Here’s a compilation of arguments for a flat earth from 1885. I laughed, but this one demands a response.
Staunch flat-earther Wilbur Glenn Voliva (1870-1942) asked: “Where is the man who believes he can jump into the air, remaining off the earth one second, and come down to earth 193.7 miles from where he jumped up?” Hard to argue with that.
Wait, but I can argue with that! There’s something wrong with his calculations: I estimate that if I were stationary relative to the rotating earth for one second, I’d only travel about a quarter mile to the west. Disappointing.
Also, if the earth were spinning at 193 miles per second, a day would only be 2 minutes long.
There might also be something else wrong with his scenario, but who cares if he can’t even get the basic math right?
Have you accounted for the orbit as well? And the solar system’s movement through the galaxy?
“the sun, and you and me, and all the stars that we can see, are moving at a million miles a day…in an outer spiral arm at 40 thousand miles an hour…”
That Newton was such a materialist. Expecting the earth to keep moving on its momentum, and even humans to do so–when they’re not natural (as we know from the IDiots).
Of course that’s where religion really lost its pretense at meaning, not with Darwin. Darwin just extended Newtonian materialism to explain the regularities–and modifications–of life.
Glen Davidson
Voliva is clearly not a man to leap out of his seat on a train.
The complete list of one hundred “proofs” is here. It starts pretty well:
which was confirmed by multiple observations during the Twentieth Century. Right?
I can hardly imagine the degree to which Glen’s remark would have horrified Newton, but the crazy old sorcerer would have had his nose too deep into his wacky analyses of the Book of Revelation to take notice or offense.
Re: Voliva – if you want to look for it, Martin Gardner’s Fads and Fallacies in the Name of Science has an absolutely brutal chapter on Voliva that details both the truckload of pseudoscientific and religious malarkey he expounded on and his sectarian control over the town of Zion (an exurb of Chicago near the Wisconsin border) back when he was alive and (not particularly) well (in the head).
Plus, the number would vary by latitude, if one could somehow negate the Earth’s spin. At the pole, the spin just makes you change orientation relative to the far field; at the equator, there is also a translational component.
But you do get this question regarding… well, usually helicopters and planes a lot on Ask an Astronomer (where I answer questions). People accept that the Earth is spinning, but don’t seem to grok that it means everything on it spins and that doesn’t go away when you lose physical contact with the ground.
(To be fair, once you tell them that, they get it.)
Yes, he kinda forgot a little something called inertia.
Then again, seems he didn’t like Newton anyway:
Hold on now. If you are standing at one of the poles, jump, and remain stationary relative to the earth, you’d land exactly where you started. Thus proving that the earth is indeed flat.
*cough* Inertia. *cough*
So Voliva made a major mathematical error? Well, that figures…
…I’ll get my coat…
I don’t know about the Earth being flat, but it must be immovable in the center of the universe.
This was proven by two famous scientists, Michaelson and Morley, in 1887. They were attempting to calculate the absolute speed of the earth relative to the luminiferous æther. They sent beams of light down and back in two tubes fixed at 90° to each other. It was assumed that, if æther existed, the light beam would be slower in one direction, giving the physicists a tangible “absolute” speed of the earth. Their calculated speed: Zero.
They were baffled by this, wondering how the Earth could be sitting in one spot, while every aspect of the conventional “wisdom” indicated that the planet must be orbiting its own sun, and therefore must be moving with an orbital velocity. Moving quickly to avoid having to admit that they were wrong, they were able to instead “infer” from their results that the æther must not exist, and that light must propagate through no medium at all (impossible for a wave by the very definition of a wave). Their inference was generally accepted by the scientific community and the “ridiculous” notion of æther was thrown out.
But light waves would still require a medium for transmission, and the actual purpose of the experiment was to determine the existence of that medium. The results speak for themselves: the Earth does not move. And even if the Earth did, the problems inherent in keeping it moving through the æther are overwhelmingly supportive of geocentric theory.
The stupid, it burns when I pee… oh, wait, ummm… :)
#13: You are being satirical, I hope. If not, attend some basic physics classes before making assertions like “waves would still require a medium for transmission”. They do not. And light is not really a wave, either…
We all know the earth is flat. As for proof that the earth is round from the moon shots, there is a simple explanation:
The sun is not a glowing sphere, it’s a flashlight. When viewing earth from a rocket ship above the plain (it cannot be in orbit), it illuminates a circle of land. That should be apparent to everyone.
Also, the poles only appear to be smaller, but really, when it’s cold, people walk faster, so they’ve measured the distance wrong.
-ap
PS: ~1530 feet/sec at the equator. Multiply by cos(Longitude) to adjust for other locations on earth. At 35 degrees N (reasonable average over US), that would be about 1250 ft/sec.
@Randomfactor #2:
Pray that there’s intelligent life somewhere up in space, ’cause there’s bugger all down here on Earth.
—
Stan
Minor Cephalopod,
‘Tis has his tongue firmly planted in his cheek.
Andrew,
you should multiply by the cosine of the latitude for meaningful results.
Well, Dhorvath, you can hardly blame Cephalopod for not knowing. You know how economists are. Almost as bad as engineers…
*ducks and runs out of the room giggling*
MinorCephalopod
I used to teach celestial navigation. As part of the introduction to the first class, I’d tell the students: “Forget Copernicus, for the purposes of navigation we will assume the Earth is fixed in the center of the universe and the sky rotates around us.” It makes the math a whole lot simpler.
“Here’s a compilation of arguments for a flat earth from 1885.”
Most of these read like the witch-burning argument in “The Holy Grail.”
Oh, I wasn’t blaming anyone, I just wanted to allay any fears.
We’re not that bad. At least I hope we’re not that bad. Are we that bad?
