Challenge Questions

"transparent" does that include the showersection of the space craft...?

80% of the sun...1,392,000 KM (865,000 MI.)

I'm sure that somebody can now easily work out what the distance should be with the above info.

so if the engineers names were Janine and Tammy, you could visualize the "problem" better?

Let's assume the radiation shielding magically lets light through. And, we'll assume John is wearing appropriate eye protection so that he doesn't have the world's smallest pupils. Even the 80% is a little ambiguous, but let's assume some reasonable numbers.

Pfft. Oh, darn, lose more astronauts that way.

A GA for you, but raise the circle. You see 50^o up, 70^o down, so the straight ahead gaze puts the disk up a bit.

GA, as anything else means you are already inside the sun. I don't see any point refining your visual detail further.

On the other hand, the challenger appears to be feeding us a gravitationally-massive red herring. The manoeuvre is not merely suicidal, but has no relevance to what might be known as orbital slingshot acceleration either.

Well, a GA in spirit to you too. I was so taken by the position less than radius problem that I didn't even see the straight in "what the heck kinda hyperbola is this" trajectory.

Hi Fyz, I agree about the suicidal part, but I think it is theoretically possible to slingshot the Sun to gain energy relative to our galactic center. Might be useful for inter-stellar probes of the future.

-J

Hi Jorrie

Please reread the challenge:
"John has to fly directly at the center of the Sun until the Sun fills 80% of John's visual field as he stands facing straight ahead on the bridge. At that moment John will change direction"
If John can do that he has no need of the slingshot effect of the sun. However the main point was that the approach described is not compatible with a slingshot method.

Even if it were, the maximum effect of the sun is not actually very large: I think you will find that the maximum velocity change that could be attained in this way (relative to the sun) is in the order of 600-km/sec. The closest stars (at around 5-light-years distance) only have relative velocities in the region of 20-km/sec, so they cannot usefully be the other part of the slingshot system. If they could, at speeds where 600-km/sec is a significant change we would still be looking at time-scales in the order of 2500-years for a single transit. Taken together with the acceleration required to escape the sun from this distance, I think we can safely conclude that John is not human, and we therefore do not know anything about his field of view

Don't forget that one form of an acceleration is a change in direction. They may not be trying to get to another star system, but just to a different part of our own solar system where a direct boost requires an inordinate expenditure of energy (or a very long time). Perhaps they're trying to catch up to a sun-grazing comet. So final speed (relative to the center of the Sun or to the center of the Galaxy) may be less important to them than final direction (relative to wherever they were originally headed).

But yeah, travelling directly toward the Sun, then changing directions at the last minute does not seem like the best way to navigate unless are very confident in their engines and don't mind passing through part of the corona.

This does propose somewhat silly astrophysics. In practice, you calculate where the sun will be, then aim off to the side in the first place, in whatever direction is closest to you so you can get there quickest.

Even that ignores the need to slingshot outside the material of the sun -- if you go too fast, then you cannot slingshot, you just change direction within a diminishing cone of possible outgoing vectors. The sun can be modeled as a point mass, but the point mass model is violated if you go below the surface. Not only do you lose speed from friction, the mass above you cancels some of the mass below you, so you also get less slingshot effect and punch through, slowly. If you have the delta-v to brake, you probably do not need to use the sun as a slingshot. I would think the density and relative placidity of planets makes them safer slingshot pivots.

And we are ignoring flares and other flying material from eruptions!

Aren't there substantial tidal stresses involved in the proximity to su much mass? Humans might not react well to these, either. (See Neutron Star, a story by Larry Niven: http://en.wikipedia.org/wiki/Neutron_Star_(short_story) Larry, sorry I bid against you for the Ring World painting at the 'con. You were behind me, I was a dunce, but art is always a good cause!)

Aren't there substantial tidal stresses involved in the proximity to su much mass

No! Tidal stresses are essentially caused by the difference in gravitational force between the near and far sides of a body from the central mass: a difference of a few meters in 0.7 billion meters is not going to be significant.

I guess it's only a problem with neutron stars -- that, and the overdue bills when you get back.

John has to fly directly at the center.....

I can't reconcile that with doing a slingshot The rocket would have to be going a tad to one side. I'm guessing that as the rocket nears the sun it sort of falls toward it, so John will be facing it, hopefully escaping at it's closest point (maybe on the other side to us).

To further confuse myself, it occurs to me that the moon has a roughly similar angular diameter when viewed from Earth......A moon rocket orbits at about 100km, diameter is 3474......angle subtended from rocket is about 173^o.....That's not so different to the proximity given in the question, and Armstrong et Al al came home. OK, so it's different masses & not slingshot,blah blah blah, but speed was never mentioned in the question.

Yes, I think there is something of a red-herring about the question......

.....not saying nuthin'....just posting a picture

No two peple have the same exact vision, so how can you measure the values of the astronaut's peripheral visual acuity with what is mentioned in the problem?

I agree. What if John has tunnel vision? And to what degree.

What are we supposed to use as guideline for field of view?

Are we missing something?

Jim, this hardly matters now, but can you clarify something for me;

"My take is that anything seen by either eye is inside the visual field. If I approximate the binocular vision region by an elipse..."

The diagram (top right) shows the grey area as "Binocular vision". In that diagram it's the same as the region of stereo vision, seen by both eyes at the same time. Your last diagram uses an ellipse that approximates the total area that can be seen be both eyes (alone and together). Should the quote not read, "My take is that anything seen by either eye is inside the visual field. If I approximate the visual field by an ellipse"? The two lines quoted don't seem consistent. The grey area wouldn't approximate to the ellipse shown (although I'm pretty sure you meant 'binocular' in the sense of 'two eyes', rather than 'both eyes at the same time')

The lack of detail in the official answer makes it almost unworth comment. Your answer seems to be in accord with what the OP intended, and is better shown. I'm just nit-picking terminology in frustration.

What I think I intended to mean by "eyes at the same time" was whatever was visible using both eyes at the same time not what was seen only by both eyes. If I had used either eye would that have meant the area not in gray? I'm not sure.

There is a line in Lewis Carroll's Through the Looking-Glass that I use often. "When I use a word," Humpty Dumpty said in a rather a scornful tone, "it means just what I choose it to mean – neither more nor less.Clearly, my use of language is similar to others but clearly not the same.

Thanks,

Jim

That Humpty quote is one of my favourites too !

I was just on a ramble because or the rubbish official answer. 'Binocular' seems to get used to mean 'stereoscopic' (the overlapping grey bit), though taken literally it only means with 2 eyes (I think !). Since, as TVP45 noted, they appear to be flying upsiode down (whichever way that is), the situation seems confused enough. If I was really going nuts I's suggest the shape should be approximated as two halves of different ellipses. A bit like this;

Hey - look at the scrappy test curves I left on that ! A bit tardy of me not to try to include the 80% Sun, but John & Tom are now merrily on their way. It'd be a horrible calculation to get the exact 80%. using 2 ellipse sections.

Did he turn left or right?

Is he left handed or right handed?

Reaction time is of essence not noting power available.

I think he burnt his hotdog.

I don't think it is possible to have a transparent (defined as allowing light in the visible spectrum to pass through) material that can shield the crew from extreme heat, radiation, etc. A large amount of the total energy emited by the sun is in the visible spectrum which (if the hull is transparent) will pass through the hull and fry anyone on board. I assume it must be partially transparent such that it filters most of the light only allowing a small amount through that will not harm the crew. As far as the distance to get 80% visual field filled by the sun an earlier post has answered that although as has been already pointed out aiming straight at the sun to that distance is not going to slingshot around it.

Yes, you would have to assume it was a light-sensitive partial reflector.
However, the acceleration required to escape from this situation (going straight at the sun and then changing direction) means that we already know that John cannot be human.

Lets assume that Tom has a full screen view of the Sun & he has normal vision. Assume that normal vision gives a viewing angle (incl peripheral vision) of about 160 deg. Then if the Sun is about 1,740,000km Dia. We can work out that the distance for the Sun to fill 80% of 160 deg vision is approx 405,700km which in my books is a little too close to the frying pan.

