I hope you mean you based your answer just on the title of the OP.
-FrL-
Well, yeah. I’m on vacation and the wine is sweet…what can I say? 
I actually work with this sort of thing. Contrary to what Orbifold says, the speed of light in a vacuum is the conversion factor, as other people have said, and there is a physical reason for it.
Let’s say you and I are standing on a plane (infinite flat surface). We have two odd compasses, and each one gives a different direction for “north”. We want to find out how far a post is from our current position. I use my notion of “north” and “east” to find out how far north and how far east from us the post is, and call those numbers n[sub]M[/sub] and e[sub]M[/sub]. You use your notions of "north and “east” to get the coordinates of the post and find numbers n[sub]F[/sub] and e[sub]F[/sub]. Since your notion of north and east are different from mine, our coordinates for the post disagree.
Now we each try to calculate the distance using the Pythagorean theorem. I calculate
d[sub]M[/sub][sup]2[/sup] = n[sub]M[/sub][sup]2[/sup] + e[sub]M[/sub][sup]2[/sup]
while you calculate
d[sub]F[/sub][sup]2[/sup] = n[sub]F[/sub][sup]2[/sup] + e[sub]F[/sub][sup]2[/sup]
What do we find? Our calculations agree! It doesn’t matter what we each call “north” and “east”: our notions of “distance” are the same.
Now when we do this in spacetime we each have a notion of “right”, “forward”, “up”, and “forward in time” (x, y, z, t). If we’re at the same point in spacetime and we want to look at some other point, we’ll each get different spacetime coordinates for that point. However, there’s something that we will agree on: the “spacetime displacement” or “squared-distance” to that point.
d[sup]2[/sup] = x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - c[sup]2[/sup]t[sup]2[/sup]
To bring time displacements into the formula, we have to convert our time measurements (in seconds) to space measurements (in meters). The conversion factor is exactly c. No matter what x, y, z, and t directions you and I pick separately, we will always agree on the result of the above formula.
The weird thing about time is that there’s a minus sign in the formula for spacetime displacement rather than a plus sign like for the spatial coordinates. That has deep consequences for the geometry of spacetime, but the formula makes clear that to consider space and time on the same footing we have to use c as our conversion factor.
How many grams in a cup?
See the problem? They’re units of measurement; metaphors, when you come down to it; not actual things. In effect they’re like adjectives, not nouns. You need a noun to modify, or your adjective has nothing to hang on. You need something to measure, not just the abstract measurements themselves.
It’s like you’re asking, “How big is red?”
“Big” and “red” only exist in relation to the thing they describe.
This is complete nonsense. A measurement is not a metaphor, and it’s not an adjective. A meter is a very well defined amount of spatial displacement, and a second is a very well defined amount of temporal displacement.
The reason “how many grams in a cup” doesn’t work is that grams measure mass and cups measure volume. Mass is not the same as volume. What Einsteinian relativity tells is is that spatial displacements are the same as temporal displacements, and the conversion factor between a spatial unit and a temporal unit is c, in those units.
How many meters in a second? c, in meters/second. How many miles in an hour? c, in miles/hour. How many light-years in a year? 1.
I have no personal expertise in this area, I just find the question interesting,
but my “life partner” happens to be a mathematician (a tenured research professor in Riemannian Geometry) who was also a Physics minor in college, who had this to say via a terse e-mail reply after I brought the subject up (Orbifold and Mathochist, please don’t shoot the messenger) and after scanning the postings so far:
I don’t understand how one could say Orbifold is being ignored. But that’s a nitpick.
I thought with Mathochist’s post I had finally found salvation, but now I believe myself as ignorant as I believed myself before. 
Please, everyone, continue your altercation for my amusement. 
-FrL-
I am describing Minkowski spacetime because I don’t want to complicate the matter with curvature when it’s not an essential part of the description. Spacetime is locally Minkowskian, and that’s what we’re talking about. Orbifold asserts that “there’s no physical reason to say that 1 second is 299,792,458 metres.” I’m asserting that there is a compelling physical reason. Technically it’s involved in measuring the lengths of tangent vectors, which do form a Minkowskian space. I alternately get yelled at for going into too much detail and get yelled at for streamlining my explanations around here. This time it’s the latter.
Still, the upshot is that you can’t just use any old conversion factor to put time and space on an equal footing in the formula for spacetime displacement. I can’t say 1s = 1m and get a sensible result. I can’t say 1s = 80m and get a sensible result. Once I choose a unit to measure lengths and one to measure times the conversion factor must be the speed of light in those units.
d[sup]2[/sup] = x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - c[sup]2[/sup]t[sup]2[/sup]
Of course there are other formulas that we’ll agree on, as Orbifold mentioned in another post
t[sup]2[/sup] - x[sup]2[/sup]/c[sup]2[/sup] - y[sup]2[/sup]/c[sup]2[/sup] - z[sup]2[/sup]/c[sup]2[/sup]
calculates the square of the “proper time” between two events, as opposed to that of the “spacetime displacement” that I’ve been talking about. But still we’re using the speed of light in a vacuum to convert. This time it’s just converting distance to time. You’d have to be a lunatic to say that the conversion factor “3” only converts from feet to yards and can’t be used to convert yards to feet. It’s the exact same thing, just run backwards.
Oh, and I am well-versed in geometry and theoretical physics, though knot theory and category theory are my stock-in-trade.
Incidentally, robardin, I’m fascinated by coincidence. I’ll try to probe this without being too public: are you in computer science yourself?
I am not in academia and so would not classify myself as being “in” computer science. I certainly don’t do research in the field or anything like that. I do, however, have an undergrad degree in CS and work as a developer and manager of software analytic tools in the financial sector.
Has the IASMAA (It’s A Small World After All) theory received a boost?
Unfortunately, no. I was thinking of a geometer I know whose “life partner” is an academic in a computer science department.
Quite frankly I think my replies to Pasta have already addressed the points you bring up, particularly my previous post in this thread.
The nit that Orbifold brought up and which my original replies were neglecting is that nature doesn’t offer any physical constraints connecting time and space measures. The physical laws are symmetric w.r.t. interchange of x and y so there is a naturalness to asserting that a meter in x is the same as a meter in y. However, no such physical symmetry exists for x and t. You write
d[sup]2[/sup] = x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - (ct)[sup]2[/sup],
with c=299,792,458 m/s. Equating 1 s with 299,792,458 m reduces the expression to
d[sup]2[/sup] = x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - t[sup]2[/sup],
but nothing stops us from equating instead 1 s with 1 m, reducing the expression to
d[sup]2[/sup] = x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup] - (299,792,458 t)[sup]2[/sup].
There’s no physical reason to pick one over the other. (Cleaniless of the math is a motivation, just not a physical one.) And no physical prediction of the theory depends on the choice.
11 Bonner Units.
(Did I do that right?)