213 lines
7.3 KiB
Text
213 lines
7.3 KiB
Text
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The formula p(x) for calculating the air pressure at a given altitude
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---------------------------------------------------------------------
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Well known is the baromertic(?) formula
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rho0
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------ * g * x
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p0
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p(x) = p0 * e
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with p0 being the airpressure and rho0 being the air density at an altitude of
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0 metres above sea level and g being the gravity constant of 9.81 m/sq. sec
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This formula can easily be derivated when you know, that:
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* the pressure difference is
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dp = - rho * g * dx
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* Boyle-Mariotte says:
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p0 : p = rho0 : rho
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Combinig the terms and changing them around I get:
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dp [ rho0 ]
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-- = - rho * g = - [ ------ * p(x) ] * g
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dx [ p0 ]
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rho0
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p'(x) = - ------ * p(x) * g
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p0
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Solving that differential equation and knowing that p(0) = p0 I get:
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rho0
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- ------ * g * x
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p0
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p(x) = p0 * e
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q.e.d.
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-------------------------------------------------------------------------------
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The problem with that equation is that it doesn't take different temperatures
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at different altitudes into account. And the inaccuracies due to it are huge.
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That's why this formula is only used in very low altitudes.
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So to get a usefull formula for FlightGear I need to extend it. And as I'm
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already 'recreating' that formula I'm taking the change in g into account, too.
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This doesn't make such a dramatic difference to the result as the inclusion of
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temperature change does, but it doesn't complicate the final formula too much.
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So I get three formulas that I'm combining in the end:
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* the change of g with the altitude:
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G * m
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g(x) = -----------
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(x + r)^2
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G is the universal gravity constant(?) and is 6.673e-11 m^3 kg^-1 s^-2
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m is the mass of the earth and is 5.977e24 kg
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r is the radius of the earth and is 6368 km
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* The pressure difference stays the same:
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dp = - rho * g(x) * dx
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* If I combine Boyle-Mariotte with Gay-Lussac I get:
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rho0 * T0 p
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rho = ----------- * ---
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p0 T
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Combining the terms again I get this time:
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dp [ rho0 * T0 p(x) ]
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-- = - rho * g(x) = - [ ----------- * ------ ] * g(x)
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dx [ p0 T(x) ]
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rho0 * T0 p(x) * g(x)
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p'(x) = - ----------- * -------------
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p0 T(x)
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This DE isn't that easy to solve as the one above, it by looking into the right
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books you'll see the general solution for:
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y' + f(x)*y = 0
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is
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x
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/\
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- | f(x) dx
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\/
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n
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y = m * e
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and P(m,n) will be a point on the graph.
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For q = n = 0 metres altitude we get y = m. As y is p(x) we know that m has to
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be p0.
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So our final formuala is
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ho0 * T0 g(x)
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f1(x) = ----------- * ------
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p0 T(x)
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x x
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/\ /\
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- | f1(x) dx | f(x) dx
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\/ \/
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0 0 F(x) - F(0)
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p(x) = p0 * e = p0 * e = p0 * e
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The only disturbing thing we've got left is the integral. Luckily there is a
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great service at http://integrals.wolfram.com/ that helps me doing it :-)
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But the f(x) is still too general so I'm substituting:
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rho0 * T0 * G * m
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f(x) = - -----------------------
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p0 * (x + r)^2 * T(x)
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but even that isn't good enough. But as I'm linearily interpolating between
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two different temperatures I can say that T(x) = a*x + b for the x inbetween
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two different stored temperatures. So I just need to integrate every pice
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independandly. But anyway, I get:
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rho0 * T0 * G * m
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f(x) = - ------------------------------
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p0 * (x + r)^2 * (a * x + b)
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Integrating that I get:
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rho0 * T0 * G * m [ 1
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F(x) = - ------------------- * [ ------------------------ -
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p0 [ (-b + a * r) * (r + x)
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a * log|r + x| a * log|b + a * x| ]
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---------------- + -------------------- ]
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(b - a * r)^2 (b - a * r)^2 ]
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To lower the computional cost I can transfere the equation.
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* I'm defining
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rho0 * T0 * G * m
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factor = - -------------------
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p0
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1
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c = --------------
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(-b + a * r)
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* now I can write
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[ c ]
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F(x) = factor * [ --------- - a * c * c * [log|r + x| + log|b + a * x|] ]
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[ (r + x) ]
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* and simplyfy it to
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[ 1 ]
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F(x) = factor * c * [ --------- - a * c * log|(r + x) * (b + a * x)| ]
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[ (r + x) ]
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-------------------------------------------------------------------------------
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The following table shows quite nicely how accurate my formula is:
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Altitude[m] | Airpressure [hPa] | Error [%]
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| Official | My formula |
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------------+---------------+---------------+---------------
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-200 | 1037.51 | 1037.24 | 0.0260
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-100 | 1025.32 | 1025.19 | 0.0127
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0 | 1013.25 | 1013.25 | 0.0
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500 | 954.59 | 955.224 | 0.0664
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1000 | 898.70 | 899.912 | 0.1349
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2000 | 794.88 | 797.042 | 0.2720
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3000 | 700.99 | 703.885 | 0.4130
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4000 | 616.28 | 619.727 | 0.5593
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5000 | 540.07 | 543.89 | 0.7073
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6000 | 471.67 | 475.731 | 0.8610
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7000 | 410.46 | 414.643 | 1.0191
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8000 | 355.84 | 360.054 | 1.1842
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9000 | 307.27 | 311.422 | 1.3513
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10000 | 264.21 | 268.238 | 1.5245
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20000 | 54.670/55.3 | 55.7971 | 2.0616/0.8989
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30000 | 11.8 | 11.3149 | 1.5441
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40000 | 3.0 | 2.74665 | 18.9703
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50000 | 0.88 | 0.753043 | 41.9183
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60000 | 0.257 | 0.221907 | 57.9802
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70000 | 0.0602 | 0.0530785 | 61.9153
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80000 | 0.0101 | 0.00905461 | 51.5725
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100000 | 2.14e-4 | 2.03984e-4 | 5.5131
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The official values are from the CINA atmosphere which assumes a air pressure
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of 1013.25 hPa and a temperature of 15 degC at sea level and a temperature
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gradient of -6.5 deg/km. The CINA atmosphere gives only values for altiudes
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up to 20 km. The values for 20 km and above are from the 1959 ARDC atmosphere.
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That's why I've got two values at 20000 metres.
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The temperature changes dramtically in the altitudes over 20 km which I didn't
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take care of in my calculations which explains the huge errors at that altitude
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range. But you can see nicely that the values are at least quite close to the
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official values.
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Using a better temperature model for the altitudes above 20 km should
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dramatically increase the accuracy there.
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