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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link=blue vlink=purple><div class=WordSection1><p class=MsoPlainText>Dear Frank<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>I am adding a little to what Kevin wrote:<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>OK... basically, a surface radiates or absorbs energy as a function of its emissivity. A "Black Body" is such a perfect radiation and absorption surface. It has an Emissivity of 1.00 Good "real world absorbers" have an emissivity in the range of about .9 to .95, while "poor real world absorbers" have an emissivity in the range of about .02 to .05. <o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Brick and concrete are about 0.93. Cheap instruments that cannot be edited are set on 0.95 to 0.93 as their default.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>> heated up by the radiant heat it then gives off. OR can you reflect heat <o:p></o:p></p><p class=MsoPlainText>>> to another surface without heating the surface doing the reflecting?<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:black'>It will heat by the amount it absorbs. 65% reflection is already pretty good. It would look shiny to you.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText>># I did an interesting experiment as follows:<o:p></o:p></p><p class=MsoPlainText>I have a flat mirror on a wall. I have a cheap "Laser Thermometer", which <o:p></o:p></p><p class=MsoPlainText>when pointed to the wall beside the mirror reads<o:p></o:p></p><p class=MsoPlainText>: 45 degree angle: 55 F<o:p></o:p></p><p class=MsoPlainText> 90 degree angle 55 F<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText><span style='color:black'>It is reacting the IR from across the room. You will get very little IR from the mirror itself as it has a very low emissivity. Probably below 0.2.<o:p></o:p></span></p><p class=MsoPlainText><span style='color:black'><o:p> </o:p></span></p><p class=MsoPlainText>>#When pointed at my hand, I read 92 degrees F<o:p></o:p></p><p class=MsoPlainText>>When pointed to the mirror, but with the reflected red dot hitting my hand, <o:p></o:p></p><p class=MsoPlainText>>I read 60 F<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>It is probably seeing something larger than your head, in the mirror. The spread of the field of vision is probably 1:8, not a dot. A good quality one like the one you have Frank is about 1:12, right?<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>>3: It measures the surface temperature of the mirror, and not the <o:p></o:p></p><p class=MsoPlainText>temperature of teh reflected surface.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>I am pretty sure it is seeing a reflected surface from across the room.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>If you put a small mirror on the side wall of a freezer and point at the mirror with the IR gun at not-90-degrees I should show you the temp of the back wall, say. If the object in the reflection is black it would be better because ice has a low <i><span style='font-size:14.0pt'>å</span></i>. In fact that experiment should be definitive if the icebox is not covered in frost.<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p><p class=MsoPlainText>Regards<o:p></o:p></p><p class=MsoPlainText>Crispin<o:p></o:p></p><p class=MsoPlainText><o:p> </o:p></p></div></body></html>