TOS Enterprise Question...

[Cary L. Brown]

CarbonCopy wrote:

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Determining the apparent rate of rotation of the planet is only for gauging how long it is taking the ship to make one lap around the planet; knowing the length of the planet's day let's you know when to say you've made one whole lap, relative to the surface of the planet. The rate of rotation of the planet doesn't determine the gravity well. Does that help?

Actually, NEITHER Of those relates to the gravity well... though, obviously, one is at least in the same general family, while the other is unrelated.

Mercury is a great real-world example of this. Most people don't know this, but Mercury doesn't rotate... AT ALL. So, one side of Mercury is incredibly hot (base-metal-melting hot),while the other side is really quite cold, heated only through conduction through the planet's mass, and radiating into space, so the "dark side" of mercury is very, very cold.

Then, there is the fact that the Earth's rotation is believed to have slowed significantly over its history... I saw in a paleontology bit ("Dinosaurs of Patagonia," on Netflix) that it is widely believed (obviously, we have no PROOF) that the Earth's rotation, at the time of the biggest dinosaurs of that period, was barely over 23 hours.

So... rotation of the planet has no relationship to an orbit, but mass of the planet does. Or rather, there's a fairly straightforward relationship between (a) planetary mass, (b) orbit distance, and (c) orbital velocity.

In fact, there is only one distance from the center of mass of the earth where geosynchronous orbit is possible. (Geosynchronous meaning "always over the same spot relative to the terrain) and even then, that's only possible if the orbital path is directly over the equator... otherwise, the orbit will oscillate from north to south, relative to the ground.

This is likely why Star Trek ships require power to maintain orbit over a landing party site... because they're in a lower orbit, which requires them to be moving at a higher orbital velocity, and thus would not be over the landing site most of the time. So, they perpetually have to provide "corrective thrust" to avoid falling from "orbit."

A REAL orbit can be maintained essentially perpetually, without any need for power at all.

Let's face it... the Enterprise in orbit over a planet will not be moving along anything we'd ever see as a "curved path," merely as a matter of scale. The modeling of the ship's "orbit" along a curved path was one of two things:

1) A production falsehood, "tricking" us without any regard for real science, or

2) The same "representational graphics" that cause us to see ships nearly brushing hulls when dialogue says that they're tens of millions of kilometers apart.

I like #2 better. The path the ship travels is curved, just as we see it. But we're not seeing the tiny pinprick of light that you'd really see if viewing that sequence in-scale. Instead, you're seeing a "computer-augmented image" with the ship expanded for easier viewing.

Just like a naval "fleet command" board uses ship models which would, in scale to, say, the pacific ocean theater, make those ships hundreds of miles long.

[DrBashir]

Actually, Mercury does rotate:

From Wikipedia

For many years it was thought that Mercury was synchronously tidally locked with the Sun, rotating once for each orbit and always keeping the same face directed towards the Sun, in the same way that the same side of the Moon always faces the Earth. Radar observations in 1965 proved that the planet has a 3:2 spin–orbit resonance, rotating three times for every two revolutions around the Sun; the eccentricity of Mercury’s orbit makes this resonance stable—at perihelion, when the solar tide is strongest, the Sun is nearly still in Mercury’s sky.

The original reason astronomers thought it was synchronously locked was that, whenever Mercury was best placed for observation, it was always nearly at the same point in its 3:2 resonance, hence showing the same face. This is because, coincidentally, Mercury's rotation period is almost exactly half of its synodic period with respect to Earth.

[Cary L. Brown]

Ah, fascinating... I had not learned this. When was this discovered?

DrBashir wrote:

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Actually, Mercury does rotate:

From Wikipedia

For many years it was thought that Mercury was synchronously tidally locked with the Sun, rotating once for each orbit and always keeping the same face directed towards the Sun, in the same way that the same side of the Moon always faces the Earth. Radar observations in 1965 proved that the planet has a 3:2 spin–orbit resonance, rotating three times for every two revolutions around the Sun; the eccentricity of Mercury’s orbit makes this resonance stable—at perihelion, when the solar tide is strongest, the Sun is nearly still in Mercury’s sky.