<sobs>
If you are standing on Niven’s Ringworld, your rotational velocity is 770 miles per second; jumping up and down in one spot there would not be for the faint of heart.
Awwwww. I was just teasing! I’m sure you’re just fine… You know, mostly… *pat pat*
Fixed.
I’m not an expert or anything on physics or gravity, so if anyone here knows, I’m curious about something. Would we feel gravity in the same way on a disk with the same mass as the Earth, assuming such a structure could exist?
Oh, please let Robert Sungenis hook up with these guys.
If you were on the rim of the disk you’d feel the same gravity (more or less…mainly less, but not a lot less). If you were at the center of the disk there’d be less gravitational attraction.
I think.
RahXephon @#28
Good point. I would think that the answer would be “no” unless there’s some freaky distribution of mass on the bottom of the plane. On a more or less spherical surface, wherever you are you’re subject to roughly the same gravitational forces. On a flat disc with the mass distributed equally, you might feel fine at the dead center. As you move away from the center say north-bound, you’ve got increasingly more mass to your south than to your north.
For that matter, the whole thing seems to be inherently unstable. Of course, if you’re in the “God did it” camp, no objections matter. Your feet are held down by the weight of your sins!
@28: In the center of the disk, it would probably feel much the same, but the closer you got to the rim of the disc, the more it would feel as if you were going up-hill, as more and more of the disc would be behind you, pulling you backwards. By the time you made it to the very edge of the disc, it would feel as if the surface were almost perfectly vertical.
@32
Ah, I see. That’s kind of what I was thinking. Would the distribution of mass matter to that? For instance, if the disk got thicker approaching the rim, would that counter the pull of the rest of the disk?
I obviously find this topic fascinating because of Terry Pratchell, by the way.
Isaac Asimov has a good essay on this type of subject, The Relativity of Wrong.
http://hermiene.net/essays-trans/relativity_of_wrong.html
No. The gravitational attraction would be toward the centre of the disk-Earth’s mass, so if you’re not directly over the centre, there will be a force pulling you “sideways” toward the disk’s centre. The contra-argument is that the disk is infinite, and has no definable “centre”.
(Don’t ask “So how does the Sun go around an infinitely wide Earth?” Everybody knows there are holes for the Sun to pass through.)
You’d feel less, because a disc is assumed to be thin. There just wouldn’t be that much mass near you pulling you down, unlike with a sphere (or “oblate spheroid” for the pedants).
In the center, you’d just feel a force downward. When off-center, you’d be pulled toward the center, and would have to walk tilted away from the center to keep your balance.
Glen Davidson
So I was wrong.
Would anyone like an analysis of what’s going to happen in Wall Street in the short term future?
This is almost as bas as when creationists bring up the 2nd law of thermodynamics and try to argue about increasing order on earth! I’m with Stan the sooner we resume space travel the greater our chances of encountering intelligent life forms in our galaxy.
“If she weighs the same as a duck, logically shes made of wood and therefore a WITCH (burn her!)
I propose a new superhero: Relativity Man. He can jump long distances and travel quickly by changing which object his motion is relative to.
It would at least partly compensate. But then you might feel rather heavy at the rim, at least if it’s very thick.
The point of gravity its force is inversely proportionate to the square of the distance between the two masses (you and earth in this case). A disc pulls you less because the mass is spread out over a much larger distance–at least that’s part of the explanation.
Glen Davidson
@39: Of course he’d have to choose his objects very carefully :)
(I’m picturing him thinking of the Sun at the wrong time, and either being thrown into space or smashed down into the ground )
@28 RahXephon
I would guess you’d feel pulled toward the center of the disk. So, a little down and a lot sideways?
@37
Yes.
sk
Wait, a flat Earth doesn’t spin?
Guess that must mean there’s a conspiracy by the Foucault Pendulumists.
“But, since all observation fails to find any evidence of this want of parallelism which theory demands, the idea must be renounced as being absurd and in opposition to all well-known facts”
I remember watching the construction of the Verrazano Narrows Bridge from the windows of my house in NY, and one of the wonders of that project was the centerline of the tops of the towers had to be 1 5/8″ farther apart than the bases. cool!
skmarshall #43
After one of the most volatile weeks in its history, the Dow Jones Industrial Average is back where it started. The Dow jumped 213.88 points to 11482.90 on Monday, nudging it above the levels last seen Aug. 5—just before Standard & Poor’s downgraded the U.S. credit rating. After three consecutive days of gains, the Dow is up 7.1%, its biggest such move since March 2009, when markets were pulling out of their financial-crisis nadir. Those gains helped recover the steep losses incurred earlier in the week on the back of S&P’s decision and worries about Europe’s continued debt.
It appears Wall Street is bullish right now. How long that’ll stay is rather iffy.
Tis Himself, where did you teach celnav? I’m curious because I’ve taken some/will be taking more classes in the near future. Also, what would you recommend as useful beyond azimuths, amplitudes, sun lines, and LAN’s?
Where are the gravity magnets on a flat earth?
@ ‘Tis —
All I know is that I was born on an island that passes roughly under Sirius. Just give me the appropriate ocean, let me know whether I’m to the east or west of my target, and maybe add a couple of exclusively land-nesting birds on my vessel — I’ll find my way home!
Oi!!! At least engineering works…
On the question of the disc shaped earth (assuming that the disc was relatively large compared to you and of fairly even thickness), you would feel relatively little gravity at the centre as most of the force on you would be applied laterally and equally in all directions. In much the same way that the apparent gravity at the centre of the earth is 0.
Anyone care to apply Gauss’ law for a finite thin disk? Based on my experience with the math of electricity and magnetism, in an infinite flat, thin disk, then the strength of the force you feel would be inversely proportional to the height you are in from the plane.