Now if Tom were a Kielon then that would change the equation & we would not be concerned about how close he was to the Sun. In fact he should be standing on the outside of the ship to get a better view.

It's interesting that in 2120 astronauts will have reverted to dead reckoning and judging navigational distances according to how big something appears in reference to one individual's subjective visual field. Also odd that we will have lost so much knowledge about how to accomplish a slingshot maneuver. Even stranger is that we will not only have lost that knowledge, but also have lost the ability to reason through a plausible technique for slingshotting. What might have happened between now and then to make astronauts so bent on suicide? This problem paints a pretty grim picture of the future.

Well, the calculations mentioned above are concrete, but the circle should not be the sun itself but significantly larger, including the bright halo that one would see ''filling his view''.

Blink, you clearly have not seen the movie "Idiocracy." The answer to your question abou negative technical progress is that the toothless morons are outbreeding the intelligentsia by a wide margin, so that by the year 2505, everyone on earth is an idiot. But apparently, by 2120, they're still smart enough to make a transparent material that is structural enough to endure superhuman accelerations, but still able to block infra-red and ulta-violet light.

I was with you till you said Sine. I think you meant Tangent. I'm using the esteemed SOHCAHTOA technique. In your diagram the distance is the "adjacent" length and the radius of the sun is the "opposite" length with respect to your angle. So I think your equation should be:

tan 72°=695500/Distance

Distance= 695500/tan 72° = 695500/3.08 = 225,811 km

I have no idea whether or not your original angle of 72 degrees was correct, but your reasoning seemed....reasonable. I just don't understand for sure what's meant by "visual field".

What's up Roger? carel's got it right: he's looking for the x in his diagram (which is the hypotenuse) then subtracting the radius.

You appear to be looking for the length of the tangent.

His definition of "visual field" and the way the sun fills it is pretty well as valid as anyone else's.

Hi Randall,

Look at his diagram again. X is not the hypotenuse. You're flying at the center of the sun, which means your distance to the sun and the radius of the circle of the sun that you see make a right angle.

As for his visual field, I agree, his answer is as good as any. What I said was that I didn't know, so I couldn't comment on whether it's right or wrong.

Roger

Wakey wakey! The tangent is at right angles to the radius that intersect it on the circumference. x is opposite the right angle.

(Even visually it is clearly the longest of the lines - remember that the squaw on the hippopotamus is equal to the squaws on the other two hides?)

You wrote:"Wakey wakey! The tangent is at right angles to the radius that intersect it on the circumference."

In this problem, as I understand it, the sun is approximated as a circle (The flight path being perpendicular to the plane of that circle). In that approximation the distance is perpendicular to the radius.

If we did it the way you suggest, the calculated distance would be from the center of the sun, not the surface.

Roger

He calculates the distance to the centre. Then he subtracts the radius to give the distance to the closest point on the surface. Either of those distances could serve - but I've never seen anyone use the tangential distance without a special reason (and explanation).

Fyz,

See post #46 where wikipedia used tangent.

And in the post I was responding to he didn't

a. Draw a sphere
b. Subtract the radius

Meaning my interpretation and correction makes a lot more sense than yours.

I know you don't admit your wrong ever and I could care less if you do admit it, but don't do that "Wakey Wakey" stuff with me, alright.

Roger

I wish I'd read all the updates before I buggered off to doodle.....just for examples sake, I chucked in a 'full' view of the sun as being 140 degrees, then worked back to show 56 degrees.....

Not intended as 'accurate', more to illustrate where the right angle (F,D) is. Bah..... Concept doesn't work anyway - sides of eyes are shaped wrong, can't get 80% field of view by moving suns edge, etc etc.

Here's a link from wikipedia on how to calculate the Angular Diameter of the Sun from the Earth. In that equation the distance is given and they are solving for the Angular Diameter. In this problem we figure out the Angular Diameter and calculate distance. Notice in the problem below Arctan is present. That's because in our problem we are supposed to use Tangent.

