As usual I totally blew this. Instructor Greenman has told me for 3 years now 
'when you play with numbers you will get burned". 
In the equation it is the velocity not the range ( throw ) that is solved. It 
is rearranged from  R=((2*v^2/g)*sin*cos). A thousand apologies and thank yous 
to yousall

> On Feb 25, 2014, at 12:45 AM, Brad Casterline <[email protected]> wrote:
> 
> From a recent thread I learned a 2.5" Rockwood nozzle would throw a 250 GPM 
> stream a min of 50 feet. This is at the end of all 2.5 equipment with 100 ft 
> of hose. I found an old formula for Range based on nozzle angle and velocity 
> and got 62.5 . So I started thinking about NFPA 5000 and when I overheard a 
> guy in the office reading aloud the part about all areas within 30 ft of 
> spray from a hundred ft of hose---- EXCEPTION  -  calc it! I thought I'd calc 
> it.
> R=(gR/2*sin*cos)^.5. R is range or throw. sin & cos is the angle of the 
> nozzle. I think without air friction the the optimal angle would be 45 deg. 
> But in the <perfect it is 37.5
> So if I needed 50 ft 50 GPM throw from all 1.5" stuff:
> ((32.2 * 50) / (2 * .61 * .79))^.5=40.87 ft/sec
> (Actually we don't even have to know it is 1.5" yet)
> The only way to say we are getting 50 GPM is to get a pitot reading of 11.24  
> psi. This is because h=v^2/2g. When 50 GPM flows thru 100 ft of 1.5" hose the 
> Pf is 18 psi. Anyway we know we need an "end head pressure" of ~12 psi and we 
> lose 18 thru the hose so P=30, Q=50, and
> K=Q/the square root of P=5.48 so it's like a little head right there at the 
> discharge side of the valve. A 2.5 inch T, nip and valve @ 250 loses 10 psi 
> so call K in your calc Q=50, P=40, K=7.9 and calc happily on to the BOR, 
> picking up one more of those if need be!
> 
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