In the case of the disk, at very close to the center, you can approximate the field with the math of infinite thin plane. When very far away, you can treat the disk as a point and use the inverse squared of the distance relationship. What I don’t know is how the fields act when one is very close to the edge of the disk. But I speculate it is just as Clark Cox says.
Correction to my post above: In an infinite flat plane, the gravity one feels doesn’t change at all no matter how high you go. In fact, that is what you assume when you do a potential energy problem for objects near the Earth’s surface. Think about it, in reality gravity gets weaker as you get farther away from the Earth, but the Earth is so huge that you can treat potential energy as m*g*h instead of m*M*G/r when very close to the surface. That is, you approximate it as if you are in an infinite flat surface.
Hopefully you’re not purporting to be an engineer. There is still the bulk of the earth sitting below you, pulling you down. (And it would be closer to you than at any other point, so you weigh more.)
I think you’re confusing the situation with being at the centre of a spherical earth. (What an amusing concept. Spherical earth indeed.)
@53 WhiteHatLurker
I’m pretty sure that Ramel @50 got it right with respect to gravity at the center of the disc. The vectors for the vast majority of the disc’s mass point outward away from the center, leaving only a small ‘downward’ force. A sphere, on the other hand, acts as a point mass.
@33 RahXephon and @40 Glen Davidson
I’m pretty sure that excess mass at the edge of the disc would not compensate for the inwards gravitational force as all of this mass (being ‘outside’ of you) cancels itself out. See “Shell Theorem” on Wikipedia; there is no net gravitational attraction on something inside an evenly distributed shell of mass.
Perhaps rotating the disc could counter some of the inwards gravity with ‘centrifugal’ acceleration?
Ok, I may not be explaining myself clearly because I’m ill and it’s 4:30 am here.
The question specifies
So the question is how does the magnitude of the force that is holding you to the surface compare between a disc and a sphere of the same mass.
So what we are interested in is the force acting on our body in the direction of the centre of mass.
When we say disc it is generally assumed that we are describing an object where the diameter is constant in all directions and is significantly greater than thickness.
If you are standing at the centre of the disc the bulk of the material is slightly below and rim-wards of your location, and so the force it applies to you acts down and rim-wards. The rim-wards component of the force is countered by an equal force from the opposite side of the rim, so the resultant lateral is 0 leaving us with just the downward component.
The resultant force (R) applied to your mass (m) by a small element of the earth’s mass (dm) is therefore equal to:
R=(G(dmxm)/r^2)sinθ
Where r is the distance between dm and your centre of mass, and θ is the angle between r and the disc’s centre of mass.
You would have to integrate that across the volume of the disc to get the total force acting on you, but that’s a total bastard that I’m not going to manage while I’m this sleep deprived…
So the end result is similar to being slightly away from the centre mass in a solid spherical body.
http://xkcd.com/54/
@53 and @54
If the mass of the earth were distributed uniformly into a spherical shell (hollow earth) then anywhere inside the sphere all the gravity vectors would cancel out… you would float. You don’t have to be at the center. Anywhere outside the sphere and you couldn’t tell the difference between a hollow earth and a solid earth. Gravitationally that is. One could always tap on the surface and see if it sounds hollow ;-)
@54 Snap et al.,
A uniform spherical shell obeys the Shell Theorem, but a uniform circular ring does not. Not even for observation points in the plane of the ring. (I made this mistake during an oral exam for an undergrad lab class and never forgot it.)
The gravitational field near a thin uniform-density disk is going to look similar to the electric field near a uniformly charged disk… which is to say, similar to half of a parallel-plate capacitor. Fairly uniform near the center of the disk, with edge effects becoming more important near the rim. Mostly perpendicular to the surface of the disk, but not exactly. (For conductors, the electrostatic field is perpendicular, but then again a conducting disk wouldn’t be uniformly charged…)
Answering how the field strength compares to an equal-mass sphere, and strength at the rim versus the center, are going to have to wait until sleep refreshes the brain.
@ouigui
How does one apply Gauss’ law to a finite disk anyways? It is pretty easy on an infinite flat plane using cylinder. But just like the disk in E&M, I run into complications.
#45 skmarshall
In 1870, the “evolutionist”, Alfred Russel Wallace, used a telescope to sight along the Bedford Canal to prove the curvature of the Earth (and to win a 500 pound bet put up by flat-earther, John Hampden). Hampden refused to pay up, and it took 15 years of wrangling before the courts decided such bets weren’t legal. Wallace was so harassed by Hampden that he regretted ever getting involved. http://lclane2.net/bedford.html
Those religious types who think their beliefs trump the data are the Red Queens of our world.
“Alice laughed: “There’s no use trying,” she said; “one can’t believe impossible things.”
“I daresay you haven’t had much practice,” said the Queen. “When I was younger, I always did it for half an hour a day. Why, sometimes I’ve believed as many as six impossible things before breakfast.”
For Discworld, you have to take into account Atuin, the great turtle that moves through space, carrying the 4 elephants that carry the disc. The turtle obviously has the most mass. Given that the 5th Elephant hit the Disc and became mountains, their mass is hard to estimate; not so great as to break the Disc on impact, and yet, together, enough to support the mass of the Disc.
I suspect if you threw something up in the air on the Rim, it would fall to the ground slightly Hubwards. But because Atuin is not only more massive but extends out farther than the Disc, down would still seem pretty much the same wherever you were on the Disc. Interestingly, because Atuin is asymetrical, the amount of difference would vary depending what part of the Rim you were on.
Does this sound correct?
That’s all well and good, PZ, but where’s your rebuttal to TimeCube?
Forget the calculations! How can he think that the surface of the earth is rotating along with the rest of the earth, and not realize that the atmosphere and objects near the surface are partaking of the same, or nearly the same, momentum?!?
