Physics Discussion Thread

[negi]

Rsinghji's question is very similar to the example taught in class IXth(?) Physics , i.e. if you are sitting in a moving train and toss a ball up in the air vertically will it land on your palm or will it land behind it ? Many of us who traveled in 2nd class all our lives (man those were the days, as kids we used to look forward to the 36+ hour train journey to Dilli and then a 12 hr leg to Mussoorie ) have actually validated this i.e. the ball will land in your hand as it retains the train's velocity along with the vertical velocity imparted by the hand during the toss and if the air friction can be neglected it will land at exact same location . Basically in absence of air friction there are only 3 things at play here , forward component of velocity which is same as train (in absence of air friction this remains unchanged) , initial velocity in the vertical plane imparted by hand toss and 'g'. Now only if train were to accelerate/decelerate after the ball was tossed will the latter land in a different location.

[Vayutuvan]

negi: A non-accelerating train is an inertial reference frame where as a rotating frame is not inertial. In rotating frames (uniform or not), source-less forces like Corialis force arise. I understood rsingh's question to be whether a non-inertial frame is different from an inertial frame for this particular problem.

These forces determine the path that is seen by an observer who is in the rotating frame itself. If the observer is in an inertial reference frame outside, my understanding is that the Coriolis force is non-existent. In other words, Coriolis force is an artifact of the observer in a rotating reference frame (which is not inertial).

The following 3D visual is not too bad as a first approximation of earth.



Looking at the picture, say we perform a Gedanken experiment. A rocket is sent straight up at the north pole. Symmetry dictates that it should come down right on top of the launch point. Now let us move down along say the prime meridian to the south pole and our experimental rocket would fall back at the south pole. At equator, the rocket would fall back on the launch pad but this time the rocket would take parabolic path in the plane of the equator to an observer who is an inertial frame, i.e. not on earth. As the experiment is conducted at each point as one moves along the prime meridian from N. Pole to the equator (as the other case of moving from S. Pole to the equator is reflection symmetric, we need to only consider only this half of the thought experiment), one can visualize the shape of the trajectory to be parabolic for an observer in an inertial reference frame and the projection to be a straight line on the surface of the earth connecting the starting and ending points of the launch pad.

[Theo_Fidel]

Interesting arguments.

Technically a rocket firing its motor is not following a parabola right. For a pure parabola, meaning ballistic mode, would mean only gravity acts on initial velocity imparted. Hence a rocket fired straight up in hover mode will need a small forward thrust to match up with the pad during hover and then a small reverse thrust to land back on the pad. Without the horizontal thrust rocket will fall further and further behind. And without reverse thrust rocket will land forward as angular momentum accelerates it.

What am I missing.

[SriKumar]

So the resultant outward normal velocity will increase if the rocket engine is imparting acceleration (assume constant for now ...) and which opposes earth's acceleration due to earth's gravity g (again assume all gravitational forces due to all other bodies are absent as a first approximation). So when the rocket stops firing, one of the three things will happen - rocket falls back to earth, rocket continues to go away from earth, or it orbits earth.
If the rocket falls back to earth, then it should fall back on the pad. Is that correct?

The question gets more complex, let alone the solution. Assuming that the rocket goes vertically up from a launch pad, up to a certain height and the engines stop and the rocket falls back to earth, I believe that it will not fall back on the launch pad. I expect it to fall a little west of the launch pad (since earth rotates from west to east). Here is the reason (as I see it): Neglect atmosphere and its role to keep things simple. The rocket is under the influence of 2 forces- its own thrust and gravity. In addition, it has a tangential velocity of 1600 km/hr, which it gets from the surface of the earth where it launches from. Assume that this rocket goes up to a height of, say, 6000 km. At this height, the rocket's thrust turns off. Gravity pulls it towards the center of the earth. Now, the rocket has a tangential velocity of 1600 km/hr. The launch pad also has the same tangential velocity. However, when the rocket ascends to that height, it is moving tangentially in an orbit with a larger radius. This means that if it travels with the same tangential velocity as earth surface, it will 'fall behind' relative to the launch pad.

Best way is to imagine 2 concentric circles, one with radius 6000 km (this is earth surface, say) and another with radius twice that (this is the height to which the rocket ascends before it falls back to earth). For the same angular sweep from center of circles, the arc on the inner circle is smaller than on the outer circle. So, any object moving on the outer circle has to move farther and faster than a moving object on the inner circle (=earth) that it wants to stay on 'top' of. I think this makes sense but I could be wrong...this is just a guess based on some very basic geometry.

