system
June 20, 2012, 3:59pm
1
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.
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,
pylon
June 20, 2012, 4:26pm
2
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).
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.
system
June 20, 2012, 7:11pm
3
You may also want to try:
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);
system
June 21, 2012, 2:42pm
4
pylon:
float tempC = 25.7;
float temp = (tempC -32) * 5 / 9;temp)/(237.7 temp+log(RH/100))/(17.271-(17.271temp)/(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.
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:
24.22 *C 75.59 *F 43.69 %RH 11.09 dew point *C 18.66 Humidex
Can you confirm if I'm right?
Reinderien:
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.
thanks for your help,
system
June 21, 2012, 2:59pm
5
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.
from - Arduino Playground - DHT11Lib - the NOAA reference algorithm and the fast variation
// 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;
}
system
June 28, 2012, 1:17am
7
I took out the recurring calculation of the temperature in Fahrenheit and made it before the dewpoint calculation.
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