2004-03-15T22:51:21Z")
> [@RM Mentock](#): > > The coriolis effect is zero at the equator, right? So, it does have a latitude dependence. Wrong! If you are at the equator you have a tangential velocity that is equal to the angular velocity of Earth (2_pi/(24_60\*60) rad/s times the radius of Earth (6366 km) = 460 m/s. If you move towards south or north, your distance to the axis of the Earth will be the radius of the Earth times the cosine of latitude. A lesser radius means a smaller tangential velocity, so you must provide an acceleration in the east-west sense in order to keep over the same meridian. This acceleration is due to Coriolis effect. ---
**Author:** ![z1221z](https://avatars.discourse-cdn.com/v4/letter/z/ee7513/32.png) [@z1221z](https://boards.straightdope.com/u/z1221z)\ **Post date:** [March 15, 2004, 11:45pm UTC](https://boards.straightdope.com/t/hemisphere-question/234295/38 "2004-03-15T23:45:46Z")
> [@Sérgio](#): > > Wrong! If you are at the equator you have a tangential velocity that is equal to the angular velocity of Earth (2_pi/(24_60\*60) rad/s times the radius of Earth (6366 km) = 460 m/s. > If you move towards south or north, your distance to the axis of the Earth will be the radius of the Earth times the cosine of latitude. A lesser radius means a smaller tangential velocity, so you must provide an acceleration in the east-west sense in order to keep over the same meridian. This acceleration is due to Coriolis effect. Sorry, but you´re wrong again. Once you move towards the north or south, you are no longer at the equator, and the coriolis force takes effect. A cannonball fired due north from the equator will be deflected by the coriolis force; I think we are in agreement about that. But the coriolis fore is “stronger” the higher the latitude is. Also artillery fire aimed due east or west is deflected by the coriolis force as well. If you fire in those directions from the equator, the coriolis force is nil. But if you fire from, say Oslo, due east or west, your shells will be deflected, and might miss. My “jump straight up from the turntable” examle was really not very good. But it is still a fact that the coriolis force is independent of the direction of the moving object. And it does vary with latitude. Your explanation of the coriolis force is simply wrong. It´s like the snopes model, and it fails to acknowledge that that damned force deflects _ALL_ moving objects, regardless of their direction (sorry for being repetitive). Since you seem to like equations, why don´t you take an undergrad textbook on mechanics, look up the coriolis force, and tell me where exactly the direction of the moving object comes in? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [March 16, 2004, 12:21am UTC](https://boards.straightdope.com/t/hemisphere-question/234295/39 "2004-03-16T00:21:51Z")
> [@](#): > > Since you seem to like equations, why don´t you take an undergrad textbook on mechanics, look up the coriolis force, and tell me where exactly the direction of the moving object comes in? Would you settle for a graduate physics textbook? From Fetter and Walecka, pg. 39 > [@](#): > > The term _-2m_ω x _(d__r_\*/_dt_)[sub]body[/sub] , called the _Coriolis force_, vanishes unless the particle moves in the rotating frame along a direction different from _ω_^. In other words, if you move straight east or west, you won’t feel a Coriolis force. The direction dependence comes from the fact that you’re taking the cross-product of your velocity with the rotation pseudovector, and as every undergraduate physicist knows, the cross product depends on relative direction. Your example of hopping along a rotating surface is irrelevant, here, because hopping includes vertical motion as well as horizontal, and the vertical motion will be subject to a Coriolis force. But if you stay at a single height and walk east or west, then Coriolis won’t have an effect. Meanwhile, exactly at the Equator, horizontal motion won’t move you towards or away from the axis, so exactly at the Equator, the Coriolis force on a horizontally-moving object is zero. Again, if you move vertically, then there will be a Coriolis force, even at the Equator. ---
**Author:** ![Sergio](https://avatars.discourse-cdn.com/v4/letter/s/b77776/32.png) [@Sergio](https://boards.straightdope.com/u/Sergio)\ **Post date:** [March 16, 2004, 12:21am UTC](https://boards.straightdope.com/t/hemisphere-question/234295/40 "2004-03-16T00:21:57Z")
> [@z1221z](#): > > Yes I do understand the coriolis force. You don´t have to keep it simple for me. And yes I also understand what you´re trying to say. But your expamle also holds true if you´re stepping in a tangential direction. Imagine your rotating turntable is in space (no atmosphere rotating with you etc.), and your “steps” are mini jumps. If you jump off in a tangential direction, you will _NOT_ land at the same distance from the axis as you took off. > And it is dependent on the latitude (or distance from the center of rotation in the turntable example). If you jump straight up at the edge of the turntable, your “takeoff” point will be farther away from your “landing” point than if you jumped staight up nearer to the axis of rotation. The jumping example is a bad one. All that matters is the angular speed that is the same in any point, the difference in tangential velocity (in your example of the jumps it will certainly happen) and the time it takes for this increase in velocity happen. In the surface of the Earth, there is no increase in tangential velocity if you jump, since you will always be over the same paralel. When you go up you will be farther from the axis and you will lag relative to the vertical, but when you come down you catch up and you will land in the same spot. Of course, since there is air resistance, that acts both ways decreasing your speed, you will fall a little behind, but this has nothing to do with Coriolis. [Previous page](https://boards.straightdope.com/t/hemisphere-question/234295.md?page=1) [Next page](https://boards.straightdope.com/t/hemisphere-question/234295.md?page=3)