equatorial physics
December 10, 2007 6:50 PM Subscribe
Are there any easily demonstrable physical effects to being on the equator?
So I'll be standing on the equator in a week or two and was hoping to do something more entertaining than watching other tourists being taken in by various scams.
If I'm visualizing the celestial mechanics correctly the sun won't be even near to directly overhead this close to the solistice, so building a linear sundial won't work. I'll do that another day.
I'm pretty sure a pendulum would need a bob with more mass than I'm willing to carry to show anything about centripetal force (and frankly I'm not sure what it would show).
At this point I'm pretty much reduced to taking pictures of the 0.0000 latitude on my GPS.
Hope me, Metafilter!
So I'll be standing on the equator in a week or two and was hoping to do something more entertaining than watching other tourists being taken in by various scams.
If I'm visualizing the celestial mechanics correctly the sun won't be even near to directly overhead this close to the solistice, so building a linear sundial won't work. I'll do that another day.
I'm pretty sure a pendulum would need a bob with more mass than I'm willing to carry to show anything about centripetal force (and frankly I'm not sure what it would show).
At this point I'm pretty much reduced to taking pictures of the 0.0000 latitude on my GPS.
Hope me, Metafilter!
Can you balance eggs there or is that a lunar deal?
Either way don't fall for the manufactured coriolis effect.
posted by sanka at 7:14 PM on December 10, 2007
Either way don't fall for the manufactured coriolis effect.
posted by sanka at 7:14 PM on December 10, 2007
Response by poster: Can you balance eggs there or is that a lunar deal?
You can balance eggs any time, any place.
Well, maybe not on a 45 degree slope during a hurricane, but still.
posted by tkolar at 7:23 PM on December 10, 2007
You can balance eggs any time, any place.
Well, maybe not on a 45 degree slope during a hurricane, but still.
posted by tkolar at 7:23 PM on December 10, 2007
Best answer: I don't know if this is dramatic enough for you, but at the equator, sunrise and sunset are always 12 hours apart, any day of the year. Actually, you'll be there about the best time of the year to demonstrate that, right around the southern ("winter") solstice, when daylight hours are shortest in the northern hemisphere and longest in the southern hemisphere.
posted by DevilsAdvocate at 7:58 PM on December 10, 2007
posted by DevilsAdvocate at 7:58 PM on December 10, 2007
Best answer: Holy cow yes, regarding what DA just said -
I was at the equator a few years ago and the thing that absolutely blew me away was how rapid the sun rises and sets.
One minute you're standing in broad daylight, then, boom. Night.
The sun seems to simply get sucked below the horizon. Impressive, if, like me, you're from a northern region.
posted by Baby_Balrog at 8:15 PM on December 10, 2007
I was at the equator a few years ago and the thing that absolutely blew me away was how rapid the sun rises and sets.
One minute you're standing in broad daylight, then, boom. Night.
The sun seems to simply get sucked below the horizon. Impressive, if, like me, you're from a northern region.
posted by Baby_Balrog at 8:15 PM on December 10, 2007
Would a bath tub/sink/toilet NOT have a whirlpool effect in either direction when draining?
posted by Kevin S at 8:23 PM on December 10, 2007
posted by Kevin S at 8:23 PM on December 10, 2007
Response by poster: I don't know if this is dramatic enough for you, but at the equator, sunrise and sunset are always 12 hours apart, any day of the year.
Huh, I'd be interested to know if that's true or not. I would figure the axial tilt would throw the hours off the same way it throws the path of the sun off.
posted by tkolar at 8:27 PM on December 10, 2007
Huh, I'd be interested to know if that's true or not. I would figure the axial tilt would throw the hours off the same way it throws the path of the sun off.
posted by tkolar at 8:27 PM on December 10, 2007
Response by poster:
Would a bath tub/sink/toilet NOT have a whirlpool effect in either direction when draining?
You should probably read the scams link.
posted by tkolar at 8:28 PM on December 10, 2007
Would a bath tub/sink/toilet NOT have a whirlpool effect in either direction when draining?