Lady Vimes is partly right about Discworld mechanics. The Discworld is not very thick, so a large proportion of its gravity must come from Atuin. Although Atuin is not 360 degrees symmetrical, for all practical purposes it could probably be taken as a sphere located some distance below the centre of the disc. There is also the mass of Cori Celeste to take into account – a mile-high mountain at the hub must have some effect.
The Discworld rotates, so one must assume that centripetal force helps to counter the pull towards the centre. Also many flat earth scenarios envisage a shallow upturned saucer rather than really flat, which also helps.
The rotation of the discworld is at an unknown speed – the day is caused by the sun orbiting the assembly, but this orbit also rotates, so the pair could be set to allow the disc to rotate at whatever speed is need to achieve the desired balance.
OT – have the deep dwarves ever broken through the bottom?
Ouch. But I giggled.
OT (sorry, it’s a habit):
Teh gheys’ unatural…
You guys are silly. The Earth is, of course, perfectly flat!
… in a spherical coordinate system. :D
And yet almost half of all Americans believe the Earth is flat!
(Yes, yes, I know the actual belief held by 45% of Americans is that the Earth is only 6000 years old. This is just as insane a belief as that of the Earth being flat. I am merely highlighting the ignorance of the average American. I seriously don’t understand why they get given a pass for their insane beliefs, yet flat earthers are ridiculed.)
off topic poll alert http://www.gympietimes.com.au/story/2011/08/05/about-500-chaplains-have-this-week-issued-a-warnin/
Should chaplains be allowed in Queensland schools?
[OT]
DanDare, it is off-topic, but worthwhile.
I note the disclaimer therein, too:
I have been following this story for awhile on pharyngula. I think this poll needs pharyngulated!
The Bill O’Reilly of the 19th century.
WG Voliva “The earth is flat.”
PZ Myers “Get a pump.”
I don’t know about the Earth being flat, but it must be immovable in the center of the universe.
Obviously false. New York City is the center of the universe. Duh!
Actually, if I understand cosmology correctly (and there’s a good chance I do not), every point in the universe is expanding away from every other point equally so ever point is essentially the “center”. So the earth is as much the center as anywhere else.
The Disc takes 800 days to rotate, although there are two of each seasons per year, so for agricultural purposes, a year is treated as being 400 days.
http://en.wikipedia.org/wiki/Discworld_(world)#Calendar
I wasn’t aware the sun’s orbit also changed.
Reading that list of “proofs”… Am I the only one who kept mentally replacing the final phrase of each with, “… So I can clearly not choose the wine in front of you/me…”
It’s both round and flat. Like a pizza.
Back, and to the left….
My IT degree came from the university’s College of Engineering. I’m not sure if that makes me an engineer. I’m pretty sure the rest of the engineers on campus don’t think of us as engineers, so I guess I must be alright.
Has anyone else noticed that the comments indicator isn’t working correctly?
On the main site I’m seeing “7 comments” on this thread, but when I actually go and look there are 79. I’d think it were just omitting the second number, except the last thread shows as “3 comments” and actually has 28. Maybe it’s rounding even numbers up, odd down, and then truncating the tens place? I don’t know. Anyways, something needs fixing.
Also, I am very disappointed that the new server has pop-up ads seemingly all over the place. They definitely come up every time I click on a comments thread.
Also, PZ, if you get around to reading this – maybe you could say something on the main site to the effect of “Hey, go check the other site because it finally works”. This site had been showing the same outdated posts for probably the better part of two weeks, then there was a weekend of absolutely no activity in either location, then all of a sudden this blog starts up again Monday without any indication that it’s even working correctly now on the old blog.
I will note that after going back to the main site and refreshing the comments to see what my comment did to the current count, the “80” and “28” are now showing correctly.
I’m curious. For all of you who love solving these kinds of puzzles: What is the actual distance travelled by someone jumping up and landing one second later at the same spot (presuming they’re not standing on a train :snicker:), taking into account the earth’s spin, orbit, the galaxy’s spin etc.? In honour of Voliva, say that the jumper is in Zion, Illinois.
Zero. They come down again in the same spot. :)
But seriously, in which reference frame are you talking? There’s no privileged “stationary” one about which everything else rotates, and so you could get practically any answer you like by deciding on a suitable frame of reference.
@Snoof
You have to use the Universal Prime Meridian. (The rotational axis of Gallifrey at 3:45 am March 12th 40321AD)
The answer to his challenge would have been extremely simple even in his day. When you jump on the deck of a moving ship, where do you land?
#82 – first approximation would be the same exact distance travelled if they had not jumped at all. I suspect you want a specific number, however.
It all depends upon your frame of reference and the configuration of the earth, sun, milky way at the time of measurement. Imagine the milky way humming along, the sun whipped on a spiral arm in that direction, the earth coming around the sun in that same direction, and you on the earth doing the same. I think it will be night time, but my brain hurts right now. Assuming that maximal peak, here are the relevant data:
1. At earth’s equator: .46 km/sec
2. earth around sun: 30 km/s
3. sun around milky way: 216 km/s
4a. milky way relative to CMB: 552 km/s
4b. milky way relative to Great Attractor: 600 km/s
#1 is too small to worry about for our calculations, let’s ignore it. Also, the total velocity will be large, but not large enough to worry about relativistic velocity effects (adding these speeds should produce a good approximation).
So, speed in reference to the CMB: 798 km/s (552+216+30)
Speed relative to the Great Attractor: 846 km/s (600 + 216+30)
In the second it took you to jump up and land, you will have travelled somewhere around 800 kilometers. Better stick that landing, otherwise you might get really hurt.
PS: @19 Philipe Legge – “latitude” – facepalm. Thanks.
On your arse?
Snoof:
Nuh-uh. The great scientist James Michener proved all you have to do is add up the speed of every frame of reference:
(From Space, by James Michener).
As this argument dates, according to our esteemed host, from 1885, we might want to cut Wilbur Glenn Voliva a slight bit of slack for having been only 14 or 15 at the time.
