Freethinker Posted June 30, 2004 Report Share Posted June 30, 2004 Originally posted by: TeleMadYep, I goofed in the details (but the point I was making is still valid).Still a qualifier in there, but I'm glad to see you can admit error. Perhaps we can have some good posts together. Quote Link to comment Share on other sites More sharing options...
BlameTheEx Posted June 30, 2004 Report Share Posted June 30, 2004 Telemad "So an object in space gravitationally attracted to the Earth has potential energy: it just needs to fall closer to the Earth in order to convert that potential energy into kinetic energy" Just so. However The object doesn't just have the potential energy of a fall to earth. It can potentially fall into ANY gravitational field. The potential energy available MUST approach that of its entire mass/energy, as such gravitational fields are known to exist. This potential energy is there regardless of whether it goes anywhere near any gravitational field (unless you care to propose that it knows its future, or can gain potential energy at some point). If this potential energy is not in fact its own mass/energy then I am at a loss to figure out where you propose it is stored. Are you proposing a store of energy, equivalent to the objects mass, attached to it, but undetectable? Potential energy does not just exist. It has to be somewhere. The force that we have most understanding of, the electromagnetic, follows this rule. An electron releases energy when approaches an unpaired proton. This is also the basis of chemistry (OK, I am simplifying a bit here). However this potential energy IS got from somewhere known. The combined mass of proton and electron drops. We know less about the forces that bind an atom together, but the same applies. When protons and neutrons bind together energy is released, but this energy is identical to the energy of (mass of the separate neutrons and protons) - (mass of resulting atom). We may not understand all about electromagnetic and atomic binding forces, but we can weigh the particles concerned accurately. We have plenty of good evidence that the energy released is equal to mass/energy lost. This is why the principle of conservation of mass/energy is so widely respected. Why should it be different for gravity? Regarding this concept of gravitational binding energy. All it really says is that once a gravitational body is created, it takes energy to separate its constituents again. True enough. The energy releases in its creation has to be returned. But energy is just mass/energy. This is just another way of saying what I have been saying. Energy was liberated in the creation of the body, AND THAT MEANS IT LOST MASS. As the energy was lost as heat, and the quantity of matter remained constant we must conclude that the mass energy of matter in a gravitational field is less that it was outside. But this is a separate issue from that of further objects falling onto the body. Those objects have there own mass/energy to loose. Quote Link to comment Share on other sites More sharing options...
Tormod Posted June 30, 2004 Report Share Posted June 30, 2004 Originally posted by: BlameTheExSorry.... seem to have created a duplicate of last message. I can't seem to remove it completly so I have replaced it with this message. Any chance of adding a delete button to the options? No, sorry - the option to remove posts is not available since it could render debates useless if someone accidentally (or intentionally) deleted their posts in a thread. So just do as above - usually we just write "*sorry - double post*" (if applicable) or "*post deleted by myself*" or someting. No big deal. Tormod Quote Link to comment Share on other sites More sharing options...
Freethinker Posted June 30, 2004 Report Share Posted June 30, 2004 Originally posted by: Tormod Tormod HE'S BACK! Welcome back. Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 1, 2004 Report Share Posted July 1, 2004 BlameTheEx: Regarding this concept of gravitational binding energy. All it really says is that once a gravitational body is created, it takes energy to separate its constituents again. True enough. The energy releases in its creation has to be returned. But energy is just mass/energy. This is just another way of saying what I have been saying. No it's not...not at all. I'll finish the relevant part of your quote before explaining... BlameThEx: Energy was liberated in the creation of the body, AND THAT MEANS IT LOST MASS. As the energy was lost as heat, and the quantity of matter remained constant we must conclude that the mass energy of matter in a gravitational field is less that it was outside. . Do you see any mention of the speed of light having to be reduced in anything you just stated? Nope. Yet here is what you said that I have been countering... BlameTheEx: So how does it work out? An object falling into a gravitational field gains energy as velocity. This kinetic energy is balanced by a loss of energy from its mass. That is calculated as E=MC2. The C is the speed of light. Inside a gravitational field C is lowered. The best way to think of gravity is that mass/energy reduces the speed of light. The formula for this reduction is clear enough, in non-relativistic situations, but it would be brave to extrapolate it further without evidence. Remember all my comments about "no need to monkey around with the speed of light" to account for the changes in energy/mass? So actually, what you just said in your last post is what I have been saying...and it contradicts what you said. You can't have your cake and eat it too. Pick a position. Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 1, 2004 Report Share Posted July 1, 2004 BlameTheEx: Potential energy has to be stored somehow. Any object gravitationally attracted to Earth, but not in physical contact with the Earth, has potential energy simply from the fact that it is gravitationally attracted to Earth. Take the moon Io: it has potential energy. We don't realize this because Io is more strongly bound, gravitationally, to other objects in our solar system. But imagine removing the Sun, Jupiter, Saturn, etc. - leaving only the Earth and Io. What would happen? Io would fall in towards the Earth (and vice versa, to a lesser degree) converting gravitational potential energy into kinetic energy. The potential energy is there...an object does not have to be physically hoisted away from the Earth's surface in order to store gravitational potential energy in it. THERE IS NO NEED TO MONKEY AROUND WITH THE SPEED OF LIGHT. BlameTheEx: Worse our object falling into a gravitational field has GAINED energy. You mean because it converted some of its gravitational potential energy into kinetic energy? Yep, just as I've been saying. BlameTheEx: To balance the books you have to consider the storage of negative energy, or to put it another way you have to look for an already existing supply of energy that can be tapped. Yeah, the gravitational potential energy I keep telling you about. BlameTheEx: You can't just assume that it happens. There has to be a mechanism. Which I've explained, and, unlike you, in a manner that doesn't require breaking any law of physics. BlameTheEx: Right or wrong, I have given a plausible mechanism. No, you haven't. Your mechanism violates the special theory of relativity. That's not a plausible mechanism. Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 1, 2004 Report Share Posted July 1, 2004 BlameTheEx: The force that we have most understanding of, the electromagnetic, follows this rule. An electron releases energy when approaches an unpaired proton. This is also the basis of chemistry (OK, I am simplifying a bit here). However this potential energy IS got from somewhere known. The combined mass of proton and electron drops. And when a free proton and a free electron bind, gravitational binding energy is also released and contributes to the total reduction in mass. BlameTheEx: We know less about the forces that bind an atom together, but the same applies. When protons and neutrons bind together energy is released, but this energy is identical to the energy of (mass of the separate neutrons and protons) - (mass of resulting atom). And when free nucleons bind, gravitational binding energy is also released and contributes to the total reduction in mass. BlameTheEx: We may not understand all about electromagnetic and atomic binding forces, but we can weigh the particles concerned accurately. We have plenty of good evidence that the energy released is equal to mass/energy lost. This is why the principle of conservation of mass/energy is so widely respected. Why should it be different for gravity? Who said gravity is different? You? That's because you are confused. Any binding that involves free objects becoming more tightly bound physically releases gravitational binding energy, resulting in a decrease in mass. It's not usually noticed because the gravitational force is terribly feeble compared to the strong and electric forces, which are about 10^40 times stronger than gravity. Imagine you fuse a huge amount of protons and neutrons – enough to have one gram of mass converted into energy because of the strong force. You could easily measure a 1-gram difference in mass. But how would you measure the change in mass caused by gravitational binding energy? For the same experiment, that difference would be only about 0.0000000000000000000000000000000000000000001 grams. Quote Link to comment Share on other sites More sharing options...
BlameTheEx Posted July 1, 2004 Report Share Posted July 1, 2004 Telemad You do leave me with a lot of things to reply to. Please forgive me for limiting myself to your last statement. "Any binding that involves free objects becoming more tightly bound physically releases gravitational binding energy, resulting in a decrease in mass. It's not usually noticed because the gravitational force is terribly feeble compared to the strong and electric forces, which are about 10^40 times stronger than gravity" Now we are getting somewhere. We can agree on that. The mass is indeed reduced. There's no reason to suppose that any of the atoms, electrons ect disappear. However while the gravitational force is relatively weak, we are considering situations where there is enough mass to make it very strong indeed. The reduction of mass for matter on the surface of some know astronomical bodies will be a good proportion of the total. How do you feel that mass can be gone? Are you imagining that a proportion of the atoms disappear, or that the are just lighter? If some disappear, then where do they go. If they become lighter then what is the formula? I can't imagine where the atoms would disappear to if they did, but I can tell you the formula if they get lighter. It is proportional to the time dilation! Gravitational time dilation is DEFINED that way. It is equal to the ratio of frequency of a photon from when it is emitted from the surface, and when it is received outside the gravitational field (Ok,Ok mostly outside.There is no end to a gravitational field), or the for that matter the ratio of frequency gain if the photon goes the other way. This ratio is identical to the ratio of energies for the photon, and naturally enough it's also identical to the ratio of mass/energy (including kinetic) of an object dropped to the surface. It doesn't matter how a lump of mass/energy gets to the surface. The proportional increase of mass/energy must be the same. The dropped object gains kinetic energy in exact proportion to its gravitational time dilatation. Coincidence? Quote Link to comment Share on other sites More sharing options...
