Interaction between voltage, current, and temp

I have a question regarding the interaction between solar panel voltage, current, and temperature. My confusion
comes from examining the current-voltage charts for a Kyocera KC85T panel.

For discussion, lets assume that the charge controller is trying to equalize at 15.5V. My assumption is that the battery system voltage (set via the charge controller) will determine the operating voltage produced by the panel.

The STC specs for a KC85T panel are as follows:

Vmp = 17.4
Imp = 5.02
Voc = 21.7
Isc = 5.34

Temp Coefficient Voc = -8.21 x 10 -2 V/deg C or (-.0821)V/degC
Temp Coefficient Isc = 2.12 x 10 -3 A/ deg C or (.00212)A/degC

(This was a surprise for me. I had never seen a temp coefficient specified for Vmp)
Temp Coefficient Vmp = -9.32 x 10 -2 V/deg C or (-.0932)V/degC


Using the temp coefficient with the STC Vmp, I can calculate the voltage that will be generated at other
temperatures.
50C would be: 17.4 - (.0932 * 25) = 15.07V
75C would be: 17.4 - (.0932 * 50) = 12.74V

Here's the problem. Using Kyocera's current-voltage graph from their specifications, my calculations don't appear to match. I enlarged their graph, and best I can estimate, the voltage at 50C is about 14.5. I know this is close to the 15.07, but it's obviously less than 15. For 75C the chart value appears to indicate about 12V. Why isn't the temp coefficient more accurate? I notice that the graph is a curve, and not a straight line. As I recall from algebra many years ago, a curved line is defined by a non-linear equation, and a straight line by a linear equation. The temp coefficient as specified by Kyocera is linear. Is this the reason for the error? If so, why doesn't Kyocera give you the "real" equation to define the graph? Or could it be that the graph is just poor quality. So, what am I doing wrong?

My ultimate goal is to predict the current that will be available at a given temperature. I'm assuming in my calculations above, that the current is still 5.02A at the calculated voltage. In order to bring the voltage up to the desired 15.5, the current is going to have to drop significantly. Is there a way to determine this more accurately than just estimating it from the chart? My first thought was to calculate the watts from the temperature reduction, and then "back into" the current for the desired voltage:

12.74V * 5.02A = 63.95W

63.95 / 15.5 = 4.13A

Ok, can you tell I know just enough to get into trouble? So everybody stop laughing, