NavList:

In the attached email below, I have an operator problem.

When Marcel gave his step by step procedure, Q was given as

Q = (P/1010)(283.15/T)

My eyes and brain saw that as

Q = (P/1010) / (283.15/T)

When clearly Marcel meant

Q = (P/1010) * (283.15/T)

Whooops!

Q = (P/1010)*(283.15/T) is of course is the equivalent of Q = (P/1010) / (T/283.15) and f = 0.28 * P / (T+273) being careful with T's units

The error was entirely due my own faulty vision and understanding. I needed to set this record straight.

Regards
Brad

On Apr 8, 2013 7:29 PM, "Brad Morris" <bradley.r.morris@gmail.com> wrote:

Hi Marcel

Now that I too have found the relevant section, I can see that the refraction correction is indeed as you thought, astronomic. The 'H' in R0=0.0167/tan(H+7.32/(H+4.32)) is the altitude of the body after dip is taken out.

This nominal refraction R0 is then adjusted for non-standard T & P by f=0.28*P/(T+273) where P is in millibars and T is degrees C

The refraction for conditions is then R=R0*f

'f' therefore is perfectly synonymous with 'Q' in the adjustment of the refraction portion. We would adjust the refraction portion of our Dip by f or Q.

++++++ I solved for f under the 4 extrema of 880mb, 1082mb, Death Valley 57.7C and Antarctica -89.2C

f=0.28*880/283 = 0.870671 f=0.28*1082/283=1.071519 f=0.28*1010/(56.7+273)=0.857749 f=0.28*1010/(-89.2+273)=1.538628

I then resolved for Q1=(P/1010)/(283.15/T)

Q1=(880/1010)/1= 0.871287 Q1=(1082/1010)/1=1.072277 Q1=1/(283.15/(56.7+273))=1.164400 Q1=1/(283.15/(-89.2+273))=0.649125

I then tried Q2=(P/1010)/(T/283.15)

Q2=(880/1010)/1=0.871287 Q2=(1082/1010)/1=1.072277 Q2=1/(56.7+273)/283.15)=0.858811 Q2=1/(-89.2+273)/283.15)=1.540533

I think it clear that both Q1 and Q2 agree with f for pressure. From a practical standpoint, the first two digits after the decimal point are sufficient. It is also clear that only Q2 agrees with f for temperature. Q1 does not, Q1=1/f for temperature.

I'll leave it at this point for you to draw your conclusions. I've drawn mine

Brad
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