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[on:]

Hi,

In the old forum, I found this topic:
http://www.arduino.cc/cgi-bin/yabb2/YaBB.pl?num=1267245927

Indicating how to convert the values ​​in% RH, dewpoint and humidex.
But the function to convert dewpoint, I get an error, which is a value added and multiplied, what is the correct formula to get it?

Code:

dewpoint = (237,7*(17.271*((temp-32)*5/9))/(237.7+*((temp-32)*5/9)+log(RH/100))/(17.271-(17.271*((temp-32)*5/9))/(237.7+*((temp-32)*5/9)+(log(RH/100)));
// the numbers are constants, RH =relative humidity, temperature must be in C

Thanks,

[Reply #1 on:]

This code is syntactically incorrect. I simplified it a little and added the necessary braces at the end, although I don't know if this calculates the correct value (I haven't studied the wikipedia article).

Code:

  float tempC = 25.7;
  float temp = (tempC -32) * 5 / 9;
  float RH = 35.9;
  float dewpoint = (237.7*(17.271*temp)/(237.7*temp+log(RH/100))/(17.271-(17.271*temp)/(237.7*temp+(log(RH/100)))));

This at least compiles. I took out the recurring calculation of the temperature in Fahrenheit and made it before the dewpoint calculation.

[Reply #2 on:]

You may also want to try:

Code:

const float
tempC = 25.7,
temp = (tempC - 32) * 5/9,
logRH = -1.024433, // ln(35.9/100)
a = 17.271,
b = 237.7,
y = a*temp/(b + temp) + logRH,
Td = b*y/(a - y);

[Reply #3 on:]

This code is syntactically incorrect. I simplified it a little and added the necessary braces at the end, although I don't know if this calculates the correct value (I haven't studied the wikipedia article).

Code:

 float tempC = 25.7;
  float temp = (tempC -32) * 5 / 9;
  float RH = 35.9;
  float dewpoint = (237.7*(17.271*temp)/(237.7*temp+log(RH/100))/(17.271-(17.271*temp)/(237.7*temp+(log(RH/100)))));

This at least compiles. I took out the recurring calculation of the temperature in Fahrenheit and made it before the dewpoint calculation.

For this formula, the result I get is 0.00.
I used this formula, but I can not confirm if the values ​​are correct.

Code:

float aux = (log(RH / 100) + ((17.27 * temperatureC) / (237.3 + temperatureC))) / 17.27;
  float dewpoint = (237.3 * aux) / (1 - aux);

The values ​​that I'm getting are:

Code:

24.22 *C   75.59 *F   43.69 %RH   11.09 dew point *C   18.66 Humidex

Can you confirm if I'm right?

You may also want to try:

Code:

const float
tempC = 25.7,
temp = (tempC - 32) * 5/9,
logRH = -1.024433, // ln(35.9/100)
a = 17.271,
b = 237.7,
y = a*temp/(b + temp) + logRH,
Td = b*y/(a - y);

In this calculation, it seems to me you were to ignore the value of the logarithm of %RH.
But I'm not sure.

thanks for your help,

[Reply #4 on:]

In this calculation, it seems to me you were to ignore the value of the logarithm of %RH.

In your code, %RH was a constant, so log(RH/100) is also a constant. Since log is an expensive operation, it might as well be pre-computed.

[Reply #5 on:]

from - http://arduino.cc/playground/Main/DHT11Lib - the NOAA reference algorithm and the fast variation

Code:

// dewPoint function NOAA
// reference: http://wahiduddin.net/calc/density_algorithms.htm
double dewPoint(double celsius, double humidity)
{
        double A0= 373.15/(273.15 + celsius);
        double SUM = -7.90298 * (A0-1);
        SUM += 5.02808 * log10(A0);
        SUM += -1.3816e-7 * (pow(10, (11.344*(1-1/A0)))-1) ;
        SUM += 8.1328e-3 * (pow(10,(-3.49149*(A0-1)))-1) ;
        SUM += log10(1013.246);
        double VP = pow(10, SUM-3) * humidity;
        double T = log(VP/0.61078);   // temp var
        return (241.88 * T) / (17.558-T);
}

// delta max = 0.6544 wrt dewPoint()
// 5x faster than dewPoint()
// reference: http://en.wikipedia.org/wiki/Dew_point
double dewPointFast(double celsius, double humidity)
{
        double a = 17.271;
        double b = 237.7;
        double temp = (a * celsius) / (b + celsius) + log(humidity/100);
        double Td = (b * temp) / (a - temp);
        return Td;
}

[Reply #6 on:]

I took out the recurring calculation of the temperature in Fahrenheit and made it before the dewpoint calculation.


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