Sorry about the simplicity, I figured I would gloss it over HARDCORE.

the 0 represents the hub, the ) the brake rotor/drum, the | the
outside of the tire.

STOCK
distance of force from hub:
0--5"----//---12.5"

direction of force:
0--up-------down

illustration:
0--)-----//----|

lets say the car is rolling along, and the torque needed to stop the
wheels abruptly (ignoring/subtracting any sort of rotational inertia,
and friction, etc) is 5000in-lb.

that means, we would have to exert a force of 5000in-lb/12.5in = 400lb
(at 12.5" out from the hub) to stop the wheels abruptly.

now, we go in more to where we can stop the wheels, the brakes. at 5"
out from the hub, that 400lb force needed is now (400lb*12.5in) =
(xlb*5in), or x = ((400lb*12.5")/5"), so

x = (5000in-lb)/5" = 1000lb

SO... to stop the wheel abruptly, we need 1000lb of force at 5" out
from the hub, and that's the torque our stock brakes can put out.

when we increase the tire size to a 28" tire:

OVERSIZE
distance of force from hub:
0--5"----//---14"

direction of force:
0--up-------down

illustration:
0--)-----//----|

SO NOW WE GO BACKWARDS!!!! lol~
due to our original assumption of 1000in-lb,
we have just increased the diameter of our tires to 28".

this means that our 1000 lb force (5000in-lb) that was enough to stop
the 25" tire will not be enough to stop a 28 inch tire by the
following equation.

5000in-lb/12.5in = xin-lb/14in, or xin-lb = (5000in-lb*14in)/12.5in

x=(5000*14)/12.5 = 5600in-lb is our new stopping torque needed.

our stock brakes can only exert 1000in-lb of torque, so we won't be
able to stop the wheel abruptly anymore.

hopefully y'all are following me still...
we just increased the amount of force needed to stop the wheels, and
we see that our stock brakes are not enough to do so.

NOW, we need to calculate the size of brakes we need to stop the tire
abruptly (ignoring all other factors).

we take the known stopping torque, and our original brakes (10" brakes
LOL), and see how that compares with our new 5600in-lb
x-in/5600in-lb = 5in/5000in-lb

x = (5*5600)/5000 = 5.6in

so we'd need 11.2" diameter brakes to put out the 5600in-lb torque
needed to stop the wheels abruptly.

which is another inch diameter of brake needed.

I hope my simple explanation (with calculations to show i wasn't just
pulling numbers out of the air) was satisfactory. :)

Sudhir