The original reason astronomers thought it was synchronously locked was that, whenever Mercury was best placed for observation, it was always nearly at the same point in its 3:2 resonance, hence showing the same face. This is because, coincidentally, Mercury's rotation period is almost exactly half of its synodic period with respect to Earth.

[CorporalCaptain]

CLB, I don't believe you understand what I'm saying. Hopefully, this will clarify things to everyone. (Yes, Mercury does rotate in the manner described by DrBashir.)

I'll use a particular example with fixed numbers, then please reread everything I've said to make sure you see that this is what I've been saying.

Suppose you know that the planet's sidereal day is 10 hours. Suppose you are orbiting the planet in the same direction as the planet's rotation, in the plane of the planet's equator, and you arrive back over the same spot every 6 hours. That information allows you do deduce your orbital period exactly.

If you are orbiting to the East, then by assumption the planet rotates in the same direction, but it appears to pass under you moving to the West. Measure angular speeds to the East as positive. Let x be your angular speed. The sidereal day of the planet has an angular speed of 36 degrees per hour. Transforming your orbital angular speed to rest transforms the angular speed of the planet to negative 60 degrees per hour. Therefore,

36 - x = -60

So, your orbital angular speed must be 96 degrees per hour. That means your orbital period must be exactly 3 hours and 45 minutes, which is the exact amount of time it takes to go around 360 degrees at the rate of 96 degrees per hour.

So, while of course the rate of rotation that you observe has nothing causal to do with your orbit, how fast you observe the planet to be rotating beneath you constrains the parameters that your orbit can have.


ETA: I see I have a minor typo in what I wrote before:

Are you taking into account apparent the rotation of the planet and the length of its day?

should read

Are you taking into account the apparent rotation of the planet and the length of its day?

[Cary L. Brown]

CarbonCopy wrote:

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CLB, I don't believe you understand what I'm saying. Hopefully, this will clarify things to everyone. (Yes, Mercury does rotate in the manner described by DrBashir.)

I'll use a particular example with fixed numbers, then please reread everything I've said to make sure you see that this is what I've been saying.

Suppose you know that the planet's sidereal day is 10 hours. Suppose you are orbiting the planet in the same direction as the planet's rotation, in the plane of the planet's equator, and you arrive back over the same spot every 6 hours. That information allows you do deduce your orbital period exactly.

If you are orbiting to the East, then by assumption the planet rotates in the same direction, but it appears to pass under you moving to the West. Measure angular speeds to the East as positive. Let x be your angular speed. The sidereal day of the planet has an angular speed of 36 degrees per hour. Transforming your orbital angular speed to rest transforms the angular speed of the planet to negative 60 degrees per hour. Therefore,

36 - x = -60

So, your orbital angular speed must be 96 degrees per hour. That means your orbital period must be exactly 3 hours and 45 minutes, which is the exact amount of time it takes to go around 360 degrees at the rate of 96 degrees per hour.

So, while of course the rate of rotation that you observe has nothing causal to do with your orbit, how fast you observe the planet to be rotating beneath you constrains the parameters that your orbit can have.


ETA: I see I have a minor typo in what I wrote before:

Are you taking into account apparent the rotation of the planet and the length of its day?

should read

Are you taking into account the apparent rotation of the planet and the length of its day?

Ah, that makes a bit more sense now... and is how I understood it all along. I thought you were saying something a bit different than it seems you were actually saying. Given that, I believe we are in agreement.

I'm still surprised that I somehow managed to miss the re-evaluation of Mercury's rotation. I've found several online refs mentioning it, but nothing (so far) telling me when this error was determined. I can say for certain that back when I was in college, (granted, more than twenty years ago) this was the way things were. ( Actually, I graduated college in 1988, so it's more like 23 1/2 years now.)

Clearly, this was discovered at some point between then and now, and I somehow managed to miss the re-evaluation.