That is using the distance to a plane through the centre of the Sun. Note that the angle subtended by the Sun's surface is slightly larger than δ - but when you are as far away as the Earth the difference is small (about a part in 10⁵) - especially in comparison with the precision with which we can estimates of the position of the surface (even if we had an ideal definition for it).

No, it clearly calls it the distance to the Sun, not the distance to the center of the sun in the wiki article. I assume they did it that way because as you noted, at that difference the precision makes little difference.

I know you can get myoptic in your fervor to never be wrong so step back a moment. This started because I corrected a post that drew a diagram with the sun represented as a circle, indicating that in his calculation he should have used Tangent. I was absolutely correct, the way he drew it up, and the math he used, it should have been tangent. Notice in his equation he never subtracts the radius of the Sun.

However.....ok....ready for it?....you are absolutely right (as usual) that the sine version of the calculation should be used at these distances since the actual physical size of the Sun in these calculations, is not trivial.

Now I know there is no chance in hell that you'll admit "you know, I looked at his equation and applied my thinking to it rather than actually read what he said so I corrected you (roger) as if you were correcting me, not him and what he wrote", but that is what happened.

Actually, I'm wrong here. Although he drew it as a circle, he did in fact subtract away the radius of the sun at the last minute as if it were a sphere. I didn't see that.

Fyz, for what its worth, I apologize. I guess when I said that you hate to be wrong what I meant was that I hate to be wrong.

Hi Roger

Apology accepted (I owe you one as well - see below).

We both hate to be wrong. But!!
We have both admitted we were wrong when we realise we were - because (I believe) we both hate to be wrong more than we hate to appear wrong.

Apologies that the "wakey wakey" caused offence. I both thought you would find it funny (perhaps a cultural difference) and that it would concentrate your mind - the latter on the basis that you would realise that I considered any mistake to be at a level that I knew you would not make if you were concentrating.

Regards

Fyz

Hi Roger

I wanted to say "look again" way back, but I see you have looked again.

I also should have said "from the center of the sun" and not just "from the sun" It is confusing. I am sorry.

Kind regards.

Carel

Carel,

I think you were clear enough. I didn't read the last sentence. I can't blame you when I don't read the whole thing.

Roger

Distance= 695500/tan 72° = 695500/3.08 = 225,811

Wouldn't that put the ship inside the orb of the sun by 469,689 km? Might as well go on through.

ed

There are some obvious questions that might be red herrings or might be real solutions, and there are too many for me to sort out, but...

We don't have a good definition of visual field. My ergonmics chart (Henry Dreyfuss Assoc) shows a deifferent number than some I find on line.

They'll come near the sun without a doubt. Will they be going fast enough WRT the sun to narrow the light angle?

They're coming in normal to the sun and they need to have a velocity that is tangent. If they could make infinite accelerations (or nearly so), why bother slingshotting?

How much radiation pressure will there be on this impervious spacecraft and how does that affect approach velocity?

To close. The gravitational pull of the sun would be to strong and they would be pulled into the sun.

diameter of sun 865000 miles, if this = 80% of view, 1/0.8 = 1.25 so diameter of view = 1.25*865000 = 1081250

although most SLR cameras come with 50mm lens, I am told a 35mm lens best represents a human eye.

The resulting isoscoles triangle with a base of 1081250 miles and an included angle of 35 degrees may be split down the middle, giving a right angled triangle of 17.5 degrees and a vertical height of 1081250 miles. The base is the distance to the sun, so this length is given by 108125 divided by tan of 17.5 degrees (0.315298etc)

= 3,429,287 miles. (we are about 91 million miles at the moment)

but I am probably completely wrong. Comets do this trick all the time. The way to fly into the sun is to reduce orbital velocity and let the sun's gravity take over, I can't imagine how John will change direction.

you forgot to halve the vertical height, it should be 540625/tan(17.5 deg) = 1,714,643 miles.

Why are there so many discussions regarding ancillary data which is not related to the root of the problem?