You should probably read the scams link.
posted by tkolar at 8:28 PM on December 10, 2007
Where are you going? If it's Ecuador, there's a whole load of things to do at the Mitad del Mundo - some hokum, others real.
posted by gottabefunky at 8:29 PM on December 10, 2007
posted by gottabefunky at 8:29 PM on December 10, 2007
Response by poster: Yes, Ecuador. I've heard about a lot of the hokum at Mitad Del Mundo, but nothing about any real demonstrations...
posted by tkolar at 8:40 PM on December 10, 2007
posted by tkolar at 8:40 PM on December 10, 2007
Best answer: Huh, I'd be interested to know if that's true or not. I would figure the axial tilt would throw the hours off the same way it throws the path of the sun off.
posted by tkolar at 8:27 PM on December 10
This is true because both the equatorial line and the terminator are great circles. Geometry says that all great circles bisect each other.
The proof of this is that great circles divide a sphere in two and thus the two intersection points are a diameter and a diameter divides a circle in two.
posted by vacapinta at 9:23 PM on December 10, 2007 [1 favorite]
posted by tkolar at 8:27 PM on December 10
This is true because both the equatorial line and the terminator are great circles. Geometry says that all great circles bisect each other.
The proof of this is that great circles divide a sphere in two and thus the two intersection points are a diameter and a diameter divides a circle in two.
posted by vacapinta at 9:23 PM on December 10, 2007 [1 favorite]
Response by poster: (spends a few minutes with two rubber bands and a tennis ball)
Okay, I'm a believer now. And seeing that geometric fact demonstrated on such a grand scale is exactly the sort of thing I'm looking for.
posted by tkolar at 9:33 PM on December 10, 2007
Okay, I'm a believer now. And seeing that geometric fact demonstrated on such a grand scale is exactly the sort of thing I'm looking for.
posted by tkolar at 9:33 PM on December 10, 2007
tkolar,
I agree with Baby_Balrog - the sudden rise and set of the Sun were what I noticed most.
Vacapinta is right. It's exactly 12 hours if you ignore the bending of sunlight (refraction) by the atmosphere. Refraction adds about 7 more minutes to the length of day.
You weigh about 0.5% less at the equator than you do at the poles, since the Earth has a bit of a middle-aged spread and thus you're farther from the center of the planet and feeling a tad less gravitational pull.
Have fun in Ecuador!
posted by lukemeister at 9:43 PM on December 10, 2007
I agree with Baby_Balrog - the sudden rise and set of the Sun were what I noticed most.
Vacapinta is right. It's exactly 12 hours if you ignore the bending of sunlight (refraction) by the atmosphere. Refraction adds about 7 more minutes to the length of day.
You weigh about 0.5% less at the equator than you do at the poles, since the Earth has a bit of a middle-aged spread and thus you're farther from the center of the planet and feeling a tad less gravitational pull.
Have fun in Ecuador!
posted by lukemeister at 9:43 PM on December 10, 2007
At equinox on the equator, the water goes down the plughole in a figure 8.
posted by flabdablet at 11:44 PM on December 10, 2007 [2 favorites]
posted by flabdablet at 11:44 PM on December 10, 2007 [2 favorites]
The sun is only directly overhead a couple days out of the year at the equator... the length of days still varies. The total variance is smallest at the equator, but still there.
posted by TravellingDen at 12:15 AM on December 11, 2007
posted by TravellingDen at 12:15 AM on December 11, 2007
TravellingDen,
The Sun is only overhead on two days (the equinoxes), but the length of day is the same every day of the year.
posted by lukemeister at 5:33 AM on December 11, 2007
The Sun is only overhead on two days (the equinoxes), but the length of day is the same every day of the year.
posted by lukemeister at 5:33 AM on December 11, 2007
The length of the day is the same at the equator year round, 12 hours and 7 minutes, measured from when the upper edge of the sun touches the horizon. However, the time of sunrise and sunset relative to the clock move forward and backward by about 18 minutes throughout the year because of the eccentricity of the earth's orbit. For example, at the prime meridian on the equator, sunset is as early as 17:47 GMT in November and as late as 18:18 GMT in February. This is because the earth is revolving about the sun faster at perihelion in January and slower at aphelion in July. This causes sunrise and sunset to either lead or lag the clock.
posted by JackFlash at 8:57 AM on December 11, 2007
posted by JackFlash at 8:57 AM on December 11, 2007
Response by poster: You weigh about 0.5% less at the equator than you do at the poles
BTW, I've been given sufficient reason to doubt this conclusion. The basic gravitational equation assumes a sphere with the center of mass at its center, which doesn't really apply to the earth.