[DrBashir]

Cary L. Brown wrote:

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Ah, fascinating... I had not learned this. When was this discovered?

Radar observations in 1965 proved that the planet has a 3:2 spin–orbit resonance, rotating three times for every two revolutions around the Sun; the eccentricity of Mercury’s orbit makes this resonance stable—at perihelion, when the solar tide is strongest, t

[Mysterion]

Wingsley wrote:

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Lots of neat ideas in this thread for what those "thingies" might be.

Here's another one to add to the mix: subspace radio transceivers. We never really knew where Uhura's extraship comm traffic was conducted through, did we? Why not there?

I like this idea.

[blssdwlf]

Just a follow up to this post to confirm that you'd see very minimal change in the background if the camera was moving in a straight line motion as suggested by CarbonCopy. You'd see the planet recede slightly, but the difference would be difficult to detect in the original FX. So the camera's speed relative to the planet could range from stationary to moving at slightly over 15,000 ft/s.

CarbonCopy wrote:

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Using the parameters suggested by blssdwlf, I have plotted a theoretically real orbit for the camera, according to Newtonian mechanics. This solution is offered "as is", but I'll explain my work, so that it can be checked. Note that other alternatives are possible; I am describing only one proposal.



The displayed units are SI, so distance is measured in meters. The planet's surface is red, the camera's orbit is green, and the Enterprise's orbit is blue. The center of the planet is market by a cross, and the point of closest approach between the camera and the Enterprise is marked by a circle on the left. Both the Enterprise and the camera orbit counterclockwise in the diagram. This diagram is to scale.

The planet is assumed to be a sphere that can function as a point mass. The radius of the planet is 6371000 meters and its mass is 5.9736*10^24 kilograms, both Earth values. The value of Netwon's gravitational constant G that I am using it is 6.67384*10^(-11) in SI units. The mass of the Enterprise is assumed to be zero (which is a reasonable assumption for otherwise if its mass were not vanishingly small relatively speaking, the Enterprise would disrupt every solar system it visited).

The Enterprise orbits at an altitude of 6600 miles in a circular orbit, as suggested by blssdwlf. There, it will orbit the planet in 6.122994278617534 hours. It travels at a constant speed of 4843.676699517414 meters per second in the Newtonian inertial frame.

The orbit of the camera that I am plotting is in the same plane as the orbit of the Enterprise. For simplicity, I have chosen the point of closest approach to be the apogee of the camera. At this point of closest approach, the camera, the Enterprise, and the center of the planet lie in the same line, and the camera is 417 feet further from the center of the planet than the Enterprise (which is one ship width, or about that depending on your source; this is done just so that the ship and camera do not collide). To completely determine the rest of the camera's orbit, we need only know its speed at the instant when it reaches apogee. For this value I choose 357 feet per second slower than the Enterprise, in accordance with what blssdwlf has suggested. Using the vis viva equation and slugging things out using the properties of the ellipse and elliptical orbits, we determine that the eccentricity of the camera's orbit is 0.04441833885412092, its perigee has an altitude of 5701.960117206670 miles, and its period is 5.736632021918684 hours. So, although the camera is moving slower than the Enterprise during the hypothetical stock footage in question, it will fall in towards the planet soon and outrun the Enterprise on the next lap around the planet. The camera will even survive that lap easily, in that it will come nowhere near burning up in the atmosphere.

Now for the part which may come as some surprise to some. Over the course of about 4 seconds, the trajectory of both the camera and the Enterprise may be very accurately approximated by constant speed straight line motion. I will leave it as an exercise for the reader to run calculations that will convince him- or herself that this is so in the case of the Enterprise; these calculations are straightforward since the Enterprise orbits the planet in a circular orbit at constant speed. Convincing oneself that this is so in the case of the camera is somewhat more involved which I will now partially describe, hopefully adequately enough to convince the reader.