Not enough slack to keep him from sitting in the corner with a conical paper hat for an afternoon; not even enough to prevent his ridiculing 126 years later; but at least a few of us might refrain from posting ‘shopped pictures of him eating a foot-long corndog.
Gravity shmavity…
The Big Questions™ you’re all avoiding are: if the Earth is a disk, is it compact or vinyl? And if it’s the latter, is the Celestial Scenester who buys it wearing Wayfarers or Aviators?
Skepgineer, rusty knife of a thousand porcupines says:
16 August 2011 at 1:55 am
You can only debunk something that’s wrong.
TimeCube is not even wrong.
we’d be rather disconcerted as we were immediately launched into space.
Glen Davidson
Wikipedia is helpful as always:
Where would I have been without the informative “were incorrect”?
Glen Davidson
May I suggest for those interested in the history of this subject to consider reading “The First Copernican”, about Georg Joachim Rheticus.
http://www.amazon.com/First-Copernican-Joachim-Rheticus-Revolution/dp/0802715303
I found it interesting, how close theo world came to having a much later effective exponent of heliocentricity.
Only tangentially (heh) related, but it seems like the place to dump an applied math challenge:
Two point masses (i.e., of negligible size), each of mass m, lie a distance
d apart in deep space (that is to say, assume all external gravitational effects on the system are negligible). The two objects are stationary with respect to each other at time t = 0.
Can you set up and solve a differential equation that will express the amount of time t, in terms of d and m (along with any associated constants), that it will take for these two objects to collide, under no forces other than than their own mutual gravitational attraction?
On my feet!
Voliva made sure that the “truth” was taught in his town:
http://www.theologyweb.com/campus/showthread.php?82998-The-World-was-flat-in-Zion-Illinois
OK, I don’t see any way of checking whether this information is true, but I can say that it rings true. The Wedge strategy is similar:
So why wouldn’t Voliva do the same at a time when he could do so?
Glen Davidson
It’s funny that every argument against science that creationists throw out is satisfactorily explained by the Flying Spaghetti Monster…
If flat Earth, then gravity cannot exist, otherwise it would necessarily be spherical. Thus, since we assume the Earth is flat, we must figure out why we keep falling down on it. And the answer is obvious: the Flying Spaghetti Monster, his saucy Noodliness, Touches all of us to push us down and keep us connected to terra firma.
Ramen.
BZZZT ! wrong.
The extraneous ‘than’ in my #95 is provided free of charge.
Freerefill:
Well, no. I have read *ahem* modern flat-earthers* (I’ll look for a link when I’m not so rushed) take of gravity– it is illogical to think that someone could fly off into space (or smack their head on the dome of the sky) without an outside force. Gravity, in other words, is unnecessary.
Think about it this way: If you set a teacup on a table, it stays on the table on its own, right? There you go, no other explanation needed!
The stupidity is mind boggling.
*And I will admit that it is damn near impossible to tell a serious flat-eather from someone who is toying with Poe’s Law.
But the earth IS flat. If you live in Kansas.
http://improbable.com/airchives/paperair/volume9/v9i3/kansas.html
Gravity would exist with a flat earth, but the strength of the gravitational attraction on the surface is going to depend on the mass beneath the surface.
Any analysis of the Discworld’s gravity has to take the magical field into account, I think: we know that the ocean extends to the edge of the disk, and water (and fish, and anything else in or on the water) falls off the edge of the Disc and is not seen to return. (Remember that Rincewind was lowered over the edge in one of the early books.) The simplest hypothesis is that at least the seawater is being returned to the oceans, magically, possibly in a way that is fatal to ships and their crews.
That isn’t explicable just by noting that Great A’Tuin is much more massive than the Disc and the elephants: that would give us water gradually collecting all around the turtle. It’s not described as there, as far as I recall, but that it isn’t mentioned doesn’t prove it isn’t there. But either the water is magically returning to the oceans, or new seawater is continually produced in appropriate quantities.
these two “parallel” blogs are a little confusing I have not figured out how to check for the latest post yet so I do not know when I am reading what where.
the question of jumping and moving that was the starting point reminds me of a short story by Asimov in which the scientist inventor was forced to make a device that would make something stop all its motion. He did it and demonstrated it by banking a billiard ball into the field it generated at a particular time and sending the ball through the pool table the “bad guy” the wall and the earth at such a tremendous speed that it seemed to be instantaneous. very funny surprising ending
uncle frogy
Ah, applied maths. Engineering for people that wouldn’t know which end of spanner to lick, let alone how to establish a technological civilisation that could make one.
And economics? I suppose it is showing some faint signs of approaching ‘almost science’ status after decades of being nothing better than homeopathy. One day it might even be usable for something better than TV blather-fests.
Hopping while on a train! The newest adrenaline rush!
Yep, pretty damned flat.
I get so sick of that trope.
re 105:
As I recall he was challenged to reduce the mass of an object to zero. The demonstration was on a pool table, shooting a billiard ball into the mass reduction field. It was this that demonstrated that all objects with zero mass move at lightspeed. The billiard ball rolled into the field “aimed” at the man who was killed by it.
@PZMeyers, you wrote:
” I’d only travel about a quarter mile to the west.”
Actually, you’d have traveled 2/10th of a mile to the EAST!
Best regards
How do you figure that? That would mean going faster than the Earth’s rotation, rather than slower.
Naked Bunny with a whip:
More exciting than planking!
tim Rowledge #106
Don’t forget to collect your decaying porcupine on your way out, timmy. You know what to do with it.
