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'Windspeed circuit?'
2000\01\14@112056 by Peter L. Berghold

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Hi there folks,

Anybody out there know of plans on the web for a wind speed indicator?  I
would assume that it would involve some sort of pulse counting scheme...

--
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Peter L. Berghold                        spam_OUTPeterTakeThisOuTspamBerghold.Net
"Linux renders ships                     http://www.berghold.net
NT renders ships useless...."

2000\01\14@122516 by Wagner Lipnharski

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You can develop any kind of sensor, by wind pressure, differential
pressure, weight lifting (wing), rotational blades (most common), or any
other. Some will generate a linear voltage (requires an ADC conversion),
some will generate pulses (that should be accounted by time).

In any case, the most difficult task is to calibrate the unit. Even if
you install the unit outside your car and drive in a not windy day, who
ensures that your car's speedometer is calibrated?  If you want to go
for some accuracy you will need to use a professional laboratory or wind
calibrator.

Is there any practical method to measure air speed (low accuracy)? as
for example, imagine a 12x12 inches 1/8" aluminum plate hang by two
nylon wires 3 ft high, wind directly over the plate, how much will be
the deflection of the plate at 10mph? 20mph? is there anything like that
for practical purposes? as is the original knots measurement in
boats...  (I am curious about that).

Wagner.

"Peter L. Berghold" wrote:
>
> Hi there folks,
>
> Anybody out there know of plans on the web for a wind speed indicator?  I
> would assume that it would involve some sort of pulse counting scheme...

2000\01\14@134154 by Smith, Clay

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What you describe is called an lvdt.  Run an AC current thru a transformer
or inductor and move a spring loaded ferrite core in against a calibrated
spring and rectify the output.  You will have a voltage proportional to the
force on the lever
CSS

               {Original Message removed}

2000\01\14@135026 by James Paul

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I believe lvdt stands for "Linear Variable Differential
Transformer".

                                       Regards,

                                          Jim




On Fri, 14 January 2000, "Smith, Clay" wrote:

>
> What you describe is called an lvdt.  Run an AC current thru a transformer
> or inductor and move a spring loaded ferrite core in against a calibrated
> spring and rectify the output.  You will have a voltage proportional to the
> force on the lever
> CSS
>
>                 {Original Message removed}

2000\01\14@140926 by Smith, Clay

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One method for wind speed measurement is to connect a permanent magnet motor
to a wind vane.  The motor will act as a generator and the CEMF generated is
proportional to the speed of the motor.  May be a 100v dc if your gearing is
to step, but a higher max speed is better to get out of the IR drop RPMs
CSS

               {Original Message removed}

2000\01\14@165252 by Dave Johnson

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Peter L. Berghold wrote:

>Anybody out there know of plans on the web for a wind speed indicator?
Look at http://www.alphalink.com.au/~derekw/ane/anemain.htm, the site has
very good mechanical plans, though they want to sell you a kit with a
pre-programmed PIC.

Dave Johnson

2000\01\15@224556 by Robert A. LaBudde

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<x-flowed>At 02:02 PM 1/14/00 -0500, Clay wrote:
>One method for wind speed measurement is to connect a permanent magnet motor
>to a wind vane.  The motor will act as a generator and the CEMF generated is
>proportional to the speed of the motor.  May be a 100v dc if your gearing is
>to step, but a higher max speed is better to get out of the IR drop RPMs
>CSS

This is one of the standard methods used in commercial air velocimeters,
although it will be difficult to find a motor with the sensitivity of those
used in the meters. You can also use optical rotary encoders instead of
actual voltage, and dispense with the motor to improve sensitivity (just a
propeller and jeweled mounts).

The other standard method (used in aircraft) is the 'hot wire anemometer'.
A platinum wire RTD is held at a constant temperature in the wind and the
current required is measured. This is very accurate, and a well-worked-out
theory is available, assuming a small diameter probe. You could substitute
nichrome wire for a homebrew application.