Truth Posted July 1, 2004 Report Share Posted July 1, 2004 Simplicity is the best answer. I am sorry to say, that gravity is a falling effect and everything in the universe is in actuality, falling in expansion.In other words falling and expanding in equalibrium.So any place where you have matter that is larger or greater the density the greater the falling effect. A blackhole is an extreme example of that. Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 2, 2004 Report Share Posted July 2, 2004 BlameTheEx: We can agree on that. The mass is indeed reduced. There's no reason to suppose that any of the atoms, electrons ect disappear. Agreed. “Missing mass” caused by binding is not missing particles…it’s mass that has been converted into energy (via E = mc^2) and liberated from the particles that bound. However while the gravitational force is relatively weak, we are considering situations where there is enough mass to make it very strong indeed. The reduction of mass for matter on the surface of some know astronomical bodies will be a good proportion of the total. How do you feel that mass can be gone? Converted to energy via E = mc^2. Why do stars expand into red giants when the core of the star contracts? Yes, when the core contracts, the star’s size increases, dramatically. Where does the vast amount of energy needed to push all of that enormous amount of matter out millions of miles come from? From the release of gravitational potential energy as the core contracts. In the process, a huge amount of mass in converted into energy…a HUGE amount. You still haven’t given us any valid reason to look beyond mainstream theories/laws to explain any of this – there is no need to monkey around with the speed of light, and very good reason not to do so. BlameTheEx: I can't imagine where the atoms would disappear to if they did, but I can tell you the formula if they get lighter. It is proportional to the time dilation! Gravitational time dilation is DEFINED that way. I went back to Wikipedia and the equation they have for gravitational time dilation doesn’t include mass. ”Gravitational Time Dilation.…Time dilation around a black hole may be described using the following equation:To = Tf (1 – Ch/Co)^(1/2) Where t0 is time for the object undergoing dilation, tf is time for an observer outside the system, Ch is the circumference of the event horizion, and C0 is the circumference of the object's orbit about the black hole.” (http://en.wikipedia.org/wiki/Time_dilation) So can you support your statement that gravitational time dilation is DEFINED in relation to “missing mass” being proportional to time dilation. PS: And even if you do, you still haven’t supported the REAL issue – You still haven’t given us any valid reason to look beyond mainstream theories/laws to explain any of this – there is no need to monkey around with the speed of light, and very good reason not to do so. 2) Are you confusing time dilation – which is a direct consequence of relativity - with reduction in the speed of light – which is a direct contradiction of relativity? BlameTheEx: [Gravitational time dilation] is equal to the ratio of frequency of a photon from when it is emitted from the surface, and when it is received outside the gravitational field (Ok,Ok mostly outside. There is no end to a gravitational field), or the for that matter the ratio of frequency gain if the photon goes the other way. This ratio is identical to the ratio of energies for the photon, … Okay, you’re discussing standard gravitational red shift. BlameTheEx: … and naturally enough it's also identical to the ratio of mass/energy (including kinetic) of an object dropped to the surface. I assume you’ll have no problem providing support for that. None of your post even addresses the real issue between us – you still haven’t given me, or anyone else, any valid reason to look beyond mainstream theories to explain any of this – there is no need to monkey around with the speed of light, and good reasons not to do so. Quote Link to comment Share on other sites More sharing options...