The speed of the camera is controlled by Kepler's second law: it must sweep out equal areas of the ellipse in equal amounts of time. Although a closed form solution for the motion of the camera does exist, and is within the ability of a very advanced student of first semester integral calculus to grasp, presenting that solution and mathematically proving the point will not really be helpful in this forum. Therefore, I will simply present the following observation: if the motion of the camera is approximately at a constant speed and in a straight line over the four seconds of the shot, then the area swept out by the triangle accumulated along that line of motion between the center of the planet will be equal to the fraction of the area of the whole ellipse that is four seconds divided by the period of the orbit. This is so to 16 digits of accuracy, if one takes the line of motion and speed to be the values the camera has at apogee.

Furthermore, over this four second interval, if the camera is pointed in a fixed direction and not rotating according to the Newtonian inertial frame, then the celestial sphere (on which the stars lie) will appear to turn at most .06532751490506261 degrees, which is hardly at all. However, by tweaking the camera zoom setting, I suppose some apparent stellar motion could perhaps be seen. I say these things only to indicate how rapidly the stars might appear to move; naturally, the camera is free to pivot and point in any direction during the shot, in order to frame the ship in a cinematic manner.

So, therefore and in other words, straight line motion, against a nearly fixed stellar background, is what the camera will appear to see, over the four seconds of the shot.


ETA: The lower bound for how much the celestial sphere will appear to turn in four seconds is .06385944919511107 degrees. The actual amount is in between this value, and the upper bound value given above, but it is almost exactly equal to this lower bound given here.

The lower bound is the value assuming that the camera orbits (hypothetically and non-gravitationally) at its tangent velocity at its apogee. The upper bound is taken according to the orbital velocity of the Enterprise itself, which at all times during the four seconds of the shot is moving faster than the camera. As you can see, these values are very close to each other; the lower bound is .0224723948566697 percent less than the upper bound.

[CorporalCaptain]

blssdwlf wrote:

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Just a follow up to this post to confirm that you'd see very minimal change in the background if the camera was moving in a straight line motion as suggested by CarbonCopy. You'd see the planet recede slightly, but the difference would be difficult to detect in the original FX. So the camera's speed relative to the planet could range from stationary to moving at slightly over 15,000 ft/s.

Excellent. Thank you for checking this.

[J.grey]

Hi, new to the board but not to Trek. I'm also interested in what the 4 glowing things are. Having read the relevant parts of the thread my take is that TOS was pretty poor in terms of continuity and design. I prefer to consider the later works as building on that canon which means that whatever those things are they have to have some kind of relevant evolution in later incarnations of starship. The most obvious conclusion is that they serve the same function as the lighted areas on the refit. The top and bottom have openings directly beneath the bridge and on top of the lower sensor dome which are lit. In semi-canon literature is usually identified as the sensor array. So these could well serve the same purpose, especially as a subspace array where the concept of direction is irrelevant in that the sensor is digging into a completely separate domain which doesn't follow the same dimensional rules as this level of the universe. Making them sensors makes a lot of sense having the additional planetary sensor dome at the bottom being lit in the same way. Perhaps sensors need to be lit for some reason, either the light is a shield of some kind to deflect harmful radiation or particles or sensors in TOS period simply give off light as waste energy. Talking of energy, they could also be part of the shield system or even something to do with the transporters considering that no other external detail deals with this function when everything in TNG did have some kind of detail to represent this important system.
I like to think of them as sensors because it's the only thing that makes enough sense. The dome under the saucer and the four squares lit (I believe they were meant to be lit, just like there was meant to be a grid which was only penciled on) strongly indicate that they function as some kind of sensor network. I also like they idea of them serving as a high energy transmitter, perhaps part of the shield working in tandem with the domes which also nicely explains why the weapons are fired so close from the dome. You already have a high energy power source there and it would be easier to fire energy through directly along the plane of the shield transmitter. The only argument against this is that there is nohigher energy system than the deflector and engines and they don't glow at all (although they were intended to but couldn't due to budgetary restraints. To deal with this to a degree, in TMP they only glowed when functioning at warp as a kind of suggestion that they were glowing on the older version but we just never saw it.) So that kind of upsets the theory of a shield network because they're always glowing and high energy systems aren't glowing at all. The only thing in TOS that always glow is the sensor dome so that's where my vote goes.

