Imagine you're stood on the Earth facing the sun at 12 noon.
Bear in mind that 24h is one turn of the Earth, its important to remember this part for later on in the question.
Now imagine 6 months passes by and the Earth is now on the other side of the sun.
Assuming you're still standing where you were 6 months ago at 12 noon  you will now be standing on an Earth where you're facing away from the sun, but it is 12 noon. It should be 12 midnight, not 12 noon.
Watches and clocks don't account for this. A digital watch would still be reasonably accurate after 9 years and 6 months (assuming its battery holds out) without it ever having to "shift 12 hours every 6 months".
You can easily demonstrate this with 2 coins pretending the number on say a 10p is where you're stood. No cheating keeping yourself facing the sun! If you were to make 24h face the sun every turn, it wouldn't be 24h and it also wouldn't be one turn (it would need to be slightly more on an Earth turning anticlockwise around the sun).
The best answer I can muster up is "Each day really lasts 24h plus 1/360th of a day" (1 day +4 minutes) which sounds like total bunk because watches don't measure time that way and would need to account for it. No watchmaker in the world would sell a watch that did this, after only three days you'd be turning up 12 minutes early everywhere you went. There's also the slight problem that only you would have this watch while every other watch and clock is normal. Can't work can it.
We're magically gaining 4 minutes a day on our watches.
Has anyone got an answer to this?
Here's a puzzle for you...
Here's a puzzle for you...
Only weird bikes are interesting anymore.
Re: Here's a puzzle for you...
Good grief. Have you even tried googling ' How long is a day'.
Re: Here's a puzzle for you...
freeflow wrote:Good grief. Have you even tried googling ' How long is a day'.
Its 24 hours, not "24 hours plus 4 minutes", hence the question.
I love these questions because all you get is silly answers.
Only weird bikes are interesting anymore.

 Posts: 463
 Joined: 9 Jun 2011, 10:34pm
Re: Here's a puzzle for you...
the "silly answer" you criticise would no doubt get you to the truth. 24 hours is the length of the day, which is not the same as the time the earth takes to revolve on its axis, which is presumably 24 hours minus 4 minutes if your maths above is right. Strangely the astronomers and clockmakers haven't been getting it wrong for a thousand years
Re: Here's a puzzle for you...
Yes but a simple answer as to "why they never get it wrong" would be if there were 24 hours 0 minutes in a day and not 24 hours and 4 minutes (or 23 hours and 56 minutes, same difference as far as it being 12 hours out after 6 months).
I think I worked it out  all our clocks and watches display 24 hours a day exactly to the second, but they are in fact timing that as 24 hours and 4 minutes a day, but displaying it as a reduced figure of 24 hours and 0 minutes to keep everything in check 6 months later?
Something still doesn't sound right about that though.
Its like the one about "How does a plane maintain altitude without flying into space and without tipping its nose down all the time?"
The only way you could know the answer is if you knew how the software worked that is now used on planes to guide them. It would have to be programmed to continually tip the nose down otherwise it would fly into space lol.
Quote:
"A plane flying at a typical 35,000 feet wishing to maintain that altitude at the upperrim of the socalled “Troposphere” in one hour would find themselves over 200,000 feet high into the “Mesosphere” with a steadily raising trajectory the longer they go. I have talked to several pilots, and no such compensation for the Earth’s supposed curvature is ever made. When pilots set an altitude, their artificial horizon gauge remains level and so does their course; nothing like the necessary 2,777 foot per minute declination is ever taken into consideration."
It sure is a poser.
I think I worked it out  all our clocks and watches display 24 hours a day exactly to the second, but they are in fact timing that as 24 hours and 4 minutes a day, but displaying it as a reduced figure of 24 hours and 0 minutes to keep everything in check 6 months later?
Something still doesn't sound right about that though.
Its like the one about "How does a plane maintain altitude without flying into space and without tipping its nose down all the time?"
The only way you could know the answer is if you knew how the software worked that is now used on planes to guide them. It would have to be programmed to continually tip the nose down otherwise it would fly into space lol.
Quote:
"A plane flying at a typical 35,000 feet wishing to maintain that altitude at the upperrim of the socalled “Troposphere” in one hour would find themselves over 200,000 feet high into the “Mesosphere” with a steadily raising trajectory the longer they go. I have talked to several pilots, and no such compensation for the Earth’s supposed curvature is ever made. When pilots set an altitude, their artificial horizon gauge remains level and so does their course; nothing like the necessary 2,777 foot per minute declination is ever taken into consideration."
