*I have been asked the question how the deformation value in mm relates to the ovality %. For example, I have a deformation (mm) 0.54 and an ovality % 0.41 for position A on an uphill tyre measured on a 5.6m dia. kiln. Can you also tell me how the Delta Temp is worked out. Thanks*

%Shell Flexing (Ovality) is taken as a % movement of the nominal diameter of the shell. Then mm deformation is the "delta H" in the Rosenblad formula and where t is the shell plate thickness at the point of measurement e.g. 25mm: {4/3 x (KilnDiameter+2xt)^2 x deltaH}/KilnDiameter*100%= %ovality. In your case: {4/3 x (5.6+.050)^2}xdeltaH/5.6*100%=0.41%. This calculation is valid for an ovality instrument of 1 meter length. The amount of shell flexing per rotation is therefore: 0.43/100*5600mm= 24mm diametric change. Delta T is simply the average shell temperature around the tire less the average tire temperature.