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OK, I understand, Φ is based on Θ in this figure where they calculate hydraulic radius: http://www.ajdesigner.com/phphydraulicradius/hydraulic_radius_equation_pipe.php[^]. If I follow you, Φ = (Π - Θ)/2.
Since it's building codes that are specifying the method to use, then you have little choice but to do that, unless you have a licensed engineer come up with a good reason to do something better. Manning's equation with the empirical friction factor equivalent is probably conservative for the purpose, and that's OK.
I'd guess that the iterative solution is your best bet.
CQ de W5ALT
Walt Fair, Jr., P. E.
Comport Computing
Specializing in Technical Engineering Software
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Walt Fair, Jr. wrote: I'd guess that the iterative solution is your best bet.
we are all converging now. 
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I made this, but it's not an analytical answer..
static double Solve(double y)
{
double x = y;
double error = double.MaxValue;
double prevError = double.PositiveInfinity;
while (error != 0 && Math.Abs(prevError) > Math.Abs(error))
{
x -= (x - Math.Sin(x) - y) / (1 - Math.Cos(x));
prevError = error;
error = y - (x - Math.Sin(x));
}
return x;
}
It doesn't work for zero, and gives infinity for y < 0.00000001054, but otherwise it seems alright.
The input is supposed to be your value for 4R/D and the output will approximate Φ.
I can't guarantee that it will always terminate, but it did for every value I tested (which included negative numbers, and insanely big numbers, and very small numbers)
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do I detect a Newton-Raphson smell around here?
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Probably, because that's what it is 
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Thanks! That should be a good start. I may have a shortcut to make it converge faster, though. I'll have to think about it.
The idea is to model (one situation) a sewer line that is existing, to which a new customer - a car wash, for instance - wants to connect. I'd like to then force a change in gpm inflow, then see what happens to everything else. Since the gpm is a function of the area and velocity of flow, that's one equation. Then the other can be used to provide a bound for the next guess at Φ until the two equations converge to a single value, +/- some tolerance value.
My head hurts; I think I'll bang out something easy - a database program to inventory the entire electrical system maybe - while I'm cogitating. 
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
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That went right over my head..
Maybe I'll think about it some more but I have other things to do as well.
Good luck 
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Before you start optimizing things, may I suggest you observe the current behavior. In my experience, equations that model macroscopic physical phenomena are well-behaved and tend to converge quite well. So just log each iteration, and you will probably notice a few iterations will bring your error down to less than 0.1%, which is probably much better than the accuracy of the formula anyway.
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x - sin(x) = y
x = pi/2*(y/(pi/2-1)^(1/3)
accurate to 3.8% for 0 <= y <= pi/2-1
follow this by one NR refinement (for y > 0)
x1 = x0 + (y - x0 + sin(x0))/(1 - cos(x0))
reduces relative error to less than 1.5 x 10^-3 over this range
two refinements gets relative error down to 2 x 10^-6
three refinements gets relative error down to 5 x 10^-12
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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That's beautiful, Peter. Thank you! God bless math wizards! 
[EDIT]
A question:
The term (2-1) appears often. That makes no sense, so I'm guessing that it's an HTML encoding error. Can you clarify, please?
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
modified on Thursday, June 24, 2010 2:56 AM
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Actually a slightly more inspired approximation is
x - sin(x) = y
let q = pi/2
calculate b = (q^3/(q-1)-6)/(q-1) approx equal to 1.384269281
then
x = ((6 + b*y) * y)^(1/3)
is accurate to 0.47% for 0 <= y <= pi/2-1
follow this by one NR refinement (for y > 0)
x1 = x0 + (y - x0 + sin(x0))/(1 - cos(x0))
reduces relative error to less than 2 x 10^-5 over this range
two refinements gets relative error down to 4 x 10^-10
re the 2-1, just evaluate is as your compiler would, so pi/2-1 = (pi/2)-1
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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How are you making these approximations?
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harold aptroot wrote: How are you making these approximations?
for x - sin(x) = y, for small y
x = (6y)^(1/3)
the first approximation is simply taking this form and making the value correct at the two ends of the interval, y in (0, ym) where ym = pi/2-1. x=0 for y=0, so we only have to make x=pi/2 at y=ym, so
x = pi/2 * (y/ym)^(1/3)
This gives about 4% error.
Noting that the scale factor is wrong at y=0, the second approximation is of the form
x = ((6+b*y)*y)^(1/3)
which is like x = (6y)^(1/3) for small y, and we choose b so that x=pi/2 for y=ym. This gives less than 0.5% error.