(The train analogy, IMHO, does not apply here. The train is moving horizontally, and not on a curve).

[negi]

^ yes it is complex for instance for relatively low ceiling height reached by the rocket the train analogy will hold , no ?

It is humbling at times to see how even such apparently simple questions get our knickers in a twist .

[Vayutuvan]

Theo and Srikumar: I think I see your point. OK back to some geometry (similar triangles?)

[Vayutuvan]

It is humbling at times to see how even such apparently simple questions get our knickers in a twist .

negi: Reason why I don't like simple Physics Too hard to get one's head around - especially when the geometry is 3D (or higher) and complex (polar or cylindrical). OTOH, it is easier to program the solution to a more general model and solve all these problems as special cases. Of course, no intuition intuition is developed.

[rsingh]

^ yes it is complex for instance for relatively low ceiling height reached by the rocket the train analogy will hold , no ?

It is humbling at times to see how even such apparently simple questions get our knickers in a twist .

If Gurus have knickers in twist what about mortals like me. This question was on back of my mind for long time. I have biology and dukandari background..........so very difficult to understand this. But I observe things in daily life and therefore have some strange questions from time to time

[ramana]

Bade, AmberG et al, Should we open this thread out of GDF now that it has enough posts and survived this long?

[SaiK]

<aap>
I am trying to understand if the rocket climbs vertically up 90* to gnd, and has no capability to continue to escape gravity and shuts down, no atmosphere aspects considered here, .. now how will this direction be tangential to the sphere ?

If this is tangential from any point, the flight path should be parabola no matter N/S or equator. ..may be I missed something big reading so many some high flying posts.
</aap>

[Theo_Fidel]

Please leave this here. It is meant for mostly non-physicists to mess about with physics.
No aspirations for the super hi funda of the maths thread.
Bade saar must get many a laugh from our fumbling’s.
To be honest, I also don’t want my ignorance to be paraded to the world.

[Amber G.]

Bade, AmberG et al, Should we open this thread out of GDF now that it has enough posts and survived this long?

Ramana - Didn't see this, so I put a similar request in feedback thread. I think this thread should be out of GDF. (BTW, As a lurker I don't even go to GDF that often so have not even seen this thread for a long time)

[Amber G.]

A follow-up question would be: will a rocket show the same behavior? A rocket is not constrained by atmosphere and can fly well out of it. The answer is 'yes'. At lift off, it has the same initial rotational velocity as earth. It will go straight up and into space, (and if it is not affected by any other force or heavenly body, of any kind ), it will come straight down where it took off.

Not really, (unless it is shot from north/south pole). A question similar to this was asked before,
Here: (few posts above and below): http://forums.bharat-rakshak.com/viewto ... cs#p517270

Look up coriolis force/effect

(This was an example when one drops a ball from high altitude it falls slightly "eastward" .. neglecting air etc - this displacement is almost negligible - about a few centimeters if dropped from a height of a Km.)

(Another way to look at it, if you take inertial frame - say center of earth- of reference, the horizontal component of velocity is rw ( r=radius of earth, w is angular velocity).. as rocket goes higher up (say (r+h) from center of earth), the rocket's angular velocity is no longer w but wr/(r+h)))
HTH

More details here:http://forums.bharat-rakshak.com/viewto ... 13#p517613

[Amber G.]

Srikumar: Yes, you are right. The tangential velocity would be whatever it was at the surface (ignore friction due to air for now). You are also right about the atmosphere having nothing to do with the problem. Bad wording though wanted to convey that as an indirect measurement of the gravitational force decreasing with increasing distance..

So the resultant outward normal velocity will increase if the rocket engine is imparting acceleration (assume constant for now - though as the fuel burns off the mass reduces and if the engine is burning fuel at a constant rate the acceleration actually increases) and which opposes earth's acceleration due to earth's gravity g (again assume all gravitational forces due to all other bodies are absent as a first approximation). So when the rocket stops firing, one of the three things will happen - rocket falls back to earth, rocket continues to go away from earth, or it orbits earth.

If the rocket falls back to earth, then it should fall back on the pad. Is that correct?

