The OP also states that the float line is 0.5km long. It is, but the actual flotation furnace isn't, most are well under 100m long and the perfectly level line beyond it is constantly drawing the glass through. There certainly isn't enough room in a 62m long furnace (which seems to be the standard length for Stewart Engineering furnaces), for curvature to have any effect. Another misinterpretation by the FE fans...well done.
Except there is ~.5mm of curvature every 6m.
That's a nice attempt at skewing the truth in your favour - even though .5mm in 6m is less than 0.01%, the magnitude of your answer is still more than 100x bigger than the true value.
I think your calculating with the earth's radius at 6371 meters, instead of kilometers.
There are a number of formulas for calculating the curve, most involve trig. But the easiest and closest approximate formula I've seen (for short distances) is:
y = - L^2 / 8R So:
Curve = 6^2/8*6371000 in meters
Curve = 36/50968000 in meters
Curve = 0.7 microns (over 6 meters)
Not perfect, but a small fraction of the glass thickness. Seeing the curve over small distance is very difficult at the human scale. That's why distances of several miles are needed.
(someone please check my math!)
A good approximation (which assumes the distance we are talking about is actually tangent to the surface of the Earth, (as though the Earth was flat) is
h = -R + sqrt(R^2 + d^2) , where R is the radius of the Earth and d is the distance along the Earth (approximate to the real distance along the Earth).
Make sure you use the same units, so if you measure the radius of the Earth in kilometres, then you need to do that for d as well, 1 metre = 0.001 kilometres. This has been used time and time again by FEers in order to skew their results, as shown above in the first quote of this post.
A more ideal answer is to take the distance along the round Earth, but the formula is more complicated for this,
h = R(-1 + sec(d/R)), where sec(x) is the secant function 1/cos(x).
At 6 metres, the first formula gives (using R=6.371*10^6, d=6) h=2.825*10^-6 which is about 2.8 micrometres.
the second formula gives the same result, and only deviates from the first after the 11th decimal place.
For longer distances however, the deviation will become more apparent and the difference approaches infinite as we approach 1/4 of the Earth's circumference.