Warp 1 equals the speed of light, c (300.000.000m/s), warp 9 equals about 1500c. As you can see it's a logarithmic scale and it's rigged so that warp 10 equals infinity.

Warp 9.9 is equal to about 4 billion miles pers second according to the Warp Factor scale here; en.memory-alpha.org/wiki/Warp_factor

The function v = c*(9/(10-w))*exp(0,64*(w-1)) produces those numbers.

195.169.213.92 19:09, January 28, 2012 (UTC)

Unfortunately the writers don't always stick to this convention, in many situations a ship will use impulse and seemingly go faster than light. At the closest it takes light (or radio waves) from Earth almost 35 minutes to reach Jupiter, I believe at least one Enterprise did it in significantly less at impulse. --StarkRG 23:03, January 28, 2012 (UTC)

If one defines "fast" as linear velocity, warp drive is not technically "fast" in that it does not produce linear velocity through normal space. If one defines "fast" in terms of absolute time, warp drive is "fast" in that measurable in terms of *objective *time elapsed is the absolute time interval between which a warp-driven vessel exists at one definable point in normal space, until it next exists at another definable point in normal space.

Warp drive has nothing to do with the propagation of the "wavicles" known as photons, which, in a hybrid fashion possess both the properties of particles and of electromagnetic waves, and which results in their linear velocity through normal space, at least as the properties of light are known to current physics, other than that they propagate through normal space and through subspace in the same fashion.

Einsteinian theory dictates that all matter subjectively experiences "time dilation", that is, an "accelerating" difference between the elapse of *objective *time versus the elapse of *subjective *time, the closer it approaches a velocity or propagation equivalent to what is considered "the speed of light".

This does not happen with certain particles, such as those of solid matter which comprise the mass of a starship, according to *Star Trek *physics. Thus, *subjective *time shipboard, and *objective *time in normal space, elapses at a parallel rate, even though for the duration a starship is at warp, it, and everything within the warp field, does not exist in normal space or time, and this greatly defies Einsteinian theory, because of these and other contradictions between Einsteinian physics and *Star Trek *physics.

Warp drive generates a warp field around an object, and it is the interaction of this warp field with and through subspace that results in the existence of a warp-driven starship at two vastly different points in normal space at two different points in time with an objectively-measurable interval and an equal, subjectively-measurable interval occurring--that is, where warp drive is not employed to create a difference in the elapse of *objective time* versus that of *subjective time*, in order to produce time travel.

According to *Star Trek *physics, the term "warp factor" does not express a particular level of intensity of a warp field, measured in terms of the unit *cochranes*, although there must be some kind of measurable and calculable variance between the two. It expresses a result. A starship captain does not ordinarily speak so familiarly with *Star Trek *physics concepts in mind that he would give an order to "generate a warp field of *x* number of cochranes such that this interval of time elapses between immediate existences at these two points in normal space", although he could. Instead, the shorthand and consensually-understood meaning of the term "warp factor" is how a starship captain, conceiving velocity the way the viewer does, orders the control of the *objective* and *subjective* time interval between controlled, immediate normal-space existences of the starship, and it is up to the ship's flight officers and engineers to make it technologically and physically happen.

Thus, any warp factor does not technically equal a velocity, in terms of the "speed of light" or in terms of "multiples of the speed of light", or in like terms. These are terms of concept, of expression, of explication, and of convenience, used to avoid explanation of the true principles of warp drive, essentially ignoring both *Star Trek *physics and Einsteinian theory.

The relativistic effects of Einsteinian theory are also unaccounted-for in *Star Trek* physics as to the functioning of impulse drive.

--ProfessorTrek 23:53, January 28, 2012 (UTC)

- While you are correct that the ship does not travel through space in a continuous line as we would consider it. However you could also determine velocity by the equation ((time arrived - time departed)/(location of arrival - location of departure)), in fact this is what most people would determine velocity by. (of course it's technically average velocity) In this sense warp factors do, in fact, have velocities. Due to the possible variations in space topography a given warp factor may result in different speeds depending on the space the ship travels through.