I hope we understand what rsingh means by "going straight up"? In which frame of reference? There are two frame - the rotating launchpad frame and the non-rotating center of the earth frame. I think that makes a difference too, in that in the launch pad frame, we can ignore the tangential velocity component. Using this definition for "straight up", the rocket will fall back on the pad.

Added Later: Want to make explicit the implicit assumption I made - that launch pad is on on the equator, i.e. the axis of rotation of earth is normal to the plane of the great circle.

PS: The above in red is not required as long as the rocket is fired exactly normal to the surface of earth.

PPS: I was trying to brush up my understanding of reference frames - inertial and non-inertial. Wikipedia articles on the topic are very confusing. What is intuitive is messed up using some loose phrases like "deflect to left/right". I really hate all this visualization

To be clear, if I am not misunderstanding ... answer to all such questions eg:

If the rocket falls back to earth, then it should fall back on the pad. Is that correct?

Is NO. (even assuming no air effect ,and the launch pad is not on the north pole).

(Simple physics, though it will require a little calculus. The answer is different than what most see in a typical physics book where the rocket (or ball) goes not too high, and one can neglect the horizontal deviation due to coriolis force)

[SriKumar]

A follow-up question would be: will a rocket show the same behavior? A rocket is not constrained by atmosphere and can fly well out of it. The answer is 'yes'. At lift off, it has the same initial rotational velocity as earth. It will go straight up and into space, (and if it is not affected by any other force or heavenly body, of any kind ), it will come straight down where it took off.

Not really, (unless it is shot from north/south pole). A question similar to this was asked before,
Here: (few posts above and below): http://forums.bharat-rakshak.com/viewto ... cs#p517270

Look up coriolis force/effect

(This was an example when one drops a ball from high altitude it falls slightly "eastward" .. neglecting air etc - this displacement is almost negligible - about a few centimeters if dropped from a height of a Km.)

(Another way to look at it, if you take inertial frame - say center of earth- of reference, the horizontal component of velocity is rw ( r=radius of earth, w is angular velocity).. as rocket goes higher up (say (r+h) from center of earth), the rocket's angular velocity is no longer w but wr/(r+h)))
HTH

More details here:http://forums.bharat-rakshak.com/viewto ... 13#p517613

AmberG, I had corrected myself on this point by concluding that it will not land back on the launch pad in the following post (dated 15th January). It would land westward from the launch pad.

http://forums.bharat-rakshak.com/viewto ... 3#p1576743

[Amber G.]

Hope people do not mind, A few comments.. mine are underlined and inside "***"
(Hope this helps)

So the resultant outward normal velocity will increase if the rocket engine is imparting acceleration (assume constant for now ...) and which opposes earth's acceleration due to earth's gravity g (again assume all gravitational forces due to all other bodies are absent as a first approximation). So when the rocket stops firing, one of the three things will happen - rocket falls back to earth, rocket continues to go away from earth, or it orbits earth.
(***The simple answer - it follows a conic section - ellipse/hyperbola , center of the earth as one focus*** *** most likely an ellipse if the velocity at the end of rocket firing- is less than about 11Km/sec and hyperbola if it greater than that***)
If the rocket falls back to earth, then it should fall back on the pad. Is that correct? (*** NO, where ever the "elipitical orbit" meets the earth. if it touches the earth surface it will be some distance away from the pad and make a big splash, if it misses the earth than it will be nice satellite in elliptical orbit, if the velocity is greater than 11 Km/sec (but less than 60Km/sec it will just revolve around sun like another planet ***

The question gets more complex, let alone the solution. Assuming that the rocket goes vertically up from a launch pad, up to a certain height and the engines stop and the rocket falls back to earth, I believe that it will not fall back on the launch pad. (*** NO, see above ***) I expect it to fall a little west () of the launch pad (since earth rotates from west to east). Here is the reason (as I see it): Neglect atmosphere and its role to keep things simple. The rocket is under the influence of 2 forces- its own thrust and gravity (***NO, after rocket stops firing, only gravity, no other force ***). In addition, it has a tangential velocity of 1600 km/hr (*** about 1600 Km/hr, less if it is NOT launched at the equator ***) , which it gets from the surface of the earth where it launches from. Assume that this rocket goes up to a height of, say, 6000 km. At this height, the rocket's thrust turns off. Gravity pulls it towards the center of the earth. Now, the rocket has a tangential velocity of 1600 km/hr. The launch pad also has the same tangential velocity. However, when the rocket ascends to that height, it is moving tangentially in an orbit with a larger radius. This means that if it travels with the same tangential velocity as earth surface, it will 'fall behind' relative to the launch pad.

