> In the case of the car, that's going to be the sky (surely not that warm 
> on a cloudy day?) or the wall (presumably at equillibrium with the 
> surroundings). And yet, I got a fairly warm reading.

The objects are not 100% reflecting though, what I meant was your reading is 
going to be lower than you expected because of the reflection.

> In the case of the cooker, that ought to be the gas flames. [Now there's a 
> question - can an IR thermometer measure the temperature of a flame? Or 
> will it measure the next solid object behind it?]

It works the same way as visible light.  Can you see the next solid object 
behind a flame?  It depends how bright the flame is and how bright the 
object is behind.

> If you wanted to be technical, presumably the IR arriving at the sensor is 
> the SUM of reflected and emitted? (What the computer calculates the 
> temperature at is another matter of course...)

The computer assumes the following equation, where e is the emissivity:

sensorReadingIR = emittedIR * e + reflectedIR * (1-e)

Rearranged:

emittedIR = ( sensorReadingIR -  reflectedIR * (1-e) ) / e

It calculates reflectedIR from the "environment temperature" you program in 
to it.  You also tell it e, and of course it knows the sensorReadingIR.  It 
then calculates the temperature of the surface from emittedIR.

As you can imagine, if you program e to be 0.9, but you're actually 
measuring something with a lower e, the computer is going to get the 
calculation wrong and tell you it's at a lower temperature than it actually 
is (assuming the object is hotter than the environment).

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