BlameTheEx Posted July 2, 2004 Report Share Posted July 2, 2004 Telemad. Converted into energy is indeed right. But that wasn't quite the question I was trying to put. The energy is gone, but all the atoms are still there. We now have lower energy atoms. That was the puzzle I was asking you to consider. Your formula for Gravitational Time Dilation is based on a black hole. Like I have said before, it's hardly fair to ask me to answer an argument that STARTS from the premise that I am wrong. Like I pointed out before, I can't deny the properties of black holes, once they are formed. I just suggest that it is impossible to create them. There is probably lots of self consistent literature on fairies too. Your link points to a blank page. Apparently Wikipedia does not have an article on the topic yet. I will get back to you regarding about links to Gravitational red shift, but the ratio of energy between free space and the surface of a body is constant for photons, bricks, cosmic rays or what have you. It has to be. Gravitational potential energy is independent of how a lump of mass/energy is delivered to the surface. If it were not it would disobey the law of conservation of mass/energy. Think. In principle we could have a machine that converts a brick into photons, and visa versa. so we have 2 bricks. One is dropped to the surface, and the other converted to photons, beamed to the surface, and converted back to a brick+ additional energy due to the gravitational shift in the photons wavelength. If that additional energy was not the same as the kinetic energy of the first brick then its not difficult to imagine a system for creating as much energy as we wish from nowhere. Getting the brick back into free space should take exactly the same energy as was freed by dropping it, so we just use the less energy consuming process for the upward journey, the other for the way down, and repeat till all our batteries are charged. As for monkeying with the speed of light, well I am, and I ain't. The usual answer when dealing with time dilation. Its all a matter of the observer. Local measurements will always hold C to be constant, as Einstein maintained. However for an observer looking at events on the surface from outside the gravitational field things move more slowly. As you accept that Gravitational Time Dilation happens, how can you disagree with this? A local observer on the surface of a body with a time dilation factor of 2 will measure a photon as travelling 300 Meters in a microsecond (lets avoid a few quibbles by stating that this photon is travelling parallel to the surface of the body), but the Observer looking down notes 2 microseconds have passed. He can only have one of 2 conclusion. Ether C is half what is expected, or distances on the surface are twice the local measurement. I would hold to the former, but the latter also holds fascination because it would lead to the inescapable conclusion that any singularity would have, from an external observers point of view, infinite diameter. My apologies for not going further just now into your reply but I am tired and have rather a lot to do. Hopefully I will get back to finishing next week. Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 3, 2004 Report Share Posted July 3, 2004 BlameTheEx: Converted into energy is indeed right. But that wasn't quite the question I was trying to put. The energy is gone, but all the atoms are still there. We now have lower energy atoms. That was the puzzle I was asking you to consider. I don't see a puzzle. I gave an example: when the core of a star contracts, gravitational potential energy is released and it does work in the form of expanding the overlying shells and the envelope, creating a red giant. In the process, much mass is converted into energy via E = mc^2. BlameTheEx: Your link points to a blank page. Apparently Wikipedia does not have an article on the topic yet. It's there. For whatever reason, the ending parenthesis has been included in the link: it's trying to navigate to a page named "time_dilation)" instead of just "time_dilation". .” Here's the link again... ( http://en.wikipedia.org/wiki/Time_dilation ) Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 3, 2004 Report Share Posted July 3, 2004 BlameTheEx: I will get back to you regarding about links to Gravitational red shift… Thanks, but no need. BlameTheEx: …but the ratio of energy between free space and the surface of a body is constant for photons, bricks, cosmic rays or what have you. It has to be. Gravitational potential energy is independent of how a lump of mass/energy is delivered to the surface. If it were not it would disobey the law of conservation of mass/energy. Think. In principle we could have a machine that converts a brick into photons, and visa versa. so we have 2 bricks. One is dropped to the surface, and the other converted to photons, beamed to the surface, and converted back to a brick+ additional energy due to the gravitational shift in the photons wavelength. If that additional energy was not the same as the kinetic energy of the first brick then its not difficult to imagine a system for creating as much energy as we wish from nowhere. Getting the brick back into free space should take exactly the same energy as was freed by dropping it, so we just use the less energy consuming process for the upward journey, the other for the way down, and repeat till all our batteries are charged. Okay, but it’s not clear that is what you meant in your original statements… BlameTheEx: [Gravitational time dilation] is equal to the ratio