It sure is a poser.
Only weird bikes are interesting anymore.
Re: Here's a puzzle for you...
Try looking up "sidereal day" and "solar day".
The sidereal day, which measures 23 hours, 56 minutes, 4.0916 seconds, is the actual time it takes for the Earth to complete one rotation with respect to the distant stars and galaxies. In other words, it is the true rotation period.
The solar day is the mean time it takes for the Earth to complete one rotation, with respect to an imaginary line linking the centres of the Earth and the Sun. This day, which by definition is exactly 24 hours, is approximately 4 minutes longer because this line itself rotates by about 1/365th of a complete circle  just under 1 degree  in the time it takes the Earth to rotate once. So it needs to rotate for an extra 4 minutes to 'catch up'. Strictly speaking, the imaginary line I have referred to should be linking the Earth with something called the 'Mean Sun' which is a theoretical object which is defined as advancing by exactly the same amount each day. The true Sun's path is not completely uniform because of two factors: the inclination of the Earth's axis and the ellipticity of its orbit round the Sun.
Look up "equation of time".
Happy googling!
[edit] got my minutes and seconds mixed up! ops:
The sidereal day, which measures 23 hours, 56 minutes, 4.0916 seconds, is the actual time it takes for the Earth to complete one rotation with respect to the distant stars and galaxies. In other words, it is the true rotation period.
The solar day is the mean time it takes for the Earth to complete one rotation, with respect to an imaginary line linking the centres of the Earth and the Sun. This day, which by definition is exactly 24 hours, is approximately 4 minutes longer because this line itself rotates by about 1/365th of a complete circle  just under 1 degree  in the time it takes the Earth to rotate once. So it needs to rotate for an extra 4 minutes to 'catch up'. Strictly speaking, the imaginary line I have referred to should be linking the Earth with something called the 'Mean Sun' which is a theoretical object which is defined as advancing by exactly the same amount each day. The true Sun's path is not completely uniform because of two factors: the inclination of the Earth's axis and the ellipticity of its orbit round the Sun.
Look up "equation of time".
Happy googling!
[edit] got my minutes and seconds mixed up! ops:
Last edited by 661Pete on 7 Jun 2015, 8:18pm, edited 1 time in total.
Suppose that this room is a lift. The support breaks and down we go with everincreasing velocity.
Let us pass the time by performing physical experiments...
 Arthur Eddington (creator of the Eddington Number).
Let us pass the time by performing physical experiments...
 Arthur Eddington (creator of the Eddington Number).
Re: Here's a puzzle for you...
A planet rotates of course.
A planet is a planet, and as such orbits a sun.
The planet rotates and it's rotation is governed by the sun.
If an object was in free space .................... which it would never be .................. it would rotate as the OP suggests.
A planet is a planet, and as such orbits a sun.
The planet rotates and it's rotation is governed by the sun.
If an object was in free space .................... which it would never be .................. it would rotate as the OP suggests.
Mick F. Cornwall
Re: Here's a puzzle for you...
I can't grasp how it isn't 4 minutes over and is instead, 4 minutes under.
Imagine the sun is on the left and Earth on the right, with you facing the sun, Earth turns anticlockwise  so for one full anticlockwise turn exactly (plus roughly 1 degree of orbit) the point where you are after 24h 0 minutes hasn't quite turned enough and it will take another 4 minutes until you're "lined up" with the sun again and on a clock or watch this has to measure 24h 0 minutes, despite it taking 24h 4 minutes.
If it goes anticlockwise the exact day length is 24h 4 minutes, same if it is going clockwise.
Imagine the sun is on the left and Earth on the right, with you facing the sun, Earth turns anticlockwise  so for one full anticlockwise turn exactly (plus roughly 1 degree of orbit) the point where you are after 24h 0 minutes hasn't quite turned enough and it will take another 4 minutes until you're "lined up" with the sun again and on a clock or watch this has to measure 24h 0 minutes, despite it taking 24h 4 minutes.
If it goes anticlockwise the exact day length is 24h 4 minutes, same if it is going clockwise.
Only weird bikes are interesting anymore.