Come on, both applied mathematics and economics are far more than just homeopathy. I suspect that tim Rowledge is ignorant of both.
captainahags #47
I taught navigation, celestial, coastal and electronic, at the Annapolis Yacht Club, Annapolis, MD. The odd thing was the large number of naval officers who belonged to the yacht club but had forgotten the celestial navigation they were taught at
Canoe Uthe Academy.If you’re really going to use celestial navigation, the most important thing to study is a star finder. It doesn’t do much good to be able to take and calculate star shots if you’re not sure which star you’re sighting. How to use an almanac and how to do the math are simple and straightforward. Being able to distinguish between Mirfak (α Persei) and Aldebaran (α Tauri) is more difficult (they’re both near the Pleiades and can easily be confused for each other).
I used to know someone who taught navigation to people with yachts.
One of his little tricks was if he thought they were relying to much on the electronic navigation aids he would slip below and pull out the fuses in the box. He would then ask them to point to their position on chart. Said far too many did not have a clue.
Spin @ the North Pole = 0 kph
People are so obsessed with building machines to set speed records. Relax dude! You’re already going 67,000 mph!
Maybe some kind of rocket scientist type can answer this question since it is, well, rocket science related: when do the inertia effects of the rotation of the earth wearing off become apparent? Once you leave the atmosphere or after a certain amount of time or what? Or maybe to be more clear – when I launch my space ship, when (if ever, I guess) would I notice the earth spinning at a different speed? Or if we are both in a vacuum, maybe I need to actively change my speed, or maybe it has something to do with the radius of my orbit? What’s the story?
And as for the magic water on ringworld: http://www.youtube.com/watch?NR=1&v=y_tL5sN6-YM
Nope.
But gravity would be constant with height*.
* Assuming infinite Flat Earth.
If I’m to do the calculation for real, I’d refuse to do it outside of Jerusalem.
@sili
Do you know how to do the Gauss’ law calculation with a finite disk? I really am curious what would happen at the edge. Although there was someone who speculated up there a very plausible scenario.
“Astronomers tell us…of the Earth’s ‘rotundity…’”
I love that phrase. Henceforth, I shall refer to myself as a georotundist! What fool would question Earth’s rotundicity?
At 35786 km above the surface, the centrifugal force will cancel out the force of gravity. That’s why satellites there “hover” over the same spot without falling down to earth. Given humans aren’t particularly good at judging speed and distance in space, without your instruments, you wouldn’t notice a difference until way above 10 000 km. Using your instruments, you would most probably be able to measure deviations from the ‘no rotation’ condition just after the vibrations from your engine stop.
Matt Penfold #116
The navigator I’m most in awe of is William Bligh. After the mutiny on the Bounty, he and 18 crew were put in a 23 foot launch. Using just a sextant and a pocket watch, he successfully navigated 3,618 nautical miles (6,710 km) to Dili on Timor Island. He was less than five miles from his projected landfall when he reached Timor. TV Tropes calls that sort of thing a Moment of Awesome.
kantalope @ 118:
That all depends on several factors. For sake of simplicity assume a perfectly spherical earth (Not necessary but I had to say it :-)) and no air…
If you take off from the North or South Pole straight up you will also be spinning at the rate of 2pi radians every 24 hours. (Sorry TimKO spin is measured in radians/time or degrees/time not distance/time.) So you, in your space ship, will have the same spin as the earth and the earth will appear to be fixed but receding. The stars and everything else will appear to rotate about you and the earth.
If you take off, straight up, from the equator you will have a tangent velocity of about 1,000 MPH eastwards. For sake of discussion lets say you have a constant velocity on takeoff of 1,000 MPH. After one hour you will be 1,000 miles downrange and 1,000 miles high. or 1,000 miles X sqrt(2), “away” from where you started (about 1,414 miles). Of course , where you started has also moved along in about the same direction. But where you started doesn’t move linearly, it follows the surface of the earth because it IS the surface of the earth. After a while as you get further away you will notice it spinning at a rate of 2pi radians (or 360 degrees) about once every 24 hours.
Exactly when you begin to notice this depends on how good you are at observing.
The simple answer is the earth will appear to spin half as fast as the hour hand on a clock. So you may not notice it much at all.
If you were suspended on some sort of skyhook just above the surface of the earth and the hook was “fixed in space” the earth at the equator would appear to be passing by your position at about 1,000 MPH (pretty fast!) from west to east.
This is also why we try and launch rockets in to orbit as close to the equator as possible. We get a nice boost to the net velocity vector by doing so.
None of the above takes in to account the motion of the earth in its orbit about the sun.
So, I think David Utidjian just showed that Voliva was correct?
If I took off straight up from the equator in a vacuum as assumed – wouldn’t I still be an object in motion at 1000 mph because I had until recently been attached to the surface of the earth going 1000 mph and with nothing to counter that motion (in the vacuum of space) I would stay over the launch point the whole time? wouldn’t I? Then I am guessing that the only difference in orbital speed would be my radius as compared to the radius of the earth’s surface? Which would mean that the earth’s surface would only spin out from under me as a function of that radius?
The problem becomes more troublesome with the earth’s atmosphere because it too is spinning around at the same pace as the planet is spinning (otherwise there would be 1000 mph winds all along the equator) so I was figuring that the atmosphere would continue to push my space ship sideways until I cleared said atmosphere and only then would my final “lateral” velocity be attained and from there my space ship speed as related to ground speed begin to change?
But then maybe I’m missing something here.
Frank (or would you prefer (mr. mrs. ms?) monster?)’s answer seems more on point but I don’t think I understand the “the centrifugal force will cancel out the force of gravity.” But it looks like I would stay above my launch point at 35786 km and would orbit at a different speed if I was beyond that point because of the different orbital speeds (centrifugal or maybe centripetal – never can figure those out)while if I am below that point I would orbit at a different speed because gravity would be slinging me along?
or not. Stupid Liberal Arts don’t really seem to help with practical (or impractical) physics questions…
The answer is probably a French augmented sixth chord. Or maybe Sonata-Allegro form.
