Hi,
This is an interesting project; personally I rarely see snow here in London, but probably more than the "father" of Weatherduino (in Portugal) or the majority of users in Australia, etc. There might be some more interest if it could also support "Water Level" detection, but I accept that the design issues are quite different. So I will offer only a few "Random thoughts" on how I might deal with the issues if it were my project.
Initially, I did find one RS232 Laser Rangefinder ("LDB6") on ebay, but at US$ 75. If you consider the "risks" of buying (or more particularly selling) a quantity of 100 units at 20 Euros, and the costs of individual shipping, etc., then $75 is not completely unrealistic for a single unit (compared with the sensors in the AQM, for example). However, a further search lead me to
AliExpress where they have one module at less than US$ 40, apparently with a Minimum Order Quantity of 1.
But first I would also consider other technologies: Ultrasonics still seems like a "possible" choice, but definitely using "outdoor" transducers, for example as used for car reversing sensors. It's potentially more resistant to "interference" from falling snow, etc., but I accept that the surface of snow is difficult to simulate for testing. Also, other developers have discovered that using ultrasonics is NOT an easy way to measure the water level in a well.
Therefore, if we assume that a laser rangefinder is the "best" approach, there are still two methods: "Time of Flight" (as used by the handheld units) and the traditional "Triangulation" methods used by older cameras, etc. Basically, Triangulation is better for "Short" distances and Time of Flight for "Longer" distances, but which one is this? In either case the Laser makes the optical design much easier, but for a "Do It Myself" project I would try the triangulation method first; I think it might be done with a $1 laser (pointer) and "decoding" the CVBS signal from a (modified) <$10 CVBS (CCTV)
camera module. That's more my "expert" field, so I might try it sometime, but the optical design won't be easy so let's move on to the "off-the-shelf module" ToF method:
I'm quite surprised that Time Of Flight can be used in "Consumer" equipment, since the mm accuracy needs a time resolution of only
picoseconds, with a 1 nanosecond delay corresponding to 15 cms. So the (first) three questions I would be asking of a "domestic" rangefinder are: "Will it work below -10 degrees C", "How Long will it work for" and "How will it be powered". If there is plenty of power available then the first question can be solved with a "heater", but otherwise most consumer electronics items aren't specified to work much below zero Centigrade. It does appear that many of the Laser diodes on ebay are specified to below -20C (they would probably self-heat quite quickly anyway), but their life is quoted at around 1000 hours. That's only just over one month continuous use, which suggests the application duty cycle should be perhaps less than 1% to give a "good" lifetime (~10 years ?).
Now another question: "Might the Laser melt the snow"? This may seem a silly question and that I've been watching too much James Bond, but (high power) lasers are used to cut metal and low power laser pointers can be (instantly) dangerous to eyes. So I decided to do a "Back of Envelope" (literally) calculation; I had to go back to my "schoolboy physics" and the calculation is not too difficult, but I do need to introduce two "Physical Constants". The first is the conversion factor from electrical energy to thermal energy (heat) which is "4.2 Joules per Calorie". A Joule is 1 watt per second and a Calorie raises the temperature of 1 cc of water by 1 degree C. So the "constant" says that (for example) 1 amp flowing from 4.2 volts (= 4.2 Watts) will raise the temperature of one cc of water by 1 degree C each second.
Before going further, a "sanity check" to prove that the calculation is sensible: If we put 1 litre (1000 cc) of water into an electric kettle then we should need 4.2 x 1000 x 80 watt-seconds to heat the water from 20 degrees to 100 degrees (to start it boiling). If the heater is 1 kW then it would take 336 seconds or almost 6 minutes. In practice the heater might be nearer to 3 kW, which could take only two minutes, but in reality the time will be longer because the kettle loses some of the heat to the surrounding air. But IMHO it's a good verification that the maths is correct.
Now, to melt snow, we need another Physical Constant, the "Latent Heat of Fusion (melting) of Ice", which is 80 calories/cc. This means that the SAME amount of energy is needed to melt ice as to raise its temperature by 80 degrees C (which was the value used above). If a laser is pointing at ice, the beam is perhaps 1 mm in cross-sectional area so consider a 1 mm cube, which is 0.001 cc. The ebay lasers are specified as 5 mW, but the laser rangefinders only 1 mW (perhaps taking into account that the light is modulated) so let's assume 1 mW. That's 0.001 watt, so the melting time could be the SAME as calculated for boiling the kettle, i.e. just under 6 minutes.
Of course the laser will not be operated continuously so we need to estimate the "duty cycle", or the length of time which will NOT melt a significant amount of snow. Probably we only need to consider the time whilst the air temperature is "near to zero degrees", because at lower temperatures the air will conduct heat away (like the boiling kettle was slowed down). Conversely, at temperatures above zero the snow will be melting anyway, so I have "guessed" a total period of 1 day, which gives a duty cycle of 6/1440 minutes or 0.4%.
In practice, there are more factors to consider: Snow consists mainly of "air" to the extent that 10 cms depth of snow is equivalent to about 1 cm of ice/water. Thus snow should melt about 10 times faster (in time or depth) than ice, but snow is "White" which will initially reflect away quite a high proportion of the energy (but not 100%). However, when snow starts to melt, it can become "slush" (a mixture of water and ice crystals) which is quite "grey" and not highly reflective.
So in conclusion, my "guesstimate" is that the laser should be operated with a duty cycle of less than 1%, but I'm prepared to be proved wrong.
Perhaps the (effective) laser output power is lower, but the input power is typically 30 mA @ 3v = 90 mW, so a few mW of light output doesn't seem particularly high. Or maybe the light energy isn't (all) converted into heat, but if not, then what form of energy IS it converted into? The laser's 1 mW / mm2 extrapolates to 1 kW per square metre which is similar to peak summer sunshine, but I don't have any knowledge of how quickly snow melts in full direct sunlight.
Cheers, Alan.
![[Image: _dsc9418.jpg]](https://picload.org/view/ddairlwi/_dsc9418.jpg.html)
