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Return to top of page · Post #: 1 · Written at 3:57:31 AM on 10 January 2015.
	GTC Location: Sydney, NSW Member since 28 January 2011 Member #: 823 Postcount: 6803
Folks, Please solve this equation for x and let me know your answer: x = 240.7263/(7.591386/log₁₀(67.04/6.116441)^-1) Note: The -1 at the end is an exponent. I get a result that does not agree with the textbook, but I may be making an error.

Return to top of page · Post #: 2 · Written at 8:26:40 AM on 10 January 2015.
	Marcc Location: Wangaratta, VIC Member since 21 February 2009 Member #: 438 Postcount: 5474
Did you work out what was in the brackets first? Strange way of putting the equation? Almost looks like an escapee from a "slide rule" (got two) log base 10 is the common logarithms. Marc

Return to top of page · Post #: 3 · Written at 9:26:17 AM on 10 January 2015.
	Tinkera123 Location: Melbourne, VIC Member since 5 October 2009 Member #: 555 Postcount: 467
-32.97 Ian ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ Cheers, Ian

Return to top of page · Post #: 4 · Written at 2:26:07 PM on 10 January 2015.
	Brad Administrator Location: Naremburn, NSW Member since 15 November 2005 Member #: 1 Postcount: 7451
No result will come from me unfortunately. Equations were probably my weakest point in maths at school and maths was generally a subject I loathed apart from geometry. ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ A valve a day keeps the transistor away...

Return to top of page · Post #: 5 · Written at 3:32:19 PM on 10 January 2015.
	GTC Location: Sydney, NSW Member since 28 January 2011 Member #: 823 Postcount: 6803
Brad: Maths was never my strong subject. I had to work hard to concentrate as my mind always said: "You'd rather be doing anything else than this". Marcc: That's how the equation is given in the text, and I'm using the common log function. When I put it in Excel in the format as shown I got a different result from the text, so I executed it in single steps from the inside out, following the parentheses, and still differ. Ian: Thanks. Via Excel I get 30.49565355. The book says: 38.21 Anybody able to get 38.21?

Return to top of page · Post #: 6 · Written at 4:15:52 PM on 10 January 2015.
	Tinkera123 Location: Melbourne, VIC Member since 5 October 2009 Member #: 555 Postcount: 467
Hi GTC, I will check my calculation, but it shouldn't be a difficult equation to solve. Also did it "inside out" with Excel. Could be interpretation of formula from the text. Can you email the actual text? Cheers, Ian ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ Cheers, Ian

Return to top of page · Post #: 7 · Written at 4:24:33 PM on 10 January 2015.
	GTC Location: Sydney, NSW Member since 28 January 2011 Member #: 823 Postcount: 6803
Okay, image of text emailed. (two graphics).

Return to top of page · Post #: 8 · Written at 5:40:02 PM on 10 January 2015.
	Tinkera123 Location: Melbourne, VIC Member since 5 October 2009 Member #: 555 Postcount: 467
GTC, The equation is open to various interpretations. I intrepreted the exponent as log{(X)-1} .... and being less than 1 ... yields a minus answer. If the exponent is of the whole log function ie {log(X)}-1 then I also get 30.4956..... Treating the -1 as integer "minus 1" ..... doesn't help either. Another possibility, besides the book being incorrect (and it wouldn't be the first time), is the interpretation of "10log". I would also assume 'log base 10', but other bases to logs are used. Is the text specific about "log base 10"?? Cheers, Ian ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ Cheers, Ian

Return to top of page · Post #: 9 · Written at 6:00:48 PM on 10 January 2015.
	Tinkera123 Location: Melbourne, VIC Member since 5 October 2009 Member #: 555 Postcount: 467
Hey GTC, Just got your second email ..... the book is correct .... 32.21. The problem was with the interpretation of -1. Calculation should be Tn/[{m/log(Pw/A)}-1] ie subtract 1 from m/log(Pw/A) then divide into Tn Log is base 10 I will email my spreadsheet ... see the last calculation. Cheers, Ian ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ Cheers, Ian

Return to top of page · Post #: 10 · Written at 6:07:49 PM on 10 January 2015.
	GTC Location: Sydney, NSW Member since 28 January 2011 Member #: 823 Postcount: 6803
Much confusion for me! Many thanks for the worked example. I see you mean to say 38.21 above.

Return to top of page · Post #: 11 · Written at 6:51:10 PM on 10 January 2015.
	Tinkera123 Location: Melbourne, VIC Member since 5 October 2009 Member #: 555 Postcount: 467
Yes, 38.21. May I ask what the temperature calculation is for? Just curious. Ian ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ Cheers, Ian

Return to top of page · Post #: 12 · Written at 9:22:24 PM on 10 January 2015.
	GTC Location: Sydney, NSW Member since 28 January 2011 Member #: 823 Postcount: 6803
It's supposed to allow calculation (within 1% accuracy) of dew point temperature without recourse to the psychometric chart. I now have to see how good it is with local data.

Return to top of page · Post #: 13 · Written at 9:43:22 PM on 10 January 2015.
	Marcc Location: Wangaratta, VIC Member since 21 February 2009 Member #: 438 Postcount: 5474
The only way you get 38.21 is to treat -1 as minus 1 If you divide out that which is in the inner bracket its log is 1.0398. divide 7.59.... by that and subtract 1 you end up with 6.3 as the divisor so then you have 240.72... / 6.3 Marc

Return to top of page · Post #: 14 · Written at 10:18:29 PM on 10 January 2015.
	GTC Location: Sydney, NSW Member since 28 January 2011 Member #: 823 Postcount: 6803
Marcc, yes thanks, Ian set me straight on that in post #9 above. All good now.

Return to top of page · Post #: 15 · Written at 9:00:29 AM on 11 January 2015.
	Marcc Location: Wangaratta, VIC Member since 21 February 2009 Member #: 438 Postcount: 5474
Basically, if the formula was written properly, there would never have been an issue: Only a problem, the value of "X"

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