Best way is to imagine 2 concentric circles, one with radius 6000 km (this is earth surface, say) and another with radius twice that (this is the height to which the rocket ascends before it falls back to earth). For the same angular sweep from center of circles, the arc on the inner circle is smaller than on the outer circle. So, any object moving on the outer circle has to move farther and faster than a moving object on the inner circle (=earth) that it wants to stay on 'top' of. I think this makes sense but I could be wrong...this is just a guess based on some very basic geometry.

(*** Easiest, IMO, is to use polar coordinates, elliptical orbit equation and just let the launchpad move in a circular orbit with angular velocity of 1 rotation per 23h and 56 minute.. There is no easy mathematical way than this, IMO ***)

(The train analogy, IMHO, does not apply here. The train is moving horizontally, and not on a curve).

[Amber G.]

...

AmberG, I had corrected myself on this point by concluding that it will not land back on the launch pad in the following post (dated 15th January). It would land westward from the launch pad.

http://forums.bharat-rakshak.com/viewto ... 3#p1576743

Sorry, I did not see it before. In any case, I hope this is helpful to others.

BTW, I have seen these things confusing many.
(For example, a simple looking, but harder question, (generally not asked to freshman physics students but rather graduate students) is "One throws a ball at 40 m/s straight up at latitude x, what will be horizontal deflection?")

Practically, this math is needed for long range snipers and guns.

One famous example: in Falkland war British gunners missed because they forgot to adjust their computers attached to the guns, for southern hemisphere!!!
(Deflection for long range guns depends on latitude and if the place is in Northern or Southern Hemisphere.)

[SriKumar]

^^^^ Acutally...I am glad to have gotten confirmation. I was actually a bit unsure (having reversed myself on it). I need to absorb some of the details on the shape of the orbit (ellipse/parabola), thanks for the details on that. But I did want to clarify a couple of things...when I said the rocket was under the influence of 2 forces (thrust and gravity) I meant when it was on it way up. When the engine turns off, yes, it is only one force.... .

About the 1600 km/hr, yes that is at the equator, some of my earlier posts had the launch point at equator and hence quoted the 1600 kmph number....and I did not carry the equator part over into other posts.

At the risk of provoking another topic of discussion , the post was on trying to 'prove' that the earth's atmosphere rotates with the earth, and whether this is this true at all altitudes of atmosphere from sea level (experience suggests this is correct) to say, even at 30 km height?. Am neglecting all effects from convection or other effects of solar heating ...only whether the act of earth rotating forces the atmosphere to rotate with it.

Changing topics a bit, here is a comment from a post made by you a few months ago. Could you give some examples of 'great' or elegant mathematical theories that explained some parts of a physics phenomena but diverged from observations in other aspects.
http://forums.bharat-rakshak.com/viewto ... 3#p1526263

To me, beautiful part of physics is that no matter how beautiful a mathematical model is, if it does not fit the experimental results, it has to be discarded.

[Amber G.]

It is humbling at times to see how even such apparently simple questions get our knickers in a twist .

Negiji, not only humbling but some times very embarrassing to gunners ... for example this from history books... (where one does not take coriolis force into account).. as said before I find the reference..

>>During an embarrassing battle , British battle cruisers engaged two German warships, at a range of nearly ten miles, near the Falkland Islands, but forgot to reverse their Coriolis correction. The British gunners at first couldn't figure out why their artillery was falling astray. They had adjusted the guns. But instead of setting them off to the right to account for the left turn of the Coriolis force in the southern hemisphere, they set them off target to the left, like they did in the northern hemisphere. So, the missiles ended up missing two times more than had they not made any adjustments.. Ultimately, the British eventually won the battle with about sixty direct hits, but not before more than a thousand shells had fallen into the ocean.
>>>

[Amber G.]

Okay, Here is a "simple" problem...(looking for a quantitative answer)

A person threw a ball in exactly vertical direction in Chennai (lat 13 degrees N) with a velocity 40 m/s (about a fast bowler in cricket can achieve).

Assume there is no air resistance.. ball is a point (has no spin etc..), (consider only the spin of earth's effect)

When the ball lands back to ground,

How far it will land from the starting point? In which direction (east or west etc)?
(Assume reasonable values for "g", spin of earth, etc..)

(Problem inspired by the helicopter problem given above)

[Amber G.]

Changing topics a bit, here is a comment from a post made by you a few months ago. Could you give some examples of 'great' or elegant mathematical theories that explained some parts of a physics phenomena but diverged from observations in other aspects.

One most noteworthy example comes to my mind is Michelson Morley experiment. Considered as one of the most famous "failed" experiment. There was this beautiful theory about properties of the aether, all set to be confirmed and provide a way to measure the absolute velocity of earth in this medium... This elegant theory has to be abandoned because experimental results diverged. (And M &M had courage and honesty to report the actual results - which they thought were failures at that time)

(Of course, later Einstein came with the theory of relativity and Michelson won a Nobel instead of going into obscurity due to a failed experiment)

[Vayutuvan]

rsingh and others:

One link (among many others I am sure, but these seems to have adapted across the globe from high schools to universities - a lot of them in US) that has a lot of free textbooks on Physics and Calculus that I had known for sometime and have downloaded the PDF versions is the following:

Light and Matter site of Prof. Benjamin Crowell

books
Light and Matter -- physics for students majoring in the life sciences
Simple Nature -- physics for scientists and engineers, with a nontraditional order of topics
Mechanics -- introductory mechanics for scientists and engineers, with a traditional order of topics
Conceptual Physics
Calculus
Special Relativity
General Relativity
Purcell, Electricity and Magnetism -- an open-source edition of this classic book, which was developed under an NSF grant and is now available royalty-free

open-source software
Planet Finder - an applet that shows the locations of the planets, stars, moon, and sun in the sky from any location and for any date and time
Spotter - a program that lets students check their answers to math and science questions
OpenGrade - software for teachers to keep track of grades
When - an extremely simple personal calendar program, aimed at the Unix geek who wants something minimalistic

The site also gives a list of institutions who have adapted these books. Hard copies are available from lulu for print on demand.

Disclaimer: I do not have any relation to either Prof. Benjamin Crowell or Lulu.

[SriKumar]

One most noteworthy example comes to my mind is Michelson Morley experiment. Considered as one of the most famous "failed" experiment. There was this beautiful theory about properties of the aether, all set to be confirmed and provide a way to measure the absolute velocity of earth in this medium... This elegant theory has to be abandoned because experimental results diverged. (And M &M had courage and honesty to report the actual results - which they thought were failures at that time) (Of course, later Einstein came with the theory of relativity and Michelson won a Nobel instead of going into obscurity due to a failed experiment)

thanks. was aware of the expnt but not the theoretical constructs to derive the existence and properties of aether.

[Amber G.]

^^^ Thanks.

[SaiK]

http://www.theguardian.com/science/2013 ... sics-prize
was watching the repeat show on the breakthrough prize presentation awards on sci chi.. man they make it bigger than academy and grammies.. never any sdres could expect such tfta presentations like this.

perhaps the best way to make them real heroes!

[Bade]

^^^ It is a good thing that a large majority of the million dollar prize recipients have donated all their prize money to other younger folks struggling to do science. It does tell you a lot about people who do science unlike people who are in academy and granny races. Nothing to compare with.

In a way it is good that a multimillion dollar check is given to some prominent person, who can then decide how to spend it on science activities with a larger reach. The prize givers are in not the best situation to evaluate what is best for science.

Nobel prize money being a smaller amount, usually does not get re-distributed like this as far as I know. Maybe there were exceptions among the Nobelists too, but did not receive wide press.

[Theo_Fidel]

Seems like a small amount though. The NSF disburses $6 Billion in science research grants every year. So this is not even a drop in the bucket. Not sure what the purpose is, other than letting google give some money away.

[MurthyB]

Another ripoff in the making?

This Indian physicist disproved black holes 13 years before Hawking

A new paper released late last month in which famed British physicist Stephen Hawking contradicts his own theory and says that Black Holes - in the real sense - do not actually exist has startled the world science community.
But Abhas Mitra, a theoretical physicist at the Bhabha Atomic Research Centre (BARC) in Mumbai, is not at all surprised. "I said more than a decade ago that the Black Hole solutions found in Einstein's General Theory of Relativity actually correspond to zero mass and are never formed. This implies that the so-called Black Hole candidates must be Grey Holes or quasi-Black Holes," Mitra said, adding, "Hawking is saying the same thing now."