- Also, while a vessel does not experience relativistic time effects as we define them today, time doesn't necessarily pass at the same rate within a warp field. Once a ship is in orbit of a planet it is likely travelling at a different velocity than its origin and therefore time will pass at a different rate. This was the reasoning behind stardates and time beacons. --StarkRG 01:12, January 29, 2012 (UTC)

- I'm sorry, nobody who is correct would determine velocity through that. What you're referring to is

*x*= (*T*)_{a}- T_{d}/ L_{a}- L_{d}

- or

*x = ΔT / ΔL*

- (which perhaps is the inverse of the most accurate expression you seem to want to make)

- where Δ expresses "change" as the above "change in time divided by change in location".

- If the interval between
*T*and_{a}*T*is expressed in hours, and the distance between_{d}*L*and_{a}*L*in miles, this would result in_{d}*x*being equal to*- T*hours divided by*- L*miles, or*x*hours / mile, or*x*hours per mile.

- If the interval between

- However,

- Velocity = distance / time

- or

*v = d / t*.

- Thus, if

*d*is expressed in miles, and

*t*is expressed in hours,

*v*is expressed in miles / hour, or miles • hour^{-1}

- or, in common parlance, "miles per hour".

- You may have conceived your point one way, but you did not express it consistently with that. It seems that you may have had everything reversed. What you may have wanted to express was

*x = ( L*/_{d}- L_{a}*T*), which results in_{d}- T_{a}

*x = L*_{d - a}/ T_{d - a}

- and could be described by

*x = ΔL / ΔT*

- wherein if
*L*is expressed in miles, and*T*in hours, would indeed result in*x*miles / hour or*x*miles per hour.

- wherein if

- What I am talking about is

- where
*n*denotes the dimension of normal space

- where

- and

*s*denotes the dimension of subspace

*L*cannot vary equally with_{n}*L*._{s}

- but

*T*_{n}*does*vary equally with*T*._{s}

- Thus, even where

*V*_{n}= L_{n}/ T_{n}

*V*_{n}*cannot equal**L*/_{s}*T*_{s}

- and

*V*_{s}*only*equals*ΔL*/_{s}*ΔT*_{s}

- reckoned according to subspace physics and not normal-space physics, that is

- neither
*ΔL*nor_{s}*ΔT*represent a linear change through normal space and time, and_{s}

- neither

- where
*ΔL*and_{s}*ΔT*are reckoned, if at all, necessarily non-linearly in terms of normal-space physics, but according to subspace physics._{s}*V*probably does not describe or imply a parallel relationship with_{s}*V*._{n}

- where

- Sorry.

--ProfessorTrek 02:41, January 29, 2012 (UTC)

- You're right, I got that backwards, distance over time is what I was looking for. In any case the point I was making was that there is a velocity corresponding with warp, however the relationship between warp factor and velocity is not clear, nor necessarily constant. Additionally there would not likely be acceleration as we know it (the derivative of velocity would be infinite at the moment of engaging warp). --StarkRG 03:13, January 29, 2012 (UTC)

Yes, but my calculus is this: both propositions necessarily require ignoring the fact that upon initiation of a warp field, a starship is no longer in normal space, and thus does not and cannot propagate through normal space.

Thus, *in terms of normal space*, although *ΔL _{n}* and

*ΔT*do rationally exist, but only immediately before entry into subspace and immediately after exit from subspace,

_{n}*V*cannot exist. This is because there is no linear

_{n}*d*, wherein at any given instant, the starship can be found existing in normal space; that is, there is no

*Δd*.

_{n}Thus, warp speed, in terms of *v _{n} = d_{n} / t_{n}* , cannot be calculated, because

*Δd*does not exist.

_{n}--ProfessorTrek 03:50, January 29, 2012 (UTC)

---

- Look, I realize you're probably just a sad little man in your parent's basement or whatnot which explains why you're trying to make yourself look smart by using overcomplicated math equations to explain extremely simple concepts, but going back and adding "nobody who is correct would determine velocity through that" after I already admitted my mistake is rather spiteful.