Obviously; this is hypothetical problem, therefore the tangibles of the situation can fall outside the scope of realistic factors. This is presented as a basic word problem, where many factors not relevant to the question being posed are present. It's the observers responsibility to logically remove non pertinent data from the equation, and to ultimately focus on the information which aids in finding a solution.

It seems a great deal of the responses are either facetiously, or unknowingly wrapped around the spoke with irrelevant data presented within the challenge.

Guest is correct. The only things needed to solve this problem are some agreement on the angle that defines the astronaut's field of view and the diameter of the sun. All the discussion about the slingshot maneuver and the transparent material is spurious.

As the GAs (why so few votes?) indicate, jim35848 already dealt adequately with such meat as the question ostensibly contained.

Considering how we might construct a more sensible situation that leads to the equivalent question might be far more interesting than further discussion of the ostensible topic of the question, even if it is technically off-topic*. If you are interested in that, go for it. Otherwise, why carp?

*I don't think believe is necessarily any more off-topic than the apparent answer - because the question specifies a situation that cannot involve a human - so the "challenge" is therefore unanswerable.

....how we might construct a more sensible situation.....

I've found a new toy....my brief foray has not yet produced any slingshots _{or arrows}, but it's outrageous fun. It's 'try for 60 mins, then pay what you will'. The animation is terrific, and it's got a host of good controls.

For anyone interested ; Universe Sandbox

With some time tweaking, it might be possible to insert the galactic travellers path....

Seriously cool link. You should make that it's own post Kris.

Glad you guys like it

I'm usually a bit averse to downloading stuff, but Universe Sandbox is well worth it.

For anyone who's interested, the link gives pretty good info on what it is; a sort of cosmology design tool/toy. It set's you up with an amazing animated universe, and then you can change parameters/create new objects. It all move before your eyes, with ability to zoom/pause/change view angle etc. The thing is set to allow you 60 minutes play, then the author just asks you to pay what you feel is a reasonable amount. I haven't used more than 10 minutes yet, but it's amazingly good. I'm pretty sure I'll give the author some money for it (just for the sceptics : I have absolutely no connection to the author) - it's one of the best simulators I've seen.

I'm not qualified to vouch for the accuracy of the program, but I doubt the guy/gal made many errors on fact with that much effort. It's both a fun program (destroying planets at whim !!), and a great learning tool.

Usual disclaimers apply.

Good suggestion, Roger. I'll post it up in the general section as a discussion. It would be fun to hear more of what others think of the thing.

I've posted it up - it's a bit off-topic to discuss just within this question.

http://cr4.globalspec.com/thread/38090/Universe-Sandbox

On most of the challenge questions I would agree with you, but, on this one the "tangibles" are so vague that it's impossible to guess what the challenger is really asking.

1.) Exactly how is a "visual field" defined.

2.) How do you define 80% of it? Width? Area (in which case you need to find some software package that will "measure" the area you have defined in 1.)?

3.) How do you define the "size" of the sun? It's "visual" size is considerably larger that it's physical size.

OK lets take all the simplest options: visual field is 200°; 80% of it is 160°; the radius of the sun R is 432,685 miles. Then the space ship needs to be:-

R*((1/Cos(10°))-1)= 6675 miles from the surface.

Assuming (against all the implications in the challenge) that 'John' is human, and given that we normally prefer that astronauts have no significant identifiable visual defects, I think we can take "visual field" to be identical to "field of view" for this case.

However, although the lateral field extends to at least 200^O, the vertical field is only about 135^O. I have problems in defining % of the field - is it the image as projected onto a sphere by a logical lens at its centre, or is it the percentage of the retinal areas that are exposed? As the latter still has difficulties, I think that we have to go with the former. In which case, angular view in directions other than vertical and horizontal exceeds exceeds calculations based on a planar ellipse, we should probably take an averge angle for the field of view as slightly more than 170-degrees. Based on the spherical area definition, 80% gives an angular field of about 150^O, which gives about 24400-km (~15200 miles) above the surface, just as jim said.

Refer to my picture at post #14 showing the area of the field of view (supplied by Jim) "filled" by suns subtending 150° and 180° at the observer. The challenge does specifically say "as he stands facing straight ahead on the bridge".

I'm surprised I can't find a free software package that will measure the area of a user defined 2D object. The best I can find is one which allows you to draw round areas in google earth.

Are you taking the planar viewpoint of "fraction of field of view" for this part of the calculation? If so you are not being consistent between the conversion from 200^Ox135^O to an average angle - because on the planar definition 200^O (>180^O) x 135^O is infinite. As 0.8x∞ =∞, that would mean that John would turn just as he reaches the surface of the Sun.

My personal feeling is that it's unsatisfactory to leave the range between 200^O and 180^O undefined in that way, but I suppose you pays no money and you takes your pick...

for this part of the calculation?

For this part of which calculation?

In post #14 I'm assuming that the field of view diagram is somehow representing a solid angle, and, I've tried to superimposed on it representations (both based on the 200° horizontal field) of the solid angles subtended by two suns: one subtending 150° at the observer and one subtending 180° at the observer. I've then observed that the 180° sun probably covers more than 80% of solid angle but looks to me to be closer than the 150° sun. It is just a sort of guess I haven't tried to flatten out the surface of a sphere onto the 2D space.

Now that I look at the original "field of view" diagram more carefully I see that it is not consistent in its mapping of "60 degrees nasally" and "60 degrees above". So my attempt to extend their mapping was always doomed to failure. Perhaps if we moved the central dot, which I had sort of interpreted as straight ahead, down a bit we would get a more consistent mapping.

In post #31 I've just taken the simplest possible interpretation of "field of view", and, what the challenger might mean by 80% of it. This branch of the thread started because a guest was complaining about people going off at tangents: I was just trying to point out that the original challenge didn't contain enough information to keep us (or John and Tom) on course, but, did contain some ludicrous assumptions.

I fear we have been at crass porpoises.

The differences between you and jim are still puzzling me, so I thought I would do it myself, and show the workings as best I can. (Note that we should expect carel's answer to be different, because he has specifically worked with the linear angle)

I'll try to explain how I see it (based on solid angles) - then you can be specific about where we differ.

One way of estimating solid angle is to close-fit a cylinder around a unit sphere, and project the area within the solid angle radially ("radially" being referred to the cylinder axis) onto the sphere. If we place the centre of the visual field along the axis of the cylinder, the height at the 200-degree extremes will be (1-cos(210/2^O)=1.26 and the height at the 135-degree extremes will be (1-cos(135/2^O)=0.62. Because the field along diagonals does not fall as rapidly as might be expected, the average height would be somewhere between 0.94 (the average value) and 1.
80% of total field would then give a height on the cylinder of between 0.75 and 0.8, giving a "time to turn angle" angle that is between 2*cos^-1(1-0.75*0.94) = 151^O and 2*cos^-1(1-0.8) = 157^O.

The distance to the closest surface of the Sun = Rsun(1/sin(turn_angle/2)-1) will then be between 23000 km and 14000-km.

Can't do the arithmetic now, but, I was thinking that if you were going to try to measure a persons visual field you would probably "do it" on the surface of a vertical cylinder centred between their eyes. The mapping from the cylinder to the 2D representation would then be straight forward, but, I'd still need to work out how the cone intersected with the cylinder to map the sun onto the 2D template.

The reason my guess was so much closer (to the sun) than Jim's was because I centred my suns on that dot in the "centre" of the picture, which, kind of "wastes" a lot of the sun's "filling" capacity above the visual field.

I can't now reconcile the picture with the 60° above and 75° below statements. Of course those statements and the picture came from different places, but there is another graphic from the same site as the picture which kind of confirms the Wikipedia statements:-

My "guess" was also obviously wrong because the bottom of my 150° sun did not coincide with the bottom of the visual field.

That picture has been bugging me ; It seems to indicate the extent of peripheral vision when looking straight ahead at a focal point (there is a more correct term/phrase, but I forget what it is) Clearly, if you swivel your eyes, it's possible to see a greater range than 180 degrees. The question doesn't seem explicit about just how 'shifty-eyed' the pilot is. That picture would have been better if they'd not duplicated the numbers on the 'contour' (ie the total conic angle is shown, not the deviation from straight ahead)

I'm feeling very stupid (not for the first time) because I realised overnight that you were (all along) guiding me to consider the part of the sun's disc that does not fall within the visual field. I could perhaps try the excuse that the (irrelevant binocular-vision) greyed area in your original diagram distracted me - but although it might convince someone it doesn't convince me.

