How can I accurately weigh my head?

How would you accurately weigh somebodies head, without removing it from his or her person?

• Answer:

  This is very easy and simple. We've done this in our Physics class when we were learning about weight, density, displacement, and behaviour of fluids. Note that this is only one of the three methods that I know. This method uses the % (percentage) rise of water level of head over body, i.e. head level rise divided by body level rise, multiply this by actual body weight as measured by the accurate weighing scale, then you get the weight of head. There is a reason why we use a cylindrical tank in this case. Elements/Gadgets/Materials used: 1. Weighing scale 2. Cylindrical Water tank, preferrably the one used in hydro therapy clinics Note: the hydro tank has accurate water lever measurements 3. One person willing to strip down to just underwear (so as to get the most accurate measurement) 4. Step ladder, at least same height as water tank. Procedure (with simplified example to show the answer) 1. Weigh the person using the scale. E.g. 98 pounds. 2. Measure the water lever in the tank. E.g. 60 inches. 3. Dip the whole body of the person into the tank and measure the water level. E.g. 72 inches. For the whole body, there was 12 inches of water rise: that's 72 - 60 = 12 4. Move the step ladder beside the tank. Person will climb to the top of the ladder and bend in such a way that he will dip just his head into the tank. Measure the water level rise. E.g. 61.3 inches For the head, there was 1.3 inches of water rise: that's 61.3 - 60 = 1.3 inches. % level rise = (1.3 div 12) = 0.1083333 or 10.83% therefore, based on % level rise: weight of head = 10.83% of 98 pounds = 0.1083 X 98 = 10.61 pounds *==========================* We could also use density as the factor to measure the weight of the head, and this involves getting the accurate density of the water. In any case, this will also work. But the easiest method is the one I presented above.