posted by tkolar at 2:17 PM on December 11, 2007
BTW, I've been given sufficient reason to doubt this conclusion. The basic gravitational equation assumes a sphere with the center of mass at its center, which doesn't really apply to the earth.
posted by tkolar at 2:17 PM on December 11, 2007
tkolar,
I gave a lame explanation, but it's a real effect. It has been measured very accurately.
posted by lukemeister at 8:34 PM on December 11, 2007
I gave a lame explanation, but it's a real effect. It has been measured very accurately.
posted by lukemeister at 8:34 PM on December 11, 2007
Response by poster: lukemeister,
I want to believe, but do you have a source for the "measured very accurately" assertion? Myself and another person spent quite a bit of time scouring the web for information on this today, and all we could find were people relying on suspect mathematical models.
Frankly at this point I'd settle for finding someone making a mathematical argument that didn't assume the earth was a sphere.
Thanks.
posted by tkolar at 9:29 PM on December 11, 2007
I want to believe, but do you have a source for the "measured very accurately" assertion? Myself and another person spent quite a bit of time scouring the web for information on this today, and all we could find were people relying on suspect mathematical models.
Frankly at this point I'd settle for finding someone making a mathematical argument that didn't assume the earth was a sphere.
Thanks.
posted by tkolar at 9:29 PM on December 11, 2007
OK, here's one: let's take it as a given that the overall shape of the Earth has remained fairly stable for the last billion years or so. It might wobble or bounce a bit here or there but it's not getting steadily more or less disc-like with time. Let's further assume that the Earth can be modelled as a great big blob of very viscous liquid, uniform in composition to a first approximation, and that its overall shape is defined by the combined action of gravity and centrifugal force due to its rotation.
If the net inward force experienced by a surface-bound observer (i.e. the observer's weight) near the poles was in fact greater than that experienced by a surface-bound observer of equal mass at the equator, then there would exist a net flattening force on the Earth, and it would be becoming more disc-like over time. This isn't happening. Therefore, ignoring small local gravitational variations due to high-speed effects like continental drift, we can be pretty sure that a given observer's weight is pretty much the same everywhere.
In short, the Earth has already settled into a shape that distributes its mass in such a way that the centrifugal force generated by its rotation is exactly balanced by extra gravity near the equator due to the extra mass piled up around its waistline.
If it were not spinning at all, it would be pretty much perfectly spherical. If it were spinning hella fast, it would be pretty much pancake shaped. And in all cases, provided the shape wasn't changing over time, an observer's weight would be pretty much the same regardless of where on the surface it was measured.
posted by flabdablet at 5:31 AM on December 12, 2007
If the net inward force experienced by a surface-bound observer (i.e. the observer's weight) near the poles was in fact greater than that experienced by a surface-bound observer of equal mass at the equator, then there would exist a net flattening force on the Earth, and it would be becoming more disc-like over time. This isn't happening. Therefore, ignoring small local gravitational variations due to high-speed effects like continental drift, we can be pretty sure that a given observer's weight is pretty much the same everywhere.
In short, the Earth has already settled into a shape that distributes its mass in such a way that the centrifugal force generated by its rotation is exactly balanced by extra gravity near the equator due to the extra mass piled up around its waistline.
If it were not spinning at all, it would be pretty much perfectly spherical. If it were spinning hella fast, it would be pretty much pancake shaped. And in all cases, provided the shape wasn't changing over time, an observer's weight would be pretty much the same regardless of where on the surface it was measured.
posted by flabdablet at 5:31 AM on December 12, 2007
The "closer to the centre" arguments quoted in the article linked by lukemeister don't really hold water. Yes, there is a formula that says that the acceleration due to gravity is inversely proportional to the square of the distance between the centres of mass, but this formula is really only applicable for well-separated masses. The formula for gravitational acceleration experienced by a small mass on the surface of a much larger one is a complicated solid integral.
For an intuitive understanding, consider that if you're sitting on the ground, there's a bunch of mass pulling you to the left that's cancelled by an equal amount pulling you to the right, and so on for any opposing pair of sideways directions. Your net downward acceleration is pretty much due to what's directly under your arse.