Read more at: http://indiatoday.intoday.in/story/step ... 41835.html

[rsingh]

Intersting article in latest Time edition about quantum physics. Very informative but hyp (imo).

[Vayutuvan]

NIF produces 17KJoules - a skiier doing 38 MPH downhill. Two orders of magnitude more energy than input. No idea whether it is big news or just pne of those hyped results which keep popping up every now and then - and yes in Nature.

Laser bombardment yields energy milestone

[SriKumar]

So, indulge me ...to ask some kosins. Choo mantar, kaali kalanter, Jai Ho Jantar Mantar.

The following is a picture of some small step-like constructions in Jantar Mantar of Jaipur. It is called Rashi Valya Yantra (=instrument). Supposedly (per a nice little document I read on the antar-rashtriya antarjaal), what it does is to point the locations of various rashis (12 zodaic constellations, and not the 27 nakshatra constellations). The kosin is: first, is that correct, and if so, how does it work? How to use it to point to a solar zodiac constellation. This nice little document mentions that the sun goes through 6 (out of 12) solar constellations in a single day- to an observer on earth. So, when one of the 12 step-like structures does not cast a shadow (or casts a minimal shadow), that step structure would point to the zodiac constellation. Each of the 12 step-like structure is associated with a specific zodiac constellation, and each is oriented a little differently from the rest. They all point northward, generally speaking-google maps.


What it is not clear to me is, how does the step-like structure point to a zodiac constellation? If you will notice, the slope is in a direction away from the sun (look at the shadow to know where the sun is). And we know that the zodiac constellations are all 'behind the sun', i.e. along the ecliptic path. So, the direction of the slope of the step-like structures point towards a northerly direction, almost parallel to the axis of the ecliptic. This does not suggest it would point to a solar rashi constellation. What am I missing here. Also interesting is that each of these steps has its own sun-dial scale (the curved structure below the steps.). Perhaps the curved structure points to the zodiac? If so, what is the significance of the slope of the steps (they are all somewhat different from one another).

Any comments on this, or, any description on how to use these step-like structures to see stars (of the astronomy kind, not the Deepika Padukone kind) would be bahut shukriya appreciate.

[Amber G.]

^^^ I saw those structures first time when I was quite young, and has seen/observed them in details many times.. (Explained many things to many people as I am quite familiar with these - I have taken all my sons and we have spent literally hours there)

Let me answer your question.. If you are familiar with basic astronomy the following will be quite easy to understand.

These measure sidereal time. (The main big structure measure apparent solar time -- as you know solar day is 24 hours but sidereal day is 23 hours 56 minutes -- IOW knowing both sidereal and solar time one can know exact position of the sun in the ecliptic)

Actually all 12 structures combined is just one instrument. To use it:

First find out which "rashi" the sun is - You can do this by looking at:


Now you go to that particular "yantra" from above picture (one of the 12)

And look up the curved scale of that instrument -- lines up with the (1/12) part of the ecliptic that contains the constellation)

If more you want to read more just google "Rashivalaya Yantras" or something like that.

[SriKumar]

thanks. Did not know that it was associated with the jai prakash yantra. I'd have to say...detailed technical explanations on how to use them seem to be few on the web (I could be wrong...but I did look, a lot was about construction and the stone used). I did find this nice, detailed work done by 4 under-grad (?) students from the NUS, Singapore. They did a good job trying to document what each instrument did... http://www.math.nus.edu.sg/aslaksen/gem ... mantar.pdf
This document makes a concerted effort at document the complex, its history, individual structures, their function and use of all the yantras. Am reading this...the desciption of how to use rashivalaya is a bit different (in the sense that it does not associate it with jai prakash yantra).

[Theo_Fidel]

Srikumar,

Those are the birth constellations IIRC. You know Capricorn, Sagittarius, etc, not sure what the hindi names were but it is marked on each one. Also IIRC there are guides on how to use it right on the platform. Way back when, folks were allowed to walk right up the steps and check it out. Something I had the good fortune of doing. Now you can’t even get close to 100 meters for good reasons. There are lines and arcs and marked segments that allow you to trace each item of interest astronomically, through time, dates, etc. At least there used to be, though the markers were quite worn down even back then. The aim was for even average folks with simple equipment to make quite accurate measurements and predictions.