Having belatedly got that far, I agree your conclusion that a cylinder with axis 'vertical' to John's head is the most convenient measure. Before continuing, it is worth noting that a 180^O Sun projects a rectangle onto the measurement cylinder. Based on that, I think we can comfortably say that a 180^O angle for the Sun would fill more than 180/210 = 86% of the field of view. Now, as the angle shrinks the rectangular shape only changes gradually towards an oval (I'm not properly defining this shape) before it finally tends to a zero-radius circle, so we can be certain that 0.8*210 = 168^O would fill more than 80% of John's field of view. My guess is that the angle would be significantly smaller than this - but it's no longer an analytic problem, and so outside my CR4 range.

Well, since I'm concentrating on the irrelevant, it occurs to me that as John approaches the sun, he will begin to see the corona stretching out God knows how many solar radii. Since the visible limb defines the size of the sun, we now have a sun with variable size?

Since the eyes together can see almost 180 degrees through peripheral vision, accounting for 'field-of-vision' can be ignored. However, not knowing what the level of light intensity is where human eye can discriminate light from dark when in such close proximity to a 380(10)E24 Watts source might prove irrlevant. They would be unable to see it. It was said that the craft was built to shield from the extremes of heat, radiation, etc., but in the mention of "transparent material" it must be assumed that visible light is allowed to pass. Since it is hard to view the sunlight on earth (which is even filtered) they would both be blind too. But, for the sake of the argument that all of this is 'accounted for' then I must go with the first response of 80%.

I know that some people have greater ranges of peripheral views and therfore you would have to know the size of his specific viusal field.

I personally would like to use something a little more finite then my visual field to make such a risky maneuver. I think a field of view of a camera would do the trick.

Peripheral vision is a part of vision that occurs outside the very center of gaze. There is a broad set of non-central points in the field of view that is included in the notion of peripheral vision. "Far peripheral" vision exists at the edges of the field of view, "mid-peripheral" vision exists in the middle of the field of view, and "near-peripheral", sometimes referred to as "paracentral" vision, exists adjacent to the center of gaze.

I'm no math whizz, but playing around with some rather large (Really big) round objects, If I fill my vision to 80% approx. my nose is already touching the object. Now I know the sun is really really big, but I think it might be too late to make any course adjustments.

Now wait a cotton-picking minute! Somebody already said we were reading too much into this. How can you do a slingshot around the sun? Slingshots are done with relation to the sun, not around the sun.

A welcome contribution to my favourite forum - Carpers Rejuvenated For...

If the year is 2120 AD and John is human, it is unlikely that he would be involved in a slingshot around the Sun, because such a slingshot would first require interaction with an astronomical object that is moving towards or away from the sun.

However, it could be possible to use the sun for a slingshot: if a sufficiently large object were to pass close to our solar system in the intervening years he might use the pair to give greater exit speed into the galaxy than could otherwise be achieved. But he's still need a life expectancy of hundreds of years to reach our nearest neighbours - assuming that the solar-derived velocity was to be a significant proportion of his speed (it might just work if the main speed is due to a hugely massive high-velocity object passing by, and the sun just tickles his direction - but he'd need to use a similarly massive object to turn back).

However, the described trajectory(?) does not correspond to a slingshot - it is merely suicidal.

Back to Mr. Lariton's high school physics class, I don't think it would be possible to get so close as would be needed without burning up.

Correect me if I'm rong.

I have beeen nown to error in the pass.

Ken Leigh

It's all very hypothetical, and (perhaps) a bit of fun. The highest-melting materials known at present melt at around 4500K, somewhat below the 5700K surface temperature of the sun. (I don't really like the term "burning" in this context as it implies oxidation, which isn't a risk in this case).