================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS  e-mail: .....ralKILLspamspam@spam@lcfltd.com
Least Cost Formulations, Ltd.                   URL: http://lcfltd.com/
824 Timberlake Drive                            Tel: 757-467-0954
Virginia Beach, VA 23464-3239                   Fax: 757-467-2947

"Vere scire est per causae scire"
================================================================

</x-flowed>

2000\01\15@225429 by Graeme Zimmer

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> >One method for wind speed measurement is to connect a permanent magnet
motor
> >to a wind vane.  The motor will act as a generator and the CEMF generated
is
> >proportional to the speed of the motor

I saw a rather devious variation recently. If you have a vertical shaft
anemometer (the kind with half ping-pong balls going around) and you make
one half-sphere larger than the others, so that the contraption rotates with
a kind of "wobble", then the variation in angular speed can be used to
figure out the wind direction as well.....

..................... Zim

2000\01\18@145424 by Dennis Gearon

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Hint---------- 4 strain gauges

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2000\01\18@162017 by Allan West

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I must have missed most of this thread.  But why 4?
2 would be enough surely
----- Original Message -----
From: "Dennis Gearon" <.....gearondKILLspamspam.....OIT.EDU>
To: <EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU>
Sent: 18 January 2000 19:52
Subject: Re: Windspeed circuit?


{Quote hidden}

2000\01\18@163734 by Don Hyde

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Bias it with springs and maybe only 2 strain gauges, but strain gauges are
still expensive, need lots of gain, and are a lot of trouble to calibrate.
Do you know where to find them at reasonable prices?

> {Original Message removed}

2000\01\18@182103 by paulb

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Don Hyde wrote:

> Bias it with springs and maybe only 2 strain gauges, but strain gauges
> are still expensive, need lots of gain, and are a lot of trouble to
> calibrate.

 Perhaps a different perception of strain gauges here.  Strain gauges
*do* indicate the direction of the force as well as its magnitude, so
there will be no need to bias them.

 (Consider:  Strain gauge on surface; is compressed if surface is bent
upward (concave), stretched if bent downward (convex).
--
 Cheers,
       Paul B.

2000\01\18@190722 by Wagner Lipnharski

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Some extra information: http://www.irdam.ch/theory.htm

2000\01\19@114140 by Dennis Gearon

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Strain gauges, I don't keep catalogs much any more, or data books for
that matter. But I did have a cheap source and if one were to do a
search on the web, surely something would turn up.

Don Hyde  wrote:
>
>Bias it with springs and maybe only 2 strain gauges, but strain gauges
are
>still expensive, need lots of gain, and are a lot of trouble to
calibrate.
>Do you know where to find them at reasonable prices?
>
>> {Original Message removed}

2000\01\19@114410 by Dennis Gearon

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Temperature compensation and wind direction.......anyone figured it out?

Allan West wrote:
>I must have missed most of this thread.  But why 4?
>2 would be enough surely
>{Original Message removed}

2000\01\19@145127 by Allan West

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Well only 3 are required for temperature compensation
2 for direction
1 preloaded for temperature

or you could use 1 and spin the table through a retarded mechanism thus
giving you a plot of force for relative to angle.  You then can ignore temp.
considerations.

{Original Message removed}

2000\01\19@212701 by Sean Breheny

face picon face
Several people mentioned windchill as a possible effect with the reference
temp. sensor in a hot-sensor anemometer. I was not aware that wind chill
worked on dry objects (hence the theory behind wet-bulb/dry-bulb humidity
sensors). Can someone please tell me why it is a consideration in this
case? Are you concerned that water might condense onto the sensor?

Sean

|
| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
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2000\01\19@221643 by David Lions

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>temp. sensor in a hot-sensor anemometer. I was not aware that wind chill
>worked on dry objects (hence the theory behind wet-bulb/dry-bulb humidity

Well heat is conducted from the surface into adjacent air, which carries
the heat away by convection (the wind).  Thus cooling your sensor, whether
it is
dry or not. :)

2000\01\19@221655 by Wagner Lipnharski

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Note: This is a very basic concept about heat transfer. If you know it
already, just ignore this post.

According to my point of view, not necessarily "wind chill", but just
wind, even if at the same ambient temperature, works that way:

Suppose that the ambient temperature is 80¡F, the hot-sensor is at 90¡F,
so there is a delta temp from the sensor to the air.

It transfer heat to the surrounding air molecules. The VCM (very close
molecules) will be very close to 90¡F, and it will be transferring heat
around until you will see (far) molecules at 80¡F. Without wind, the
convection system will push up the molecules above 80¡F, and it will
create a standard sensor HTR (heat transfer rate).