Add this to a simple excel spreadsheet to do a few NR iterations and calculate the errors and you are done.
I'm sure that you could improve things given enough time.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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cp9876 wrote: x = ((6+b*y)*y)^(1/3)
which is like x = (6y)^(1/3) for small y, and we choose b so that x=pi/2 for y=ym. This gives less than 0.5% error.
I'm sure that you could improve things given enough time.
Using the same b and ym as before, put
x = ((6+b*ym*(y/ym)^a)*y)^(1/3)
Run Excel's solver to optimise 'a' gives a = 0.731889437 for a relative error < 0.033%
One NR iteration reduces this to 1 x 10^-7
OK, I'm going to stop now....
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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How did you know what "form" the formula should have?
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harold aptroot wrote: How did you know what "form" the formula should have?
if you are asking where the
x = (6y)^(1/3)
comes from, it is just a first order expansion around y=0
Just to summarise for Roger, the best approxmiation so far:
let q = pi/2
calculate b1 = (q^3/(q-1)-6)
ym = q-1
x = ((6+b1*(y/ym)^a)*y)^(1/3)
with a = 0.731889437
gives a relative error < 0.033%
One NR iteration reduces this to 1 x 10^-7
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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Excellent! I have no clear idea about how it is derived, but that won't stop me from using what works.
When you refer to a first order expansion, what is it you're expanding? Taylor series?
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
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Roger Wright wrote: When you refer to a first order expansion, what is it you're expanding? Taylor series?
Just do a Taylor expansion
x - sin(x) = x - (x - x^3/3! + ..) = y
as x->0 only the lowest powers are significant, so
x^3 / 6 = y
or x = (6y)^(1/3)
The rest of the approximation stuff is guesswork, and using something like Excel to see how well you are going. It's a bit of a game that's fun to play occasionally.
Using Excel and the solve function you can get a nice tidy approximation that's probably better for implementation:
x = (ay)^(1/3) + by + cy^2
a = 5.998854
b = 0.10235
c = 0.01579
the relative error is less than 2.7 x 10^-5 over the range 0 <= y <= pi/2 - 1
If you need better than this, one NR iteration gets you to 10^-10 territory and one more should pretty well get you full double precision.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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cp9876 wrote: If you need better than this, one NR iteration gets you to 10^-10 territory
I think that's probably close enough for sewer work.!
The recent one I did resulted in an argument - I specified a slope of 1.6%, and the job foreman insisted that they never use less than 33% ! It took me a while to figure out that he was talking about % of a degree, as displayed in his laser leveling instrument. The poor clod couldn't calculate a slope in ft/ft from degrees, so I wrote him up a tutorial showing the steps. Then the dumb bastard had the audacity to submit a bill for $1000 for the "change order."
Thanks for the assistance, and for reminding me how old I'm getting. I used to do these things in my head. 
"A Journey of a Thousand Rest Stops Begins with a Single Movement"
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Roger Wright wrote: cp9876 wrote:
If you need better than this, one NR iteration gets you to 10^-10 territory
I think that's probably close enough for sewer work.!
I don't want even 10^-10 of your sewerage to end up in the wrong place!
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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How to x^2+y^2+z^2=g^2 if x^2+y^2=a^2 and x^2+z^2=b^2 and y^2+z^2=c^2?
Thanks.cheers!!
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choose a single forum and stick to it!
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ja tnx i understand.
cercare il pelo nell'uovo.
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which are the values that are given and which are you looking for?
ok, suppose x,y,z are the unknown values and a,b,c,g are given
you have 3 unknowns and 4 equatations, so this is overdetermined (may also have no solution)
as all values go squared here, lets substitute
p = x^2
q = y^2
r = z^2
write it in lines:
p + q + r = g^2 (I)
p + q = a^2 (II)
p + r = b^2 (III)
q + r = c^2 (IV)
to get p for example, substract (IV) from (I)
(I) - (IV):
p+q+r-(q+r) = g^2 - c^2
p = g^2 - c^2
then do it similar to get r and q
after this, get each two values x1 und x2 like
x1 = sqr(p)
x2 = -sqr(p)
but I believe that there is no solution in this equatation, because after having the p,q,r try setting them in the first (p+q+r=g^2) and check if it is right - I dont think so
(edit: yeah code blocks...)
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Thanks for your help and ideas.I even tried trial and error by hand for a couple of hours.I think it is a Diophantine equation and has solution but i 'm not good at this.I 'll check for it and come back.
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