So your craft would need to be of very specific construction. The surface facing the sun would need to be reflective of light (including UV and infra-red) - but let through enough light for John to see what he needed. The craft would need a rear surface that was screened from the sun, and that was highly radiative ("black"). That should solve the radiation problem, but the craft would be continually bombarded with "gas" at extremely high temperatures (albeit the gas density is extremely low), so you would need a material to transfer the heat rapidly from all surfaces to the radiative back - maybe some sort of heat pipe would do the trick. Plus, the exterior surfaces would need to be robust enough that the energetic gas did not erode it too rapidly. It's also possible that much of the hot gas could be deflected electromagnetically so that it missed the craft. That suggests that thermal control could possibly be achieved using methods that are extensions of known technology. (John would of course need to be lucky - a solar plume or flare would certainly wipe out any craft we can conceive at the moment)

However, although we could in principle control the motion of the craft under 28x Earth gravity, we have no real concept of how we could provide sufficient energy to do this over the distance needed to escape from the Sun - assuming that we had been directly approaching its surface as described. (Rocketry is probably limited at around 1/6th of the required energy - though to date we have always used planetary slings when we needed to exceed about 1/30th; ion drive is very limited as to the short-term force it can apply. Other nuclear-based concepts may be possible - but I think the techniques would need to be different in kind from anything that has been described up to now.)

Maybe you need to find a latter-day Mr Lariton and ask him where he sees the worst of the problems?

The surface facing the sun would

Maybe I'd best go read again*, but that seems to be one point not argued over so far; A comet appears to be always facing the sun because of it's tail. A slingshot rocket doesn't do that _{(takes huge gamble, both feet in mouth...).} At launch from <wherever>, the pilot faces the sun, but he steers to one side. At maneuver point, the sun fully occupies one eye + 60% of the others field of view.....

* Pah,,,,,I'm in my own little world, and my own version makes more sense to me

Now that I'm warming up to what's wrong with this challenge, there is another question. Humans begin getting tunnel vision, i.e., a drastically reduced visual field around 4 g. I can't really figure out what this craft is doing, but I suspect the traditional visual field doesn't count anymore.

"The field of vision is typically around 200 degrees but the acuity over most of that range is poor. For examination of intricate detail, the light must fall on the fovea, and that limits the acute vision angle to about 15 degrees. The total absence of rods makes the fovea a second blind spot since cones have very low light sensitivity."

If John wants to be accurate, he'll have to know when he can actually see the edge of the sun to start with. Anything over 15 degrees, and he's not sure he's seeing the edge. To then get 80% of the (circular) area in view, he'd have to be looking at r x 0.9283.

So.....a half angle, 7.5 degrees, would initially have him (r/tan(7.5) - r) away from the surface. I make that about 6.6 radius from the surface. When the area of the Sun has reduced to 80%, he'll be 6.12 r away.

It's no good John measuring from when he can see the sun out the side of his eyes, or when he can see it with binocular vision. He can't sit around figuring 80% of that overlapping shape, anyway.

He should be able to notice the moment when he can clearly see the edge of the sun, as per the quote. I haven't worked out the angle when he gets to '80% in view', but he could make himself a handy tool;

281200.6 MI

42 Adams units

280796 KM. to the suns surface sorry good thing we driving slow

The Answer will be posted right here on CR4 on May 26th

I hope John and Tom get an answer soon

Here's a transcript of their last transmission:

".....Tom, the Sun looks awfully big from here................so you're sure Tom about 80%?..........no, not miles!...the computer uses kilometers, you converted right?!.......(signal starting to break up)......Tom, you converted right?!!..(starting to lose signal).......No! I said kilometers......oh no, we're scr..........(static)......"

We should regain radio contact with them any moment now.....any moment now....

LOL - right now they're probably wondering how the heck to use the crappy answer they've been given (or cursing Mission Control for lack of clarity that resulted in a fryby) ; "Was that square kilometers or kilometers square?" "Shadup and cut up some more graph-paper, I've got terrible looking piles that I just can't describe". Must be where the expression, "flying by the seat of your pants", comes from.