- In any case there is a
*d*, the distance between the origin and the destination, that the ship didn't exactly pass through the intermediate space is immaterial. The distance between is finite and non-zero, the time is also finite and non-zero, thus the velocity is calculable. --StarkRG 03:57, January 29, 2012 (UTC)

Yes, subjectively (as it pertains to you) immaterial. But certainly not irrelevant.

I wrote that *before* you admitted your error--and I cognized and recognized the difference between what you most likely *meant* and what you *said*. It's impossible to derive the quality of "spite" from that.

Actually, I am neither sad, nor do I reside with my parents. I stand 6'4" tall. You might want to check with a psychiatrist, because "realizing" such a conclusion could only result from a reality perception deficit on your part, because it is impossible for you to have such ideas about me arising soley from the discourse herein, or any assessment of probabilities thereto, on its basis.

Yes, there is a *d*. But there is no *Δd*.

--ProfessorTrek 04:02, January 29, 2012 (UTC)

Anyone who bothers to check the history would see that, in fact, you added that *after* I admitted my error. --StarkRG 04:04, January 29, 2012 (UTC)

**Look, there is no***Δd*. There is a*d*, but it is only*subjective to the individual observing from normal space and where that individual ignores any instantaneous check as to the existence of the ship at any and all points on the shortest vector between those two points in space, demonstrating its lack thereof*. There is no*d*to either (1) the external observer conceiving the existence of subspace, the starship's presence in it, and checking its instantaneous normal-space linear presence; nor (2) to the internal observer._{n}

- Asserting that there is a
*d*ignores the fact that there is no*Δd*--and*d cannot exist*where there is no*Δd*. The only thing that can exist is the two defined and definable points in space, and hence the distance, d, between them. But there is no*d*, as in*V*, because_{n}= d_{n}/ t_{n}*there is no*Δd.

- Asserting that there is a

Sorry.

--ProfessorTrek 04:14, January 29, 2012 (UTC)

How can you say there is no Δd? The ship was in one place, now it is in another, that seems like a change to me. --StarkRG 04:16, January 29, 2012 (UTC)

- It is a subjective and contrived change--the perception and assertion of which ignores vital and essential facts. The existence of
*Δd*necessarily requires instantaneous existence at any and all points on the vector, that is, through measuring a*continuous and not continual*progression of instances, which*cannot*occur, because there is no progression to measure. That is why there is no*Δd*.

- It is a subjective and contrived change--the perception and assertion of which ignores vital and essential facts. The existence of

- While it is a change of position to an observer in normal space, it is not, to an observer in normal space, a traversal along the shortest vector between the two points. That is why to the observer there is a
*d*, but there is no*Δd*, and it is because there is a "why" that I can assert this.

- While it is a change of position to an observer in normal space, it is not, to an observer in normal space, a traversal along the shortest vector between the two points. That is why to the observer there is a

Sorry.

--ProfessorTrek 04:20, January 29, 2012 (UTC)

---

Look, first of all stop going back and altering your previous answers. I could just as easily go back and fix my earlier mistake but I don't because that would be disingenuous.

Seconly, my point was that as there is a change in position over a finite and non-zero time you can calculate an equivalent velocity. --StarkRG 04:26, January 29, 2012 (UTC)

- Yes, I would agree with that. However, "equivalent velocity" is not "velocity", and thus one cannot describe "how fast the ship went".
**The quality of velocity does not accrue to the starship, because it does not propagate through normal space. The quality of "equivalent velocity" accrues only in the mind of the conceiving observer in normal space.**

- Yes, I would agree with that. However, "equivalent velocity" is not "velocity", and thus one cannot describe "how fast the ship went".

As Data would say: "Could you please continue the petty bickering?" Thanks.

This is not petty. This is not bickering. This is simply an expression and assertion of the logical veracity of concepts.

If you ignore the existence of the laws of *Star Trek* physics, and observe a ship at point 'A' and then some time later observe a ship at point 'B' and can be assured that it traversed the shortest vector between point 'A' and point 'B' in the shortest amount of time, then your conception is valid. Otherwise, it is not. This why the *Enterprise-D* has MTBF (mean time between failure) units accruing to its operation, and not a meaningless measurement of how many light-years an external observer thinks that it has travelled.