 Posts: 463
 Joined: 9 Jun 2011, 10:34pm
Re: Here's a puzzle for you...
Manc33 wrote:I can't grasp how it isn't 4 minutes over and is instead, 4 minutes under.
Imagine the sun is on the left and Earth on the right, with you facing the sun, Earth turns anticlockwise  so for one full anticlockwise turn exactly (plus roughly 1 degree of orbit) the point where you are after 24h 0 minutes hasn't quite turned enough and it will take another 4 minutes until you're "lined up" with the sun again and on a clock or watch this has to measure 24h 0 minutes, despite it taking 24h 4 minutes.
If it goes anticlockwise the exact day length is 24h 4 minutes, same if it is going clockwise.
just giving the benefit of the doubt and that this isn't a (feeble) windup  a day is defined.to be 24 hours. And that's what watches are set to measure. A day (24 hours) is noontonoon ie based on a combination of earth's orbit and rotation of the earth.relative to fixed stars.
Re: Here's a puzzle for you...
Manc33 wrote:I can't grasp how it isn't 4 minutes over and is instead, 4 minutes under.
Imagine the sun is on the left and Earth on the right, with you facing the sun, Earth turns anticlockwise  so for one full anticlockwise turn exactly (plus roughly 1 degree of orbit) the point where you are after 24h 0 minutes hasn't quite turned enough and it will take another 4 minutes until you're "lined up" with the sun again and on a clock or watch this has to measure 24h 0 minutes, despite it taking 24h 4 minutes.
If it goes anticlockwise the exact day length is 24h 4 minutes, same if it is going clockwise.
Was my post not clear enough? The time it takes for the Earth to make one full anticlockwise turn is not 24 hours. It is 23 hours 56 minutes (plus a few odd seconds). If, for instance, I set up my telescope, on a fixed mount, and point it at a certain star, centring it exactly in the eyepiece, at, say, 10pm one night  then, if I leave the telescope untouched for the next day and go to look through it the following night, the same star will be centred in the eyepiece, not at 10pm but at 9:56 pm. And the night following that, it would be 9:52, and so on.
That is why different constellations appear in the evening sky as the year advances. In winter Orion dominates the sky; in spring it is Leo; in summer it is Cygnus and the 'summer triangle'. In autumn it is Pegasus, Cassiopeia and Andromeda.
You seem to have a genuine interest in this topic. Why not seek out a local astronomy group? You may well find it rewarding.
Suppose that this room is a lift. The support breaks and down we go with everincreasing velocity.
Let us pass the time by performing physical experiments...
 Arthur Eddington (creator of the Eddington Number).
Let us pass the time by performing physical experiments...
 Arthur Eddington (creator of the Eddington Number).
Re: Here's a puzzle for you...
I got another one, you're going to hate me...
You weigh the same all over the Earth, a 200lb man on the equator weighs 200lb at the North pole (or, the difference is so minimal, it is hard to measure).
The Earth has a centrifugal force to it because it spins on an axis.
The question is: How does gravity manage to keep everything the same weight all over the Earth in such a way that it perfectly counterbalances the centrifugal force?
Gravity doesn't just need to "keep things stuck to Earth" because there is the centrifugal force to take into account, so this means gravity has to keep objects nearer the equator "stuck more firmly" than objects nearer the North pole, how?
The best answer you'll get for this (even from a top scientist) is "it just does".
Well in science with no proof, actually it just doesn't, not without something to explain it anyway. You can't just change the laws of physics on a whim to fit a theory. People believe this stuff without evidence and most importantly without thinking about it, because they don't think there's any need to, some brainy scientists have done it all for you. Of course there is a need to check into these things and get answers  because you live here!
"If you're alive and you don't know where you're alive... this is kind of important"  Mark Knight
You weigh the same all over the Earth, a 200lb man on the equator weighs 200lb at the North pole (or, the difference is so minimal, it is hard to measure).
The Earth has a centrifugal force to it because it spins on an axis.
The question is: How does gravity manage to keep everything the same weight all over the Earth in such a way that it perfectly counterbalances the centrifugal force?
Gravity doesn't just need to "keep things stuck to Earth" because there is the centrifugal force to take into account, so this means gravity has to keep objects nearer the equator "stuck more firmly" than objects nearer the North pole, how?
The best answer you'll get for this (even from a top scientist) is "it just does".