[Robert Comsol]

I do feel the issue to be a little more complicated (but definitely favor the skylight interpretation).

In my post # 664 here I illustrated the differences of the upper side of the saucer hull (11' VFX model) between Pike's Enterprise from "The Cage" and Kirk's Enterprise in "Where No Man Has Gone Before" (unfortunately the VFX footage of the revised upper side didn't make it into the episode but we - now - know what was there!).

It think it's safe to assume that the EVA platforms (like in TMP) would have been these rectangular, yellow patches (which the 11' VFX model kept at least near the impulse engines' cover hatches).

The Pike Enterprise had no illuminated rectangular windows and lacked the four upper side white / illuminated rectangles.

The moment the windows were illuminated those four rectangles were added. Considering that these windows apparently generate some "milk glass" effect (i.e. you can't see what's behind them but you probably can look out through these), I'd assume the same would apply for the skylights.

According to the TMP novelization the new rec room seen in the film on the starboard stern of the saucer used to be the sports area on the TOS Enterprise and probably included the gym from "Charlie X".

The gym, however, is a rectangular room that would look odd in the room arrangement of the saucer hull. To assume the white rectangle to be a skylight not only corresponds with the nearby location of the gym but would also justify to have a rectangular room at this location.

Bob

[jayrath]

Lots of very interesting thoughts above, and my thanks to all.

For my part, I would slow down and look at what we do absolutely know. These are areas that either generate light (and also possibly other parts of the EM spectrum), or allow light (and other parts of the EM spectrum) to pass through from beneath. There are three. A fourth is apparently held in reserve or is seldom used.

So they could be skylights, yes (spacelights? ) but what else do we know of from canon that generates energy. Sensors and communications have been mentioned, but both seem to depend on the subspace spectrum, and I'm loathe to think that ordinary photons play a part. We also have deflectors, shields, transporters and tractor beams. Unfortunately, I don't see why any of these would need three big, glowy squares only on top of the primary hull, but all would fit in with holding a fourth "in reserve."

Or maybe the business ends of transporters, for example, are beneath the primary hull, and we're seeing the butt-ends peeking out on top? Just thinking out loud.

Maybe they're art.

[Robert Comsol]

One of my friends who's a lot in jet fighters came over last night and I confronted him with the issue. He told me that jet fighters (like other planes) have position lights but over enemy territory these are (understandably) switched off. However, to avoid collision with your other jet fighters in formation flight they have lateral, flourescent "lightsheets" that are only visible to the wingman next to you.
Of course, leaving the lights on in these rooms with "skylights" (or "spacelights") might serve the same purpose, IMHO.

As for the sensor theory I have the impression these are within the hull and indicated by the three segmented circular lines (two at the saucer's bottom stern, one at the saucer's upper bow - at least on the 'pilot' VFX models), that look like visual warning stripes to me, something along the lines of "Warning! Sensor Emitter Radiation".

Bob

P.S.
I should also point out that there's a similar white and illuminated rectangle on the bow of the pilot Enterprise's elevated bridge dome structure. Personally, I'd like to believe this one to be a panoramic window of the original circular briefing room from "Where No Man Has Gone Before" - which unfortunately we couldn't see in the episode because the camera was facing / shooting the other way.

[FalTorPan]

The squares are push buttons for Apollo's giant green hand thing. Pressing one makes the ship play a sound clip.

[Robert Comsol]

FalTorPan wrote:

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The squares are push buttons for Apollo's giant green hand thing. Pressing one makes the ship play a sound clip.

That's an interesting theory, Greg, but I remember Apollo's fingers to be rather big and the squares apparently don't stick out far enough to be pushed.

Considering that many Star Trek designers had very fond memories of Nemo's Nautilus from "20,000 Leagues Under the Sea" (including Andrew Probert and Greg Jein), we might rather be looking at an electro-shock mechanism that prevents interstellar life forms from attaching themselves to the upper side of the saucer - and blocking the main screen view.