It is not that you are incorrect in your assertion. Do not think of yourself as being told by me that "you are wrong". It is simply that you are not sufficiently correct in your assertion.

--ProfessorTrek 04:37, January 29, 2012 (UTC)

- Re-adding my comment which seems to have been deleted for some reason:
- How it gets there is not the issue, one definition of "speed" is distance over time. There is distance covered, and there is time, therefore there is speed. You seem to be arguing semantics now, insisting on a particular definition whic, in this case, doesn't fit. Additionally, if you continue to edit your previous posts (beyond returning to add your signature, which I will occasionally do) I will stop participating in this discussion. --StarkRG 04:37, January 29, 2012 (UTC)

All right. I will try to put it in your terms, from immediately above: there is no "distance covered". You insist that there is "distance covered", I show you why no distance is being "covered". And yet you insist that there is "distance *covered*" Why?

---

- Again, you're insisting on using a definition of speed which clearly doesn't fit. I said it was in one place now it is another, that is the distance travelled, what route it took to get there (whether through a wormhole or a pocket of space caused by warp fields) is immaterial. Take that distance and divide it by the time it took (either from the point of view of an observer on the origin, destination, or the ship itself) and that is the speed d. If I gave you a picture of a boat leaving a dock, then gave you a picture of the boat 1km away from the dock that was taken 5 minutes later you'd surmise that it travelled an average of 12km/hr. You don't know how it got there (maybe there was a wormhole or a teleportation device) and, in fact, it doesn't really matter to this question. The question was how fast was its journey. --StarkRG 14:07, January 29, 2012 (UTC)

I already defined "fast" as linear velocity, and not as elapsed time. According to these facts, even according to *Star Trek* physics, elapsed time simply means elapsed time, and a "derived velocity" cannot be derived based on it, as you insist.

Even according to physics as we know it today, when an object exists at point 'A' at one time and then exists at point 'B' at a subsequent time, that does not necessarily mean that the object has traversed the shortest vector between points 'A' and 'B'. There could have been any number of points of inflection constituting intermediate points, along other than the shortest vector between point 'A' and point 'B'. Even ignoring that possibility, or controlling for it, *Δd*, meaning "change in position", does not refer to the sole difference between an object's existence at point 'A' and an object's existence at point 'B' at a subsequent time, as you insist.

What you are referring to is simply the difference between two distinct and measurable points within a three-dimensional Euclidean plane, which can be expressed as *d _{B} - d_{A}*.

*Δd*, by definition, requires the *continuous existence* of the object *at any and all points* along the shortest vector between *d _{A}* and

*d*.

_{B}According to *Star Trek* physics, a warp-driven object that before entering subspace exists at point *d _{A}* and then subsequently exists at point

*d*after exiting subspace has no

_{B}*Δd*, because it does not exist at any and all points along the shortest vector between point

*d*and

_{A}*d*, nor along

_{B}*any*vector, for that matter.

I cannot make it any clearer than that. According to *Star Trek* physics, *Δd* ≠ *d _{B} - d_{A}* because, on these facts, and according to

*Star Trek*physics,

*Δd*

*cannot*equal

*d*. This is because

_{B}- d_{A}*d*≠ the length of

_{B}- d_{A}*d*.

_{A}→ d_{B}That said, you *can* say that a warp-driven object "takes" *x* number of hours, minutes or seconds to "get" from *d _{A}* to

*d*. You just cannot quantify the warp-driven object's velocity,

_{B}*v*, during the time it exists at

*d*until the time it next exists at

_{A}*d*, because there is no

_{B}*d*.

_{A}→ d_{B}Sorry if any insult is perceived by you immediately hereinabove.

--ProfessorTrek 01:13, January 30, 2012 (UTC)

As a physics professor, I've always wondered what "F" students in science and math do after they drop out. Thanks to the above, now I know. Oh yes, sorry if any insult is perceived by you immediately hereinabove.

- You're right, I shouldn't even try.
- Such power over of the fate of others. But the reality is that their failure is your failure.