Hi, guys, sorry for my English but must react to nonsense presented in your article. You simply cannot measure thermal power production of any HDD just with thermometer. What you have been actually comparing was not the total emitted thermal power but thermal conductivity between the HDD case and outside air. Let me to explain the difference: if you would use an older Seagate HDD with plastic jacket I am sure that the temperature on the jacket will be lowest of all your tests but not because of extremely low power production of such HDD but because of low thermal conductivity of plastics!!! Actually you can have HDD with very low power dissipation but high surface temperature (if the HDD's surface to air thermal resistivity is high) but because of the low power produced such a HDD can be actually cooler in some PC (with a given self-cooling abilities), than other which has lower surface temperature in the open air (because of better cooling) but high total amount of thermal power produced. Why ? Because despite of the temperature of the HDD itself all thermal energy must be emitted out from the PC case and for a given PC is usually more important the total amount of heat which has to be emitted than individual temperatures of components inside. Better method would be to measure simply the power consumption of a given HDD and compare it to other, actually almost all electricity will be sooner or later converted to unwanted heat inside you HDD (or CPU or RAM or complete PC) and it must be than emitted out from the case (if you don't want to keep inner temperatures increasing). Or very simplified in one sentence: the lower will be power consumption of your HDD (and the same for CPU,RAM,VGA, etc.) the lower will be the temperature (in case that your are using the same cooling).