If you're on the equator, the column of Earth under your arse is longer, and therefore more massive, than it would be if you were sitting on the pole. Therefore, the gravitational attractive force you experience at the equator is greater than what you'd experience at the pole, simply because there's more Earth right under you, and is enough to exactly balance your increased tendency to fly off the Equator at a tangent.
posted by flabdablet at 5:46 AM on December 12, 2007
For an intuitive understanding, consider that if you're sitting on the ground, there's a bunch of mass pulling you to the left that's cancelled by an equal amount pulling you to the right, and so on for any opposing pair of sideways directions. Your net downward acceleration is pretty much due to what's directly under your arse.
If you're on the equator, the column of Earth under your arse is longer, and therefore more massive, than it would be if you were sitting on the pole. Therefore, the gravitational attractive force you experience at the equator is greater than what you'd experience at the pole, simply because there's more Earth right under you, and is enough to exactly balance your increased tendency to fly off the Equator at a tangent.
posted by flabdablet at 5:46 AM on December 12, 2007
The fact that gravity is greater at the poles than at the equator is not a new discovery. It was measured by Jean Richer in 1671. He observed that the period of a precise pendulum decreased when he went from Paris (45 degrees north) the Cayenne Island (5% north). He didn't know the reason but Isaac Newton explain 15 years later that is was due to the decrease in gravity.
Gravity over the surface of the earth has been measured to great precision and is 9.83 m/s2 at the poles and 9.78 at the equator. This precision is used for everything from calculating the orbit of satellites, corrections for the GPS system and prospecting for oil.
There are two components of the gravity difference -- the fact that a person at the poles is 21 km closer to the center of the earth (the center of mass) and the centripetal force due to the rotation of the earth. A rough calculation of these two effects is shown here.
Yes, there is a formula that says that the acceleration due to gravity is inversely proportional to the square of the distance between the centres of mass, but this formula is really only applicable for well-separated masses. The formula for gravitational acceleration experienced by a small mass on the surface of a much larger one is a complicated solid integral.
That statement is not true. Every freshman physics student knows that it doesn't matter how far away an object is from the center of the earth, the formula is the same. You can do the complicated integral that you describe and what you find out is that through symmetrical cancellation, the gravity at the surface of a radially symmetric sphere is exactly the same as if all the mass were at the center of mass. The fact that the earth is not exactly a sphere changes this by only a tiny amount.
If the net inward force experienced by a surface-bound observer (i.e. the observer's weight) near the poles was in fact greater than that experienced by a surface-bound observer of equal mass at the equator, then there would exist a net flattening force on the Earth, and it would be becoming more disc-like over time. This isn't happening. Therefore, ignoring small local gravitational variations due to high-speed effects like continental drift, we can be pretty sure that a given observer's weight is pretty much the same everywhere.
The oblate earth forms an equipotential surface. This means that the gravity vector is everywhere perpendicular to the surface. We know this is true or else the water in the oceans would flow toward the non-perpendicular vector. This equipotential surface should not be confused with equal gravity force. The two are not the same. Gravity is not the same over the equipotential surface. The fact that the earth is oblate is in fact proof that gravity is not the same at every point. Gravity can only be be the same over an equipotential surface if it is a sphere.
Also keep in mind the Principle of Equivalence described by Einstein. The gravity which we measure at the surface of the earth is a combination of the attraction due to mass and the centripetal force of acceleration due to the earth's rotation. You really can't separate the two and they are indistinguishable.
posted by JackFlash at 10:13 AM on December 12, 2007
Gravity over the surface of the earth has been measured to great precision and is 9.83 m/s2 at the poles and 9.78 at the equator. This precision is used for everything from calculating the orbit of satellites, corrections for the GPS system and prospecting for oil.
There are two components of the gravity difference -- the fact that a person at the poles is 21 km closer to the center of the earth (the center of mass) and the centripetal force due to the rotation of the earth. A rough calculation of these two effects is shown here.
Yes, there is a formula that says that the acceleration due to gravity is inversely proportional to the square of the distance between the centres of mass, but this formula is really only applicable for well-separated masses. The formula for gravitational acceleration experienced by a small mass on the surface of a much larger one is a complicated solid integral.
That statement is not true. Every freshman physics student knows that it doesn't matter how far away an object is from the center of the earth, the formula is the same. You can do the complicated integral that you describe and what you find out is that through symmetrical cancellation, the gravity at the surface of a radially symmetric sphere is exactly the same as if all the mass were at the center of mass. The fact that the earth is not exactly a sphere changes this by only a tiny amount.