The HTR depends on this convection. It means that the sensor will
transfer heat to the VCM, only if those molecules have a temperature
lower than the sensor, so the sensor transfer more or less heat upon the
speed of the VCM transfer their heat ahead, or are pushed up by
convection.

When you blow wind over the sensor it speeds up this action, so the VCM
temperature will be closer to the ambient temperature than the sensor
temperature, so the sensor HTR increases.

So, if the VCM are carrying moisture, it also increases the HTR, since
water particles transfer heat better than dry air.  In a stable ambient
(no wind), the surrounding water particles will be not very close to the
heating element (sensor), it creates like a very small dry bubble around
it. The wind chill will disrupt this bubble and accelerate the HTR much
more that dry wind.

By this terms, wind chill can create an erroneous measurement in thermal
anemometers. I wonder what happens if the second sensor (without heat)
is also exposed to wind chill and gets wet, and still wet even after the
wind chill stops... the heat-sensor will dry fast, and the delta temp
between both sensors will not reflect the correct HTR.  It would need a
third sensor enclosed to check the delta between both (no heat) sensors?

Man, this can go forever.  I wonder how HTR (or fire) works in a zero
gravity ambient.

Wagner Lipnharski - UST Research Inc.
Orlando, FLorida. - http://www.ustr.net

Sean Breheny wrote:
{Quote hidden}

2000\01\19@231759 by Ken Webster

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Air is not a good conductor of heat.  An object which is above the ambient
air temperature will transfer heat to the air immediately surrounding it,
thus heating the air.  If the air is still, a thermal gradient will develop
.. the air closest to the hot object will be hot and the further away you
move the closer the temperature will be to the ambient air temperature.
Because of the high thermal resistance of air, the hot object will loose
heat very slowly.

If the air is moving, however, cooler air will be brought into contact with
the hot object and heat will be transferred away from the hot object at a
higher rate.  The faster the air moves, the faster cool air will reach the
hot object and the faster heat will transfer from the hot object to the air.

Of course in real life the air does not sit still even if there is no wind.
Convection will cause heat loss in the absence of wind but the flow of air
due to convection will generally be small compared to the flow of air at
wind speeds which are interesting to measure.

Evaporative cooling is an entirely different effect.  It has to do with the
heat required for water to transition from the liquid phase to the vapor
phase (the vapor phase has higher entropy).  When water evaporates from a
body of liquid water, the remaining liquid water looses heat and its
temperature will drop below the ambient air temperature unless an equal
amount of heat is added.

Water has nothing to do with the function of a hot-sensor anemometor except
for the effects that humidity has on the heat capacity and thermal
conductivity of the air (which could be compensated for by additional
computations if the humidity is known).

Hmm ... does anyone out there have an idea how much error humidity could
introduce if you just went with this approximation for King's law:

S = A * [(P - D) / (TS - TA)] ** 2
where:
S = air speed
A = full-scale calibration constant
P = power dissipated by the airspeed sensor
D = "still-air" (S = 0) power dissipation
TS = temperature of the airspeed sensor
TA = ambient temperature
** = "to the power"

The formula came from the article here:
devel.penton.com/ed/Pages/magpages/may2598/ifd/0525ifd.htm
(scroll down past the first article to the one titled "Low-Power Thermal
Airspeed Sensor")

Better yet, does anyone have King's law (full version .. not approximation)
handy?

Ken


>Several people mentioned windchill as a possible effect with the reference
>temp. sensor in a hot-sensor anemometer. I was not aware that wind chill
>worked on dry objects (hence the theory behind wet-bulb/dry-bulb humidity
>sensors). Can someone please tell me why it is a consideration in this
>case? Are you concerned that water might condense onto the sensor?
>
>Sean

2000\01\20@005611 by Ken Webster

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Wagner Lipnharski wrote:

>Note: This is a very basic concept about heat transfer. If you know it
>already, just ignore this post.

Sorry ... I couldn't resist reading it anyway :o)
(not that I actually know it already ... I just think that I might know a
little about it)

>
>According to my point of view, not necessarily "wind chill", but just
>wind, even if at the same ambient temperature, works that way:

(-- snip --)

>So, if the VCM are carrying moisture, it also increases the HTR, since
>water particles transfer heat better than dry air.  In a stable ambient
>(no wind), the surrounding water particles will be not very close to the
>heating element (sensor), it creates like a very small dry bubble around
>it. The wind chill will disrupt this bubble and accelerate the HTR much
>more that dry wind.