[Amber G.]

thanks. Did not know that it was associated with the jai prakash yantra. I'd have to say...detailed technical explanations on how to use them seem to be few on the web (I could be wrong...but I did look, a lot was about construction and the stone used). I did find this nice, detailed work done by 4 under-grad (?) students from the NUS, Singapore. They did a good job trying to document what each instrument did... http://www.math.nus.edu.sg/aslaksen/gem ... mantar.pdf
This document makes a concerted effort at document the complex, its history, individual structures, their function and use of all the yantras. Am reading this...the desciption of how to use rashivalaya is a bit different (in the sense that it does not associate it with jai prakash yantra).

Hello SriKumar, Thanks for the nus link, I will read it later. Hope it is good because there are not many good sources like this.

Few comments:

- Yes, I found that most of the "nice documents" (which one can buy) as well as "guides" at Jantar Mantar were, to put it generously, sloppy. Lot of mumbo jumbo (sansrit words, trying to give impression of deep knowledge) but scientifically vague or even outright silly/incorrect. If one is familiar with basic astronomy, knows math (basic college level - spherical trigonometry etc)and really understands the motion of sun/earth and some routine stuff, it is fairly easy to understand how these things work (or were designed to work). One only has to think logically. This is true for any scientific instrument. (For an electric engineer, it is easy to figure out how an old oscilloscope was used ).

What I found very strange is that many such guides will show-off, without understanding even the most basic principles. I knew, for example, a well respected "jyotisi" who could not even recognize a nakshatra in actual sky. (I was talking a walk with him, noticed the bright Aldebaran, and remarked about it, he had NO idea, when I pointed the name in Hindi, he knew the "rohini" name, of course, but could not identify this in actual sky!!!)

My father's hobby was to make such instruments for fun so it was easy to understand and very interesting when I visited Jaipur Jantar Mantar for the first time - My father explained just the bare minimum, I can figure out rest... Last year I visited Jaipur with my son and DIL and we had a good time. We spent a lot of time there. Again my son was able to figure out virtually every stuff . For the your yantra his remark was "Oh, so this measures the coordinates of the sun in Ecliptic coordinate system". This is basically correct. (BTW, one can go and approach these yantrans closely)

As you know, please note that sentences like "

point the locations of various rashis (12 zodaic constellations

" or "

This nice little document mentions that the sun goes through 6 (out of 12) solar constellations in a single day- to an observer on earth. So, when one of the 12 step-like structures does not cast a shadow (or casts a minimal shadow), that step structure would point to the zodiac constellation.

Makes LITTLE sense. NO they do not point to Rashis.. SUN does NOT go through 6 solar constellations in a single day" (This is actually so absurd/sloppy that it is obvious that the author has NO idea of even the basic science, and he is mistranslating some stuff). Sun spends about a month in each Rashi, and it takes a whole year for it to go through the whole circle... In fact, those 12 structures do not represent Rashis in "normal" sense. (I used rashi in quotes in my previous post, just to represent 30 degrees in a circle - nothing more)

As I said before, one can measure "sidereal time" or "longitude of the sun" (in ecliptic) just as the big sun-dial uses Equatorial system (and measures Hour angle of the sun). Because the Z-axis is fixed in equatorial system (with respect to ground) and not in ecliptic -- you need only one sun-dial, more than one (12 in this case) in the other case -- each pointing to the z-axis, depending on the time of the day..

[Amber G.]

Way back when, folks were allowed to walk right up the steps and check it out. Something I had the good fortune of doing. Now you can’t even get close to 100 meters for good reasons.

One can still walk right up to these. (I visited it last year) (Click for larger nice image)

[Amber G.]

Also found, Virtual tour of the above:
http://www.jantarmantar.org/JaipurTour_2.html

at http://www.jantarmantar.org
Which has more resources, virtual tours etc...

[Amber G.]

Dup removed

[Amber G.]

From MIT News - A story about a recent article in Physical Review Letter (probably the most prestigious journal)..
Physics of Curly Hair

And another story BBC:
Neutrino beam 'major physics experiment'

[Abhijit]

Amber ji, could I request you to send me an email at ast dot commerce at gmail? Thanking in advance.