Fuck if I know. I studied music in college. That’s why I have to hang out here.
ibyea (#51) spake thusly:
I don’t know about anyone else, but I wouldn’t. The only cases I know of where Gauss’s Law can be used are a sphere, an infinite cylinder and a Bouguer plate. There may be others, but I don’t think a disk is one of them. The crucial point in these cases is that the symmetry of the object implies a symmetry of the gravitational field. Hence you can use the flux of the field across a suitably chosen surface (which is all Gauss’s Law tells you) to work out the field at a point.
There seems to be a computation of the gravitational field of a disk here, but I confess I haven’t read it.
The answer might be market externalities caused by outward shifts in labor supply curves or cyclical variations in cost markups.
@ibyea:
Rick’s got the crucial observation. Gauss’s Law is useful for field calculations when there’s a high degree of symmetry and when the surface is cleverly chosen. For the uniform-density disk, the gravitational field isn’t perpendicular to the surface except at the axis, so a flat Gaussian surface isn’t too helpful.
Of course – insert in economist with no sense of humour. Odd how engineers are supposed to put up with trashtalk but economists are too precious. But a decaying porcupine has value – an engineer could make use of the gaseous by-products to generate energy to sell back to the grid, the final product would be processable to compost and a video of the decomposition would be sellable on the New York art market as neo-abstract destructivist performance art. And the left over spines would nicely work for a variety of interesting purposes.
Really, are you claiming economics counts as any sort of science as yet? There have been a few interesting approaches to actually trying to understand real consumers etc but most of it is little more than theology. And yes, it may be a while back but I did actually study economics, with actual Real Economist teachers that had actual real experience advising governments etc. You should have heard what they said about the intellectual weakness of the field.
Is little timmy an engineer? Do you drive a train? Or are you one of those sanitation engineers who ride around in garbage trucks?
No, asshole, I was objecting to you calling economics “being nothing better than homeopathy.” Just because it’s too difficult for your garbage truck riding brain to comprehend doesn’t mean it’s a pseudo-science.
Yes, asshole timmy, I am. Just because you misremember the Econ 101 lectures you slept through years ago doesn’t mean it isn’t.
You not comprehending economics doesn’t mean shit. You’re putting up a weak strawman and pretending it’s the entire field. I hope you’re better at garbage truck riding then you are at understanding economics or else the streets will be strewn with trash.
Guess what, asshole, I am an actual Real Economist with actual real experience in government, not just an adviser.
I know economists trash talk other economists with whom they disagree, claiming they’re intellectually weak. I’ve trash-talked Austrian economists, the Chicago Boys, and freshwater economists in general. However, I’ve given reasons for my disdain. And I don’t disparage the entire field just because I disagree with certain practitioners.
It’s really simple, asshole timmy, you shot your fucking mouth off and I told you to fuck yourself. If you don’t want to be told to fuck yourself, don’t make stupid statements. This is something even a garbage truck riding engineer should be able to understand.
I can’t answer to Sonata-Allegro as opposed to labor supply curves, my first impulse was to blame the capitalists or maybe the bourgeoisie; then I considered that rockets are typically phallic so maybe it was some dead white guys problem, but there is always the consideration for what it meant that Voliva rejects the possibility that the earth could leave him behind.
The mathy part I was trying to remember was this equation: A=rc which should have been easier to remember…A = Arc Length, r = radius, c = central angle or angle between the points measured from center of circle in radians. So, the radius would change as the ship moves away from the surface…The Arc Lengths would be the same as the orbital speed would be the same…the central angle would be inversely related to the radius.
On a side note, (note) the line between the two points on the circle…called a chord. I don’t know about the French part. But Feralboy was on the right track.
@rick
Unfortunately, I only know undergraduate first year physics, and some of the advanced ones. So when I read elliptic integral of the whatever kind and Heuman’s lambda function, my head is already ready to burst.
‘Tis Himself, pour encourager les autres says:
A possibility, certainly. But you can’t ignore the fact that if you expand the Green’s function for the diminished whole-tone scale in terms of spherical Bessel functions, you can show that playing the Minute Waltz in less than 23.745 seconds violates causality.
#31 @Improbable Joe
Hey, yeah I do. How’d you know?
#132 @kantalope
Try this thought experiment.
Draw a circle. Draw a horizontal line that just touches the top of the circle. (So they are tangential.) Get a string.
You’re spinning along on earth, and they turn off gravity right at the point the circle and line touch – you become weightless. You have the same velocity that had just before gravity switched off, but you’re not accelerating along with the earth’s surface. You follow the straight line. The surface would seem to pull away from underneath you. (On the circle.) Take the string and measure a point along the line from where you took off. Say that’s where you are. Bend the string around the circle, keeping track of the point you were at on the string. That shows where the point of dirt is that you were standing on.
If you still consider “down” to be perpendicular to your trajectory, the dirt spot you left would appear to be “behind” you.
However, that’s the wrong perspective. Earth’s perspective is such that your “down” is along the line joining you and the earth’s centre, so if you join the spot you’re at on the straight line with the centre of the circle by another straight line, that shows where you’d splat on the earth (circle) if infinite gravity switched on, and you instantaneously accelerated to infinite speed towards the ground. You should note that you would hit “behind” the spot you left.
That help?
@95 Mr. Fire:
Fun problem!
The relevant quantities are m, d, and the gravitational constant G. The only combination of these quantities with the dimension of time is sqrt(d^3/G m), so the answer needs to be proportional to this value; let’s call this T.
Energy conservation is probably the most straightforward way to approach this problem, giving a first-order DE for x(t). A couple of variable substitutions (e.g., u = sqrt(d/2x – 1) and then u = tan φ) leads to a pretty trivial trig integral.