Well in science with no proof, actually it just doesn't, not without something to explain it anyway. You can't just change the laws of physics on a whim to fit a theory. People believe this stuff without evidence and most importantly without thinking about it, because they don't think there's any need to, some brainy scientists have done it all for you. Of course there is a need to check into these things and get answers  because you live here!
"If you're alive and you don't know where you're alive... this is kind of important"  Mark Knight
Only weird bikes are interesting anymore.
Re: Here's a puzzle for you...
The question is: How does gravity manage to keep everything the same weight all over the Earth in such a way that it perfectly counterbalances the centrifugal force?
It doesnt have to balance it, it just has to exceed it. Which I think it does by orders of magnitude as we are rotating at about .0007 rpm.
Re: Here's a puzzle for you...
The simple answer is, that it doesn't. The apparent force due to gravity is not uniform across the Earth's surface. But the variations are too tiny for people to sense in the normal way of things.The question is: How does gravity manage to keep everything the same weight all over the Earth in such a way that it perfectly counterbalances the centrifugal force?
The centrifugal force  or to be more scientific, the centripetal acceleration  due to the Earth's rotation, at the equator, is tiny compared to the force of gravity  only about 0.3%. There is also a variation due to the Earth's equatorial bulge  this accounts for a further 0.2%, making a total of about 0.5% variation. Things are therefore about 0.5% lighter at the equator than at the poles.
There also exist variations due to changes in density and thickness of the earth's crust.
If you weigh yourself on a set of bathroom scales, taking them with you as you travel between the poles and the equator, you will get about a 1 lb difference, assuming you weigh as much as an average adult  and that you don't gain or lose any body mass meanwhile. Most scales are not that accurate, anyway!
Suppose that this room is a lift. The support breaks and down we go with everincreasing velocity.
Let us pass the time by performing physical experiments...
 Arthur Eddington (creator of the Eddington Number).
Let us pass the time by performing physical experiments...
 Arthur Eddington (creator of the Eddington Number).
Re: Here's a puzzle for you...
The Earth first. The world is not a globe but an oblate spheroid. The distance from pole to pole is less than the distance from one side of the world at the equator to the opposite side. So, centrifugal force, otherwise correctly known as inertia, acts differently at the pole to the equator, as does gravity. There is more planet beneath your feet at the equator than there is at the pole.
Aircraft next. They were invented long before computers, unless you count an abacus as a computer. What keeps the aircraft up is lift, what drags it down is weight and gravity. If an aircraft flied from the USA to Europe, such as Charles Lindbergh's Spirit of St Lois, he in fact, assuming for the exercise his height remained constant, in a parabola that equalled the earth's circumference. He did this by balancing the forces dragging the aircraft down with those lifting it up. There are some things that are only difficult if you make them so.
However, as you seem to like this sort of thing, try this. A cyclist is riding a bike at 20 mph. His wheels are 26" in diameter. Next to him is another cyclist also riding at 20 mph. His bike has 20" wheels. Each of them has, by coincidence, a squashed fly jammed inbetween the treads of his front tyre. The fly on the first bike will come to a dead stop at one point and then briefly travel at 40 mph before returning to zero, and this will repeat itself for every rotation of the wheel. The fly on the second cycle will do exactly the same.
Aircraft next. They were invented long before computers, unless you count an abacus as a computer. What keeps the aircraft up is lift, what drags it down is weight and gravity. If an aircraft flied from the USA to Europe, such as Charles Lindbergh's Spirit of St Lois, he in fact, assuming for the exercise his height remained constant, in a parabola that equalled the earth's circumference. He did this by balancing the forces dragging the aircraft down with those lifting it up. There are some things that are only difficult if you make them so.
However, as you seem to like this sort of thing, try this. A cyclist is riding a bike at 20 mph. His wheels are 26" in diameter. Next to him is another cyclist also riding at 20 mph. His bike has 20" wheels. Each of them has, by coincidence, a squashed fly jammed inbetween the treads of his front tyre. The fly on the first bike will come to a dead stop at one point and then briefly travel at 40 mph before returning to zero, and this will repeat itself for every rotation of the wheel. The fly on the second cycle will do exactly the same.
Pacifists cannot accept the statement "Those who 'abjure' violence can do so only because others are committing violence on their behalf.", despite it being "grossly obvious."
[George Orwell]
[George Orwell]
Re: Here's a puzzle for you...