of frequency of a photon from when it is emitted from the surface, and when it is received outside the gravitational field (Ok,Ok mostly outside. There is no end to a gravitational field), or the for that matter the ratio of frequency gain if the photon goes the other way. This ratio is identical to the ratio of energies for the photon, and naturally enough it's also identical to the ratio of mass/energy (including kinetic) of an object dropped to the surface. Were you saying: a) the ratio of the light frequencies – the starting frequency (distant from the surface) divided by the final frequency (on the surface) - is equal to the ratio of mass to energy of an object dropped to the surface. That is, Fs/Ff = Mf/Ef. Or ;) the ration of the light frequencies – the starting frequency (distant from the surface) divided by the final frequency (on the surface) – is equal to the ratio of starting mass/energy (distant from the surface) divided by the final mass/energy (on the surface). That is, Fs/Ff = MEs/MEf. Anyway, now that you actually explained what you meant, I accept your premise. Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 3, 2004 Report Share Posted July 3, 2004 Being the geek that I am, I have to point out this humorous typo of yours. BlameTheEx: A local observer on the surface of a body with a time dilation factor of 2 will measure a photon as travelling 300 Meters in a microsecond … but the Observer looking down notes 2 microseconds have passed. He can only have one of 2 conclusion. Ether C is half what is expected, or distances on the surface are twice the local measurement. Ether…c, as in the speed of light…get it? Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 3, 2004 Report Share Posted July 3, 2004 Finally found the real issue. I’ve devoted a good amount of time to explaining how you have one concept backwards. Please read through this post several times and try to fully grasp what it is saying before responding. BlameTheEx: As for monkeying with the speed of light, well I am, and I ain't. The usual answer when dealing with time dilation. Its all a matter of the observer. Local measurements will always hold C to be constant, as Einstein maintained. He maintained more than that. Einstein stated that all reference frames in uniform motion will measure the same speed for light, even reference frames that are moving relative to each other. So Einstein does not say that an “outside observer” would measure a different speed for light in another observer’s frame of reference. Both “local” and “non-local” measurements for c are identical. Let’s walk through the general concepts and math of time dilation and length contraction one step at a time. Suppose you take off in a spaceship headed to Saturn and I stay here on Earth. Your speed, relative to me and the Earth, is 0.6c (I chose this speed because it makes the math easy…also, we don’t take acceleration and deceleration into account: we assume your entire trip is carried out at 0.6c). We’ll also “adjust” the distance from Earth to Saturn a bit and call it 806 million miles (it’s close to this value only when Earth and Saturn are aligned). Finally, we’ll round off the speed of light to three significant figures and use 6.71 x 10^8 mph. The two events of interest are your leaving Earth and your arriving at Saturn. Since it is you who has a single “clock” present at both events, your measures of time and distance are called the proper time (Tp) and proper length (Lp). Note that unless explicitly stated otherwise, all observations will be considered from the Earth’s reference frame, regardless whether the observations involve the Earth’s observer (1) measuring things within his own reference frame, where he is at rest, or(2) measuring things as he sees them in the spaceship’s reference frame, which is in uniform, relative motion. For calculating both time dilation and length contraction, we have to use the relativistic gamma factor, which is 1 / ( [1 – (v/c)^2]^(1/2) ). So let’s start by solving for it, using v = 0.6c. First, let’s evaluate the denominator…[1 – (v/c)^2]^(1/2)[1 – (0.6c/c)^2]^(1/2)[1 – (0.6)^2]^(1/2)[1 – 0.36]^(1/2)0.64^(1/2)0.8 = 8/10 = 4/5 Finally, since the relativistic gamma factor is 1 over the denominator, we simply reciprocate the denominator to get the final value of 5/4 or 1.25. So when v = 0.6c, the relativistic gamma factor is 1.25 or 5/4. Now we can solve the equations for time dilation and length contraction. 1) How much time dilation would I see when observing things in your “moving” frame of reference? T = <gamma>TpT = 1.25 Tp So if you had a clock in your reference frame that ticked off one second, the same event, as viewed from my reference frame, would take 1.25 seconds. Using simple algebra, we see that equivalently Tp = 0.8 T. That is, two events that occur in your reference frame and that I see taking 1 second, take just 0.8 second from your own perspective. 2) How much length contraction would I see when viewing “meter sticks” in your “moving” frame of reference? L = Lp / <gamma>L = Lp / (5/4)L = Lp * (4/5)L = .8 Lp So if you had a meter stick in your frame of reference – where it would of course measure 1 meter – I would see it as being only 0.8 meter long (assuming it was laid out lengthwise in the direction of your relative motion). Using simple algebra, we se that equivalently Lp = 1.25 L. That is, a distance in your reference frame that I see contracted to 1 meter is 1.25 meters long from your perspective. Important Concept Quote Link to comment Share on other sites More sharing options...