If the net inward force experienced by a surface-bound observer (i.e. the observer's weight) near the poles was in fact greater than that experienced by a surface-bound observer of equal mass at the equator, then there would exist a net flattening force on the Earth, and it would be becoming more disc-like over time. This isn't happening. Therefore, ignoring small local gravitational variations due to high-speed effects like continental drift, we can be pretty sure that a given observer's weight is pretty much the same everywhere.
The oblate earth forms an equipotential surface. This means that the gravity vector is everywhere perpendicular to the surface. We know this is true or else the water in the oceans would flow toward the non-perpendicular vector. This equipotential surface should not be confused with equal gravity force. The two are not the same. Gravity is not the same over the equipotential surface. The fact that the earth is oblate is in fact proof that gravity is not the same at every point. Gravity can only be be the same over an equipotential surface if it is a sphere.
Also keep in mind the Principle of Equivalence described by Einstein. The gravity which we measure at the surface of the earth is a combination of the attraction due to mass and the centripetal force of acceleration due to the earth's rotation. You really can't separate the two and they are indistinguishable.
posted by JackFlash at 10:13 AM on December 12, 2007
For an intuitive understanding, consider that if you're sitting on the ground, there's a bunch of mass pulling you to the left that's cancelled by an equal amount pulling you to the right, and so on for any opposing pair of sideways directions. Your net downward acceleration is pretty much due to what's directly under your arse.
Thats not true at all. All the mass pulling you "to the left" is actually pulling you down and left which is two separate vectors. That is, there's no "cancelling" involved here at all. (Especially so when talking about gravitational potentials, which are purely additive.)
I don't know enough about geodetics to know how g varies across the surface of the Earth but I did want to point out that the explanations above are misleading or outright incorrect.
posted by vacapinta at 2:23 PM on December 12, 2007
Thats not true at all. All the mass pulling you "to the left" is actually pulling you down and left which is two separate vectors. That is, there's no "cancelling" involved here at all. (Especially so when talking about gravitational potentials, which are purely additive.)
I don't know enough about geodetics to know how g varies across the surface of the Earth but I did want to point out that the explanations above are misleading or outright incorrect.
posted by vacapinta at 2:23 PM on December 12, 2007
Huh. Next you'll be telling me I'm wrong about the figure-8 thing, too.
posted by flabdablet at 5:11 PM on December 12, 2007
posted by flabdablet at 5:11 PM on December 12, 2007
I do feel bound to point out, though, that there's an apparent contradiction between
Every freshman physics student knows that it doesn't matter how far away an object is from the center of the earth, the formula is the same.
and
The oblate earth forms an equipotential surface. This means that the gravity vector is everywhere perpendicular to the surface.
since the second statement implies that over most of the Earth's surface, the gravity vector does not point exactly toward the centre of the Earth.
posted by flabdablet at 5:20 PM on December 12, 2007
Every freshman physics student knows that it doesn't matter how far away an object is from the center of the earth, the formula is the same.
and
The oblate earth forms an equipotential surface. This means that the gravity vector is everywhere perpendicular to the surface.
since the second statement implies that over most of the Earth's surface, the gravity vector does not point exactly toward the centre of the Earth.
posted by flabdablet at 5:20 PM on December 12, 2007
Flabdablet, the component of gravity due to attraction of masses always points to the center of the earth no matter how close or far away you are. The centripetal component of gravity due to rotation causes the gravity vector sum to depart from the center of the earth, forming the ellipsoidal equipotential surface. There is no contradiction. They are two different things. The first is only one component and the second is a vector sum.
posted by JackFlash at 5:37 PM on December 12, 2007
posted by JackFlash at 5:37 PM on December 12, 2007
So how does that square with
The gravity which we measure at the surface of the earth is a combination of the attraction due to mass and the centripetal force of acceleration due to the earth's rotation. You really can't separate the two and they are indistinguishable.
I mean, you're undoubtedly correct overall, but you do keep making contradictory-sounding statements.
Can I also pick a little nit? For an observer on the equator, the component of force due to the Earth's rotation is not centripetal (centre-seeking), but centrifugal (centre-fleeing). If the force due to rotation were in fact centre-seeking, then it would add to the attraction due to mass and the observer would weigh more at the equator.
