Cumulus Support

Hi, due to last years heavy snow and extreme temperatures (down to around -12'C where i live) i'm wondering if there are such things to stop the freezing of my rainfall sensor. My station is an Fine Offset station.

Help would be greatly recieved.

Regards,
Adam Jones

you could build in a car lightbulb,gets pretty warm.
but you do need to power it.(12 volt)

also a ceramic resistor can be used for it,also needs power,dependent on the wattage of it.

I designed my own thermostatically controlled rain gauge heater after last winter's freeze-up but never got round to building it. Then with the recent sub-zero temperatures and snow, I checked how much heat I would need. I have a modified rain collector consisting of a 10" funnel on top of the Fine Offset gauge, held in a large plastic bucket (inverted). I found that 12v on a 33 ohm resistor was nowhere near enough so I've just designed it's bigger brother. NOTE:- I have not tested either version, so no guarantees. Using the darlington pair reduces the overall cost and component count as well as providing up to ten times the available current for the heaters - 10W ceramic cased wire-wound resistors. I'll make up and test the second circuit when I get the parts (ordered today).

This is my rain gauge modification to increase the sensitivity by about ten times with thermostatically controlled heater and thermal insulation.

Gina, that looks fantastic. Wish I had something like that as we are getting our first snow squalls off of Lake Huron right now and no precipitation reporting from my station.

Paul

I have just updated my web site with this rain gauge mod and added other info. I'll probably write a construction "how to" for the heater circuit later. Rain gauge mods are accessible from the Home page under Sensor Modifications.

With -4C last night and this morning I decided to do another test. While it was sunny and quite mild yesterday, I put a good thickness of polystyrene (mainly about an inch thick) from unwanted packaging, inside the bucket, tapered off to clear the funnel at the top. I did the diagram afterwards. The test was run with 2 33ohm resistors in parallel (68ohm not yet arrived). Half an amp of current supplied from my lab supply in a shed. So total power in the bucket was/is just over 4W and the funnel is clear of ice except at the very top. So the extra insulation helped but so did the lack of wind, I guess.

Hi Gina
Wondered if your thermostatically controlled heater circuit could be used to control one of these running off an old 12v car battery
Maplin 12v In-Car Ceramic Heater £7.49
http://www.maplin.co.uk/Module.aspx?ModuleNo=48775
PS have you done any wind vane mods yet

Hi Gina,

Just in case somebody not familiar with resistors tries to duplicate what you are doing, I believe you connected the 33 ohm resistors in series (not parallel) to achieve 68 ohms (66, but close enough). Or were you doing something different? I can't figure out a series or parallel combo that gives 0.5 A from 24 V or even 12 V? (maybe 16 V?)

Also, are you going to use the darlington as a switch, or is it actually being operated in it's (mostly) linear region?

I=V/R = 24v/68R = 0.35294amps
P = I 2 x R = 8.47053 watts

Although not as well done as Gina's, a string of outdoor Christmas lights wrapped around the gauge might do it.

You might experiment with a cup of water to see of it works near your dwelling before running power to your station.

I use the lights woven through my citrus trees and pineapple plants to get past the few very cold days we have.

Randy

Orion wrote:Hi Gina
Wondered if your thermostatically controlled heater circuit could be used to control one of these running off an old 12v car battery
Maplin 12v In-Car Ceramic Heater £7.49
http://www.maplin.co.uk/Module.aspx?ModuleNo=48775

The Maplin heater spec doesn't say how much current it uses or the power but quotes a 15A fuse so I would think it takes more current than the power darlington transistor can take (5A). There are higher power devices available though.

PS have you done any wind vane mods yet

No, not recently. I tried extending the vane some time back but it was worse. I think the best form of damping would be magnetic induction and I will be looking into this. But the problem with the F.O. unit is interfering with the reed switches. Two possible directions I have in mind - 1. Keep the F.O. vane and replace the read switches with reflective optical couplers and a Gray encoded optical disc, or 2. making a complete new wind vane, probably using 1-wire to collect the data.

If OTOH you mean heating the unit to prevent freezing up... Not yet but hopefully soon. I have components on order.

Charlie wrote:Just in case somebody not familiar with resistors tries to duplicate what you are doing, I believe you connected the 33 ohm resistors in series (not parallel) to achieve 68 ohms (66, but close enough). Or were you doing something different? I can't figure out a series or parallel combo that gives 0.5 A from 24 V or even 12 V? (maybe 16 V?)

The 33 ohm resistors are in parallel giving 16.5 ohms. Some volt drop occurs in the bell wire I'm using to connect heaters to power supply so I used the ammeter on the PSU to find out how much power the heater resistors were getting. Using Ohms law - P = I squared times R. Current was 0.5A squaring that gives 0.25 and multiplying by 16.5 gives 0.25x16.5 = 4.125W

Also, are you going to use the darlington as a switch, or is it actually being operated in it's (mostly) linear region?

It's used as a switch. The positive feedback on the op-amp ensures this.

Orion wrote:I=V/R = 24v/68R = 0.35294amps
P = I 2 x R = 8.47053 watts

Only 20volts available from bench supply so I put the resistors in parallel. But the connecting wire resistance was more than I thought - I didn't bother measuring it - just quickly strung some bell wire from rain gauge, along the fence, through the trees and into the shed ASAP coz' it was b cold! I might have done better to have put the resistors in series, but went by guesswork. Yes, 24v and 66 or 68 ohms gives around 8W. The bell wire resistance works out at 8.7 ohms. So...
Series :- 66 + 8.7 = approx 75 ohms. 20v will give 20/75=0.27A In 66 ohms that's 4.7W. (2.3W each resistor)
Parallel :- 16.5 + 8.7 = 25 ohm. 20v will give 20/25 = 0.8A. In 16.5 ohms that's just over 10W (5W each resistor).

So I was right that parallel was better.