BlameTheEx Posted July 4, 2004 Report Share Posted July 4, 2004 TeleMad Brief note only as I am in a rush. Your examples of Time Dilation are detailed, and (I ain't had much time to look) correct. Problem is, they are for time dilation caused by motion, not gravity. A clock sitting on the surface of the earth is slower than one in space EVEN THOUGH THEY ARE AT REST TO ONE ANOTHER. We have the effect of velocity, without the velocity. The lorents transform appears to apply, but the logic behind it doesn't. As the clock is moving nether towards nor away from the observer there can be no difference in the time it takes for light to reach the observer between ticks of the clock, and that is the explanation for velocity time dilation. Quote Link to comment Share on other sites More sharing options...
TeleMad Posted July 5, 2004 Report Share Posted July 5, 2004 BlameTheEx: Your examples of Time Dilation are detailed, and (I ain't had much time to look) correct. Problem is, they are for time dilation caused by motion, not gravity. A clock sitting on the surface of the earth is slower than one in space EVEN THOUGH THEY ARE AT REST TO ONE ANOTHER. We have the effect of velocity, without the velocity. The lorents transform appears to apply, but the logic behind it doesn't. As the clock is moving nether towards nor away from the observer there can be no difference in the time it takes for light to reach the observer between ticks of the clock, and that is the explanation for velocity time dilation. But length contraction in special and general relativity are two sides of the same coin. Let’s focus primarily on gravity (i.e., general relativity) and it’s affect on space (ignoring time for the moment, although gravity does also dilate time). We have two observers examining the same set of events, which occur on a massive gravitational body. Observer I is on the surface – “I”nside the system being examined by both observers - and observer O is “O”utside looking down, or into, the system, from a distance. Since O is some distance from the surface, the gravitational field will be weaker for him than it is for observer I. Note that I and O are stationary with respect to each other: no relative motion is occurring between them. Under this setup: (1) Observer O, when peering down into the system on the surface, will see lengths contracted…because gravity is stronger on the surface than out where he is. (2) The greater the difference in the strength of the gravitational field for the two observers, the greater the length contraction mentioned in (1) will be. Here’s how Brian Green explains this (his discussion spans several pages, all of which I won’t type out). ”Einstein appears to have made the next key breakthrough, a simple yet subtle consequence of applying special relativity to the link between gravity and accelerated motion, in 1912. To understand this step in Einstein’s reasoning it is easiest to focus, as apparently he did, on a particular example of accelerated motion. Recall that an object is accelerating if either the sped or the direction of its motion changes. For simplicity we will focus on accelerated motion in which only the direction of our object’s motion changes while its speed stays fixed. Specifically, we consider motion in a circle such as what one experiences on the Tornado ride in an amusement park. In case you have never tested the stability of your constitution on this ride, you stand with your back against the inside of a circular Plexiglas structure that spins at a high speed. Like all accelerated motion, you can feel this motion – you feel your body being pulled radially away from the ride’s center and you feel the circular wall of Plexiglas pressing on you back, keeping you moving in a circle. … Having used the accelerated motion of the spinning Tornado to imitate gravity, we can now follow Einstein and set out to see how space and time appear to someone on the ride. His reasoning, adapted to this situation, went as follows: We stationary observers can easily measure the circumference and the radius of the spinning ride. For instance, to measure the circumference we can carefully lay out a ruler – head to tail – alongside the ride’s spinning girth; for its radius we can similarly use the head-to-tail method working our way from the central axle of the ride to its outer rim. As we anticipate from high-school geometry, we find that their ratio is two times the number pi – [which comes to] about 6.28 – just as it is for any circle drawn on a flat sheet of paper. But what do things look like from the perspective of someone on the ride itself? To find out, we ask Slim and Jim, who are currently enjoying a spin or the Tornado, to take a few measurements for us. We toss one of our rulers to S Quote Link to comment Share on other sites More sharing options...
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