If the mass of Earth is halved and its RPM doubled, wouldn't that replicate the exact centrifugal force as Earth?
So halving its mass to 12,500 miles makes it 0.0014 rpm...
6,250 miles / 0.0028 rpm
3,125 miles / 0.0056 rpm
1562.5 miles / 0.0112 rpm
781.25 miles / 0.0224 rpm
390.63 miles / 0.0448 rpm
195.31 miles / 0.0896 rpm
97.66 miles / 0.1792 rpm
48.83 miles / 0.3584 rpm
24.41 miles / 0.7168 rpm
12.2 miles / 1.43 rpm
6.10 miles / 2.87 rpm
3.05 miles / 5.73 rpm
1.53 miles / 11.47 rpm
0.76 miles / 22.94 rpm
0.38 miles / 45.88 rpm
0.19 miles / 91.75 rpm
0.095 miles / 183.5 rpm
0.048 miles / 367 rpm
0.024 miles (1,520 inches) / 734 rpm
So now we have a ball that is 1,520" around (126.67 feet) and spins at 734 RPM.
Let's carry on... we want it football sized so we can relate to it...
0.012 miles / 1,468 rpm
0.006 miles / 2,936 rpm
0.003 miles / 5,872 rpm
0.00149 miles / 11,744 rpm
Now the ball is 94.4 inches (7.87 feet) around it, spinning at approaching twelve thousand RPM.
Gonna have to change the miles to inches now, calc is doing that "to the power of" thing and can't cope...
47.2 inches (3.93 feet) / 23,488 rpm
23.6 inches (approx 2 feet) / 46,976 rpm
So now the ball is roughly football sized at 2 feet around it.
At the end of all of that you'd have a ball thats 2 feet in circumference, with the same density as Earth (a ball of rock), that is spinning at almost 47,000 RPM.
Not only would this have a "noticeable centrifugal force", it would probably shatter the ball apart completely.
So this means Earth has to have that same pushing force on every inch of its surface. Gravity is supposedly counteracting this? If gravity is that strong, how come it never pulls the moon into the Earth?
Sorry about having a gigantic list of numbers but it shows from beginning to end, halving the mass of the Earth and to compensate, doubling its RPM.
So halving its mass to 12,500 miles makes it 0.0014 rpm...
6,250 miles / 0.0028 rpm
3,125 miles / 0.0056 rpm
1562.5 miles / 0.0112 rpm
781.25 miles / 0.0224 rpm
390.63 miles / 0.0448 rpm
195.31 miles / 0.0896 rpm
97.66 miles / 0.1792 rpm
48.83 miles / 0.3584 rpm
24.41 miles / 0.7168 rpm
12.2 miles / 1.43 rpm
6.10 miles / 2.87 rpm
3.05 miles / 5.73 rpm
1.53 miles / 11.47 rpm
0.76 miles / 22.94 rpm
0.38 miles / 45.88 rpm
0.19 miles / 91.75 rpm
0.095 miles / 183.5 rpm
0.048 miles / 367 rpm
0.024 miles (1,520 inches) / 734 rpm
So now we have a ball that is 1,520" around (126.67 feet) and spins at 734 RPM.
Let's carry on... we want it football sized so we can relate to it...
0.012 miles / 1,468 rpm
0.006 miles / 2,936 rpm
0.003 miles / 5,872 rpm
0.00149 miles / 11,744 rpm
Now the ball is 94.4 inches (7.87 feet) around it, spinning at approaching twelve thousand RPM.
Gonna have to change the miles to inches now, calc is doing that "to the power of" thing and can't cope...
47.2 inches (3.93 feet) / 23,488 rpm
23.6 inches (approx 2 feet) / 46,976 rpm
So now the ball is roughly football sized at 2 feet around it.
At the end of all of that you'd have a ball thats 2 feet in circumference, with the same density as Earth (a ball of rock), that is spinning at almost 47,000 RPM.
Not only would this have a "noticeable centrifugal force", it would probably shatter the ball apart completely.
So this means Earth has to have that same pushing force on every inch of its surface. Gravity is supposedly counteracting this? If gravity is that strong, how come it never pulls the moon into the Earth?
Sorry about having a gigantic list of numbers but it shows from beginning to end, halving the mass of the Earth and to compensate, doubling its RPM.
Only weird bikes are interesting anymore.