I'm assumung that by "dry" you mean relative humidity.  The concentration of
water molecules should be the same near the sensor as it is away from the
sensor (nature abhors a vacuum -- molecules diffuse).  The relative humidity
is lower only because the temperature is higher and warmer air can hold more
water than cooler air.

I'm not sure why wind disrupting this "bubble" will cause faster heat
transfer than with dry wind except for the differences in the thermal
conductivity and heat capacity of moist air.


>By this terms, wind chill can create an erroneous measurement in thermal
>anemometers. I wonder what happens if the second sensor (without heat)
>is also exposed to wind chill and gets wet, and still wet even after the
>wind chill stops... the heat-sensor will dry fast, and the delta temp
>between both sensors will not reflect the correct HTR.  It would need a
>third sensor enclosed to check the delta between both (no heat) sensors?

The unheated sensor is not chilled by the wind and it shouldn't get wet
unless it is raining in which case the relative humidity is 100% in which
case evaporative cooling is not an issue.  Perhaps when the humidity first
drops below 100% the cool sensor will experience a small amount of
evaporative cooling but as long as it has a small surface area and/or the
relative humidity changes gradually this should be totally insignificant.

The only thing I would worry about is the hot sensor getting hit with an
occasional raindrop.  If it is very small and hydrophobic (which most
thermistors are) then the frequency of collisions should be low and the
amount of time that the water spends in contact with the sensor should be
brief.  I don't know if the effect would be worth writing code to compensate
for, but, the beauty of a PIC solution would be that you could compensate by
filtering out fast spikes if you wanted to.


>Man, this can go forever.  I wonder how HTR (or fire) works in a zero
>gravity ambient.

Hmm... no convection.  I would imagine that fire wouldn't work very well
because the heated gas molecules would tend to hang out near it and oxygen
could only reach the combusting materials by diffusion or by the slow air
circulation patterns caused by the expanding gasses interacting with
surfaces.  On the other hand, heat transfer would be slow (except via black
body radiation) so perhaps the low rate of oxygen delivery wouldn't
necessarily extinguish the fire if its surroundings reflected back most of
the infrared and visible radiation.  If the combusting materials were in
contact with (or close to) a cooler solid object then I would expect it to
go out.  Dunno ... it sure would be interesting to try though.

Ken

2000\01\20@034909 by Robert A. LaBudde

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<x-flowed>At 02:17 PM 1/20/00 +1100, David wrote:
> >temp. sensor in a hot-sensor anemometer. I was not aware that wind chill
> >worked on dry objects (hence the theory behind wet-bulb/dry-bulb humidity
>
>Well heat is conducted from the surface into adjacent air, which carries
>the heat away by convection (the wind).  Thus cooling your sensor, whether
>it is
>dry or not. :)

This is not specifically 'wind chill', which relates to an apparent heat
loss from evaporation, which depends on the windspeed.

The effect you describe is simply thermal equilibration between the sensor
and the ambient air through convection heat transfer. Once the temperatures
match, no more heat flows.

================================================================
Robert A. LaBudde, PhD, PAS, Dpl. ACAFS  e-mail: RemoveMEralTakeThisOuTspamlcfltd.com
Least Cost Formulations, Ltd.                   URL: http://lcfltd.com/
824 Timberlake Drive                            Tel: 757-467-0954
Virginia Beach, VA 23464-3239                   Fax: 757-467-2947

"Vere scire est per causae scire"
================================================================

</x-flowed>

2000\01\20@051923 by Ken Webster

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Robert A. LaBudde wrote:
>At 02:17 PM 1/20/00 +1100, David wrote:
>> >temp. sensor in a hot-sensor anemometer. I was not aware that wind chill
>> >worked on dry objects (hence the theory behind wet-bulb/dry-bulb
humidity
{Quote hidden}

The most common use of the term 'wind chill' does simply refer to this
thermal equilibration via convection heat transfer and does not appear to
have anything to do with evaporation (unless the Internet search I just did
pulled up particularly biased results).