The result: collision time = πT/4. (Which, incidentally, is a factor of sqrt(32) smaller than the time for these two objects to circularly orbit their common center of mass.)
Any argument for a flat Earth…
Falls flat!
*hit with 3-5 servings of vegetables and 2-4 servings of fruit*
“but you’re not accelerating along with the earth’s surface. You follow the straight line. The surface would seem to pull away from underneath you.”
——–
Wouldn’t that mean that Voliva was correct? The earth cannot be rotating because you land in the same spot you jumped up from and not miles and miles away? If the earth was indeed moving, as you georotundists are claiming, when you jumped, you would see the earth move beneath you.
As I am working it in my head: It would not be that I am no longer accelerating -my and the earth’s rotation would be a constant velocity in the roundy direction (but would be acceleration in the gravity sense or maybe in terms of my changing direction around the circle?) but instead I would need to decelerate (in the roundy direction) in order to be aware of the earth rotating away.
So I don’t think that Voliva (and Aristotle for that matter) can be faulted for not accepting the movement of the earth…it does not work without accepting that an equal velocity for two objects will appear as non-motion. And I sure hope there is a cool physics term for that non-motion appearenceness.
Das Boese
At this point, debating against flat earthers is like debating against TimeCube. The same level of insanity and disregard for all known evidence is required to believe in flat earth.
Off topic, but is this PZ’s most recent post? It’s currently 2:42am EDT and this is the last post I’ve been able to access.
Yes, although there has been an annoying bug where new articles show up in my rss feed yet upon visiting https://proxy.freethought.online/pharyngula/ I get an old version of it without the new articles.
After having made a comment on a youtube video that a certain argument against flat-earth was faulty, I felt a bit stupid, as I didn’t want to be taken as one myself. So I made a video, showing how even simple optical instruments will reveal large structure on a far-away island just halfway sticking out of the horizon. A commenter on that video started calling me a globalist and trying to save his position (flat-earth) by saying ‘might not the light be bending’? (It does, but according to standard optics it should bend slightly downward as it passes through thick layers low down and thinner layers further up in atmosphere, not upward as is needed for the flat-earthers). What astounded me was that there are still such people out there, that they are able to use a computer and their complete inability to take in evidence. (PS: I know that the phenomena I described was evidence not proof and made sure to make that clear in the description.)
I always liked the time zone argument when dealing with flat earthers.
The sun rises in Los Angeles 3 hours later than it rises in New York. If the earth were flat and you assume that New York and LA were on the same side, sunrise would occur at exactly the same time for both cities.
If the two cities were not on the same side it would be a twelve hour difference.
I especially like this argument because it does not involve brain hurting math
kantaloupe @ 125:
No.
Yes. And for a short time you would appear to be above the launch pad. But after a bit you would appear to be ahead of the launch pad (Eastward), because the launch pad is fixed to the earth which is rotating “eastward and downward. Between the instant of launch and about six hours later (90 degrees of rotation of the earth) the launch pad would appear to be slowing down its eastward motion and speeding up its downward motion as that point rotates away from you. After about 6 hours (actually a bit more than six hours and if you could still see it) it would appear to be moving westward and away from you.
I am not taking into consideration gravity or anything else that would complicate the motion. Just velocities and appearances. No orbits.
If a very big very tall train is travelling along at a constant velocity 1,000 MPH going east (or any other direction) and you launch a rocket (or just jump) straight up you will stay directly above the spot you jumped or launched from. It doesn’t matter if the train is full of air or has a vacuum inside.
If the track is circular (a great big circle) and you jump straight up or launch a rocket inside you will see the spot where you started move slightly to one side and a little bit behind.
Don’t worry about gravity just yet. Get a handle on simple velocities and directions first and then you can add in the differences due to gravity.
Ouigui @136: nice work! I failed to simplify it to an energy conservation problem the way you have, and ended up doing it a much longer way.
People Are More Elegant Than Me On The Internet.
When it comes to the “problem” Voliva mentions, the previous posts here mentions Newton and his concept of inertia. Newton could certainly handle this one, but maybe we can let an even older guy field this one? I’m thinking Galileo and his theory of invariance (relativity). According to him, all systems moving with constant linear velocity according to each other would experience the same laws of physics. Now a rotating Earth is not quite a linear system. But during a 1 second jump, the displacement from a linear motion at the equator (worst case scenario) is about 1.5cm in the vertical direction (experienced as a slight modification of gravity compared to that at the poles) and about 0.4 micrometers along the horizontal direction. So, I think Galileo would say that except for a slight modification in how high you are able to jump, the experience of jumping would be virtually the same for a rotating Earth and one that didn’t. Galileo wrote about his theory of invariance in 1632, so Voliva is about 253 years too late in asking this question.
@109:
The Billiard Ball. The theoretician develops a theory for generating an anti-gravity field; the experimentalist builds a working anti-gravity device, but doesn’t understand all the implications of the theory, with fatal consequences.
Mr. Fire @145:
Same here. After I noodled on that problem, I found a solution where someone simply applied Kepler’s Third Law to a degenerate elliptical orbit to get the answer. (Of course, then the task is to derive Kepler’s Third Law, I suppose…)
#138 @kantalope
The planet’s radius is too large, or perhaps you can’t jump high enough, for that to happen in any realistic manner.
This made me see what the issue is. You’re looking at this with an Aristotelian physics perspective. The ground is still accelerating perpendicular to its velocity. It does not have a constant velocity. You do. The velocities are not the same after you floated away.
Aristotle, possibly not. Voliva had the advantage of Newtonian mechanics, and subsequent work by a number of intelligent people, to see where he was wrong. He is definitely at fault.
The velocities of the dirt patch and the observer may have the same magnitude, but they do not have the same direction. They are different.
You are correct though – if two objects do have the same velocity they do not have relative motion with respect to each other. (It just isn’t the case under discussion.)