Yes, I understand that every freshman physics student has it pounded into them that there's no such thing as centrifugal force; but for an observer whose most natural frame of reference is a rotating one, it's quite a natural concept to work with, as is Coriolis force.
From a frame of reference that doesn't rotate with the Earth, the only centripetal force identifiable is the attraction due to mass. At the poles, all of that manifests as weight; at the equator, only most of it manifests as weight, the remainder causing the centripetal acceleration required to make the surface follow its circular path. Centripetal force is not due to the acceleration implied by a circular path - it is required to cause that acceleration.
I'm still quite fond of my stabilized liquid-drop argument, though as you've quite correctly pointed out, the stability really does only depend on there being no net sideways force at any point on the surface, and this doesn't necessarily say anything about the magnitude of the net downward force at various points.
Thanks also for the reminder that the Earth is only oblate by 20km or so. Given its 13000km diameter, that makes it pretty clear that nobody will go too far wrong by modelling it as a perfect sphere.
In other news: did you know that a new Sun goes over the sky every day? They pile up behind that range of mountains over there. Every now and again people go back there and cut them up into blocks, and that's where we get margarine from.
posted by flabdablet at 6:50 PM on December 12, 2007
The gravity which we measure at the surface of the earth is a combination of the attraction due to mass and the centripetal force of acceleration due to the earth's rotation. You really can't separate the two and they are indistinguishable.
I mean, you're undoubtedly correct overall, but you do keep making contradictory-sounding statements.
Can I also pick a little nit? For an observer on the equator, the component of force due to the Earth's rotation is not centripetal (centre-seeking), but centrifugal (centre-fleeing). If the force due to rotation were in fact centre-seeking, then it would add to the attraction due to mass and the observer would weigh more at the equator.
Yes, I understand that every freshman physics student has it pounded into them that there's no such thing as centrifugal force; but for an observer whose most natural frame of reference is a rotating one, it's quite a natural concept to work with, as is Coriolis force.
From a frame of reference that doesn't rotate with the Earth, the only centripetal force identifiable is the attraction due to mass. At the poles, all of that manifests as weight; at the equator, only most of it manifests as weight, the remainder causing the centripetal acceleration required to make the surface follow its circular path. Centripetal force is not due to the acceleration implied by a circular path - it is required to cause that acceleration.
I'm still quite fond of my stabilized liquid-drop argument, though as you've quite correctly pointed out, the stability really does only depend on there being no net sideways force at any point on the surface, and this doesn't necessarily say anything about the magnitude of the net downward force at various points.
Thanks also for the reminder that the Earth is only oblate by 20km or so. Given its 13000km diameter, that makes it pretty clear that nobody will go too far wrong by modelling it as a perfect sphere.
In other news: did you know that a new Sun goes over the sky every day? They pile up behind that range of mountains over there. Every now and again people go back there and cut them up into blocks, and that's where we get margarine from.
posted by flabdablet at 6:50 PM on December 12, 2007
Response by poster: BTW, vacapinta, on reflection your great circle argument is based on a flawed premise. The terminator is not a great circle, as the earth is not a sphere.
posted by tkolar at 3:29 PM on December 22, 2007
posted by tkolar at 3:29 PM on December 22, 2007
It doesn't really matter. The argument applies to any ellipsoid, not just a sphere.
The only requirement is that the terminator "slices" the Earth in half - and it does slice it into equal night and day. Also the equator divides the Earth (or any ellipsoid) in half into a north and south hemisphere, and so their intersection line still goes through the center of the Earth. The argument is a bit more subtle but it still holds.
posted by vacapinta at 4:32 PM on December 22, 2007
The only requirement is that the terminator "slices" the Earth in half - and it does slice it into equal night and day. Also the equator divides the Earth (or any ellipsoid) in half into a north and south hemisphere, and so their intersection line still goes through the center of the Earth. The argument is a bit more subtle but it still holds.
posted by vacapinta at 4:32 PM on December 22, 2007
This thread is closed to new comments.
At the equator, the pendulum's plane of oscillation will never move. (Whereas it will rotate 360° at the North Pole over a 24 hour period, 123° at 20° north, 180° at 30° north, etc.)
(Sorry, no ideas vis a vis your primary question.)
posted by hjo3 at 7:07 PM on December 10, 2007