See "wind chill index" in this NWS glossary:

http://www.nws.mbay.net/glossary.html#chill


Also see:

http://www.princeton.edu/~oa/safety/hypocold.html

http://www.athena.ivv.nasa.gov/curric/weather/adptcty/windchil.html

http://explorezone.com/weather/windchill.htm

http://www.physics.ohio-state.edu/~aubrecht/handc.html


The only article I found which may suggest (if you really stretch it) that
evaporation plays a role is this one:

http://www.srh.noaa.gov/FTProot/FFC/html/wci.htm

(because the original experiment to emperically develop the wind chill
formula we use here in the states consisted of leaving a pan of water
exposed to the wind and measuring how much time it took to freeze)

Note, however, that the article does not mention evaporation and there is no
humidity term in the formula.

Ken

2000\01\20@094944 by Alan King

picon face
"Robert A. LaBudde" wrote:>

> >Well heat is conducted from the surface into adjacent air, which carries
> >the heat away by convection (the wind).  Thus cooling your sensor, whether
> >it is
> >dry or not. :)
>
> This is not specifically 'wind chill', which relates to an apparent heat
> loss from evaporation, which depends on the windspeed.
>
> The effect you describe is simply thermal equilibration between the sensor
> and the ambient air through convection heat transfer. Once the temperatures
> match, no more heat flows.

 Wind chill stops as well when the temperatures match.  Convection IS slow wind
chill that happens because warm air rises.  Wind accelerates this by bringing
more cool air molecules into contact with the warmer body in x amount of time.
 Yeah, you feel lots colder at the beach when you're wet,  but that's because
wind also accelerates the evaporative heat loss, not because that's what 'wind
chill' is because you're wet.  You still feel noticably colder at the same temp
(at least lower than ~85deg skin temp) when the wind is blowing vs not, even
when you're dry.  That is wind chill.
 I'll make a WAG and bet your PhD is in computers..

2000\01\20@121847 by Tom Handley

picon face
  Sean, Wind Chill goes back to Paul Siple, an Antarctic explorer in the
late 30's and early 40's. It is simply a comfort index (like Heat index)
that only applies to humans and animals. The Wind Chill index is the
equivalent temperature where the heat loss of exposed flesh is the same as
it would be in a calm wind. You normally apply it with wind speeds between 5
to 40 MPH. Below 5 MPH, your body heat warms the air around you providing
some insulation. Above 40 MPH, there is little discernable affect.

  The standard formula used by the National Weather Service is:

     Twc = 0.0817 * (3.71 * sqrt(V) + 5.81 - 0.25 * V) * (T - 91.4) + 91.4

  Where:

     Twc = Wind Chill in degrees Fahrenheit.
     T   = Current temperature in degrees Fahrenheit.
     V   = Wind speed in Statute miles per hour.

  In my PIC-based weather station, I generated a Wind Chill table using an
Excel spreadsheet and downloaded the data to a 512KByte NVSRAM in my weather
station. I did the same for Heat Index and Dew Point. The NVSRAM is a Dallas
Semiconductor DS1647 Nonvolatile Timekeeping RAM which also includes an
RTCC, as well as a lithium battery and power-fail write protect in a
standard JEDEC 512K x 8 SRAM package. It also stores hourly and daily data
from a variety of sensors for up to one year.

PSBS:

  This is one area where I have a great deal of expertise and I've been
following this discussion with a great deal of interest. The main problem
with hot-wire, acoustic, and other methods of measuring wind speed is that
they are limited to a controlled environment (ie: wind tunnel). For
meteorological use, the rotating cup anemometer is the standard used
throughout the world. It is simple, reliable, and impervious to wind
direction. Tom McGahee did an excellent job of summing up the various wind
sensors in his recent article about his work with his students on this
subject. If you missed it, check the PIC archives for the last couple of
days.

  If you are serious about doing a weather station, get your `paws' on the
"Federal Meteorological Handbook No. 1, Surface Weather Observations and
Reports". It is consistent with agreements and publications of the World
Meteorological Organization (WMO), the International Civil Aviation
Organization (ICAO), specifically WMO No. 306- Manual on Codes, and ICAO
Annex 3- Meteorological Services for International Air Navigation, and civil
as well as military weather services. You can find the on-line version of
FMH-1 at:

     http://www.nws.noaa.gov/oso/oso1/oso12/fmh1.htm

  Other related sites:

     National Weather Service (NWS)
        http://www.nws.noaa.gov/
     National Oceanic and Atmospheric Administration (NOAA)
        http://www.noaa.gov/
     World Meteorological Organization (WMO)
        http://www.wmo.ch/
     American Meteorological Society (AMS)
        http://www.ametsoc.org/AMS/amshomepage.cfm

  Finally, I'm surprised that no one mentioned Davis Instruments which is
probably the largest vendor of weather systems and sensors in the
`home to commercial' market. You can find them at:

     http://www.davisnet.com/

  - Tom

At 09:26 PM 1/19/00 -0500, Sean Breheny wrote:
>Several people mentioned windchill as a possible effect with the reference
>temp. sensor in a hot-sensor anemometer. I was not aware that wind chill
>worked on dry objects (hence the theory behind wet-bulb/dry-bulb humidity
>sensors). Can someone please tell me why it is a consideration in this
>case? Are you concerned that water might condense onto the sensor?
>
>Sean


------------------------------------------------------------------------
Tom Handley
New Age Communications
Since '75 before "New Age" and no one around here is waiting for UFOs ;-)

2000\01\20@192415 by Sean Breheny

face picon face
Thanks for all the answers. I should have stated my question better.

I understand the effects on a hot sensor, I was talking about the reference
sensor, which is not self-heating, it is just intended to reach the same
temp as the air. If I remember correctly, several people on the piclist
said, in the recent anemometer discussion, that this unheated reference
temp sensor should be inside a box to prevent wind chill from affecting it.
I don't understand why this is, since the sensor is not heated and not wet.

Sean


At 03:47 AM 1/20/00 -0500, you wrote:
{Quote hidden}

| Sean Breheny
| Amateur Radio Callsign: KA3YXM
| Electrical Engineering Student
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2000\01\21@005324 by Ken Webster

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My bad!  One of these days I gotta learn how to read :o)

Anyhoo, I guess I just missed some of those earlier posts.  Everything I've
ever seen of physics or chemistry or physical chemistry says that it doesn't
matter.  No need no box.  Unless your sensor has a built-in wetted sponge or
something.


More importantly:

I stumbled across something on the Internet that was quite interesting.
With a sensor with a low enough thermal mass it may be fairly easy to make a
thermal anemometor using only one sensor and do away with the reference
sensor entirely.  Of course I was too stupid to save the URL and can't seem
to find it again but it was a technique called "transient thermal
anemometor" or something like that.

How it works:  apply a brief pulse of current to the sensor to raise its
temperature and then track the decay.  The rate of decay is related to the
rate of airflow and the asymptote gives the ambient temperature.

Pretty nifty, eh?

Ken




{Quote hidden}

2000\01\21@122934 by Wagner Lipnharski

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Ken Webster wrote:
> Anyhoo, I guess I just missed some of those earlier posts.  Everything I've
> ever seen of physics or chemistry or physical chemistry says that it doesn't
> matter.  No need no box.  Unless your sensor has a built-in wetted sponge or
> something.

I am not sure. Suppose in a quiet wind early morning moisture is
deposited over the two sensors, the hot one will evaporate it fast, the
ambient temp sensor will will take much longer to do that, most with the
wind help. The little water molecules over the sensor (evaporating) will
not create a delta temp?  Suppose two equal thermistors, one wet another
dry, exposed to an increasing temp environment (morning sun), both will
show exactly the same resistance?

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It is interesting. Software algorithm should be much more interesting.
Imagine if the base temp is 80¡F, applying the short pulse takes the
temp instantly to 83¡F, tracking temp down it takes 1 second to get
82¡F, 2 seconds to 81¡F and 5 seconds to 80¡F, so the HTR (heat transfer
rate) is calculated resulting in a wind speed, but waiting a little
longer, it would take 10 seconds to reach 79¡F, since ambient
temperature was changed during the measurement... nasty "if then else"
routines I guess...
:)

Wagner

2000\01\21@124042 by Alice Campbell

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www.fluidcomponents.com/frameset.html

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2000\01\21@124457 by Alice Campbell

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Date sent:              Thu, 20 Jan 2000 22:47:02 -0700
Send reply to:          pic microcontroller discussion list <RemoveMEPICLISTspamTakeThisOuTMITVMA.MIT.EDU>
From:                   Ken Webster <KenEraseMEspam.....WEBSTER.ORG>
Subject:                Re: Windspeed circuit?
To:                     EraseMEPICLISTspamMITVMA.MIT.EDU
even better:

uhavax.hartford.edu/~biomed/gateway/ElectricalResistanceThermometers.html
alice
> >I understand the effects on a hot sensor, I was talking about the reference
> >sensor, which is not self-heating, it is just intended to reach the same
> >temp as the air. If I remember correctly, several people on the piclist
> >said, in the recent anemometer discussion, that this unheated reference
> >temp sensor should be inside a box to prevent wind chill from affecting it.
> >I don't understand why this is, since the sensor is not heated and not wet.
> >
> >Sean

2000\01\21@133725 by Ken Webster

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Wagner Lipnharski wrote:
>Ken Webster wrote:
>> Anyhoo, I guess I just missed some of those earlier posts.  Everything
I've
>> ever seen of physics or chemistry or physical chemistry says that it
doesn't
>> matter.  No need no box.  Unless your sensor has a built-in wetted sponge
or
>> something.
>
>I am not sure. Suppose in a quiet wind early morning moisture is
>deposited over the two sensors, the hot one will evaporate it fast, the
>ambient temp sensor will will take much longer to do that, most with the
>wind help. The little water molecules over the sensor (evaporating) will
>not create a delta temp?  Suppose two equal thermistors, one wet another
>dry, exposed to an increasing temp environment (morning sun), both will
>show exactly the same resistance?

Yes, that would cause a false reading.  I guess that's even more reason why
the transient technique would be better.  If you maintain a high enough
average temperature then moisture should not condense on the one-and-only
sensor.


>> I stumbled across something on the Internet that was quite interesting.
>> With a sensor with a low enough thermal mass it may be fairly easy to
make a
>> thermal anemometor using only one sensor and do away with the reference
>> sensor entirely.  Of course I was too stupid to save the URL and can't
seem
>> to find it again but it was a technique called "transient thermal
>> anemometor" or something like that.
>>
>> How it works:  apply a brief pulse of current to the sensor to raise its
>> temperature and then track the decay.  The rate of decay is related to
the
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The decay should simply be exponential since the rate at which the sensor's
temperature drops is proportional to the difference between the sensor's
temperature and the ambient temperature.  Thus it should follow a curve of
the form:

y = y0 * e ** (k * t)

Where y is the difference between the sensor temperature and the ambient
temperature, y0 is the initial difference between the sensor temperature and
the ambient temperature, e is the base of the natural logarithm, ** stands
for "to the power", t is time, and k is a negative number representing the
rate of decay.

If the above equation is solved for k then k could be plugged in to a simple
equation derived from the approximation to King's law (or a more accurate
equation taking into account altitude, barometric pressure, and humidity if
I can find or derive such a beast) to get the wind speed.

The nice thing about this exponential equation is that you can just sample
two points on the curve to solve for k .. there is no need to wait for the
temperature to decay all the way down until it is close to the ambient
temperature.  Simply heat the sensor up, take a reading, wait a fixed amount
of time, take another reading, then plug-and-chug with the math.

The best thing is that making a PIC do the math would be a good use of the
64-bit floating point library I wrote -- I'd hate to think I did all that
work just for one project :o)

Ken

2000\01\21@141603 by Ken Webster

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I wrote:
>The nice thing about this exponential equation is that you can just sample
>two points on the curve to solve for k .. there is no need to wait for the
>temperature to decay all the way down until it is close to the ambient
>temperature.  Simply heat the sensor up, take a reading, wait a fixed
amount
>of time, take another reading, then plug-and-chug with the math.


Oops!  I meant 3 points.

The y and y0 in the equation:
y = y0 * e ** (k * t)

are:
y = T - Ta
y0 = T0 - Ta

where T = temperature of the sensor (known) and Ta is ambient temperature
(unknown .. must solve for this).  If I'm thinking right it will take 3
points on the curve to sove for Ta and k.

Ken

2000\01\21@144716 by jamesnewton

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Ok, folks, as fascinating as this subject may be, it's off topic.

Mark future (after this one <GRIN>) posts on this thread with an [OT] (not
OT ] or [OT} or (OT) or off topic or ....)

---
James Newton (PICList Admin #3)
RemoveMEjamesnewtonEraseMEspamEraseMEpiclist.com 1-619-652-0593
PIC/PICList FAQ: http://www.piclist.com

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