Wind chill Factor calculation

kellross · December 25, 2021, 10:58pm

Hi All,
I have been working with HE for about a year. Mostly simple lighting automations. Recently I setup a zigbee plug for outdoor humming bird feeder heater to come on to keep feeder warm when the temp drops below 1.5C. I am using the OpenWeatherMap API along with my own external temperature monitor in HE to determine when to turn on and off via the Rules 5.0 machine.
Recently I want to look at calculating the "feels like" wind chill temperature based on the current wind speed from the OpenWeatherMap attributes, since this feeder is quite out in the open.

The formula for wind chill is found on wikipedia: Wind chill - Wikipedia
specifically for metric I would like to use this formula:

In trying to determine this using the Rules Engine (5.0) I wasn't able to figure out how to do the X to power of 0.16 in the formula.

Is there something I am missing? or does anyone know of a MATH app or another way to provide the calculation via Rules engine?

Thanks and best of the Holidays to you!

oldcomputerwiz · December 25, 2021, 11:26pm

Just an FYI

"Wind chill is a perceived air temperature, not a physical quantity. ... Water will not freeze with the temperature air at or above 33 degrees , regardless of how far the wind chill is below freezing. Wind chill has no effect on inanimate objects, and they cannot be cooled below the ambient air temperature"

marktheknife · December 25, 2021, 11:28pm

What is the purpose of the heater? I’m not sure wind chill applies here, since it’s just something humans that feel.

aaiyar · December 25, 2021, 11:34pm

I’ll bet that hummingbirds feel it too

lairdknox · December 25, 2021, 11:47pm

You can create a simple Groovy app that does the calculation. I’m not sure how an app can pass a value into RM but it can store the result in a hub variable that RM can then access.

You can call the app directly from RM or have it run periodically to keep the variable updated.

kellross · December 25, 2021, 11:49pm

The heater is a simple container that clamps to the bottom of a hummingbird feeder to warm the bottom (it has a 7W incandescent bulb in this plastic container) and yes it does work to keep it the sugar water from freezing (which freezes at -1.5C. It has kept it warm enough to stay liquid when the ambient temp drops below freezing (lowest is has gone here is -5C so far this year).
@oldcomputerwiz - Thank you for quoting AskTom (without a reference) https://www.chicagotribune.com/weather/ct-wea-asktom-0202-20180201-column.html
I am not debating whether wind chill would freeze the water or not above freezing, however the article does state "An object will lose heat in calm conditions, but wind will remove it more quickly"

The questions was how can this calculation be done in the rules engine and if not, what other apps/tools could provide this?
I want to calculate this for the information tracking and the dashboard.

oldcomputerwiz · December 26, 2021, 1:06am

Just trying to be helpful. You have a Merry Christmas!

marktheknife · December 26, 2021, 1:09am

If you’re using the OWM driver, then why not use the “feels like” condition?

libbythekid · December 26, 2021, 1:10am

You don't really need to reproduce the exact formula to achieve your goal.
A simplified approximation will get you close enough for practical purposes.

The accuracy of your instruments for measuring wind and temperature will probably be +/- 5% or more, so an approximation that's within a degree or two will be good enough.

Here's the formula for silly American units:
WCF = 35.74 + 0.6215T – 35.75 (V^0.16) + 0.4275T (V^0.16)
where T = degrees Fahrenheit and V = wind velocity in miles per hour.

And here's a simpler approximation that will produce results within a degree or two of the more complex formula.

Approx WCF = -12 + 1.2T -V/3 +VT/200

Remember that the formula you're looking at is specifically designed for figuring out how much heat a human body will lose under certain conditions. It's useful for figuring out how warmly you should dress to avoid frostbite, but it doesn't have any effect on inanimate objects, and would not work for birds either, as they have different metabolisms.

The wind will determine how QUICKLY the bird feeder cools off, but won't affect the final temperature it ends up at.

Having said all this, I understand the appeal of wanting to solve the original problem, and I'd like to have the ability to do exponents in rules.

nh.schottfam · December 26, 2021, 2:12am

You can do this in webCoRE.

PunchCardPgmr · December 26, 2021, 4:33am

With respect to "freezing temperatures", you're right. But a chilling wind, or rather airflow, has been shown to be a significant accelerator of water pipe freezing once temps do get low enough (which, by the way...is lower than most people realize...i.e. well below 32F)

https://www.ideals.illinois.edu/handle/2142/54757

Disregarding Wind Chill per se, OP might be able to use some of the data from this study to use a combined temp threshold AND wind SPEED to invoke his actions on actual measurements vs calculations. My application is out of the wind ...but my heat source is set to kick in at 28F; enough of a safety margin/lead time to avert a problem.

What I'm really wondering about is what the hell those hummingbirds are doing still around once it's getting anywhere close to 1.5C ! The ones I'm familiar with are long gone South by then

ogiewon · December 26, 2021, 4:38am

I was thinking the exact same thing. I live in NC and the humming birds have been gone since ~October, long before temperatures ever get close to freezing.

libbythekid · December 26, 2021, 4:49am

Yes, that's what I said previously. It's not about the final temperature, it's about how quickly that temperature is reached. Your pipes will never get colder than the ambient temperature, no matter how strong the wind is. But a strong wind will make them freeze sooner, because the heat is lost more quickly. It actually makes perfect sense if you think about it. How quickly heat transfers by conduction is proportional to the difference in temperature. Cold air takes on heat from the warmer pipes, which become cooler. But as the air gets warmer, the cooling rate slows down. However if the wind is constantly blowing the warmer air away, then the cooling rate never slows down, and the pipes reach the minimum temperature more quickly.

lairdknox · December 26, 2021, 5:49am

@kellross, here's a quick and dirty hack of a wind chill calculator. I don't have anything to monitor windspeed so you can see I set it to a value of 10 just for testing. You should be able to add any wind speed sensor you are using. It will run every five minutes and update a hub variable you setup in the config.

definition(
	name: "Wind Chill Calculator",
	namespace: "jlkSpace",
	author: "Jeff Knox",
	description: "Calculate the current wind chill and save it to a hub variable.",
	category: "My Apps",
    iconUrl: "https://s3.amazonaws.com/smartapp-icons/Convenience/Cat-Convenience.png",
    iconX2Url: "https://s3.amazonaws.com/smartapp-icons/Convenience/Cat-Convenience@2x.png",
    iconX3Url: "https://s3.amazonaws.com/smartapp-icons/Convenience/Cat-Convenience@2x.png")

preferences {
	section("Sensors") {
		input "tempSensor", "capability.temperatureMeasurement", required: true, title: "Select your temperature sensor."
	}

    section("var") {
        input "gWindChill", "capability.sensor", required: true, title: "Select wind chill variable connector."
    }
	
	section("Debugging") {	
	    input name: "debugOutput", type: "bool", title: "Enable debug logging?", defaultValue: true
	}

}

def installed() {
	initialize()
}

def updated() {
	unsubscribe()
	initialize()
}

def initialize() {
	log("••• INIT •••")
	schedule('0 */5 * ? * * *', calculateWindchill) 
}

def calculateWindchill(evt) {
	double currentTemp = tempSensor.currentTemperature
	log("Current Temp: " + currentTemp);
	double windSpeed = 10
	
	// Fahrenheit
	double windchill = 35.74 + 0.6215 * currentTemp - 35.75 * Math.pow(windSpeed, 0.16) + 0.4275 * currentTemp * Math.pow(windSpeed, 0.16)
	
	// Celcius
	//double windchill = 13.12 + 0.6215 * currentTemp - 11.37 * Math.pow(windSpeed, 0.16) + 0.3965 * currentTemp * Math.pow(windSpeed, 0.16)
	
	log("Windchill: " + windchill)
	gWindChill.setVariable(windchill)
}

def log(msg) {
	if (settings?.debugOutput || settings?.debugOutput == null) log.debug "$msg"
}

Edit: Changed the schedule from 0/5 to */5. I'm not sure if it makes a difference but RM created schedules use the */5 notation.

kellross · December 26, 2021, 1:57pm

Living on the west coast, I see them often throughout the winter. It is odd that not all of the humming birds migrate.

kellross · December 26, 2021, 2:18pm

Thank you for the suggestion.
I didn’t know about WebCoRE. Seems I am learning more every day about capabilities and community apps available for the HE. I’ll look into WebCoRE as a way to calculate this.

kellross · December 26, 2021, 2:21pm

Thanks @lairdknox. I can work with this too.
Cheers.

marktheknife · December 26, 2021, 4:07pm

I believe this is a description of heat loss by convection, which does apply to a bowl of sugar water sitting out on a calm vs. windy day.

Something like this seems “close enough” for this application. The wind chill calculation is, as far as I can tell, needlessly complex if the goal is to prevent a bowl of sugar water from freezing.

dsh1705 · December 26, 2021, 8:26pm

I love this heater idea. Did you make or buy it? Of course I am thinking about this AFTER it is freezing here, not before, and this would save me having to rotate the feeders all day. Already figuring out who I would do it…..

PunchCardPgmr · December 26, 2021, 10:19pm

Man oh man this is a bigger deal than I ever imagined. Don't tell the mid-Continent hummers about this, I'm fine with them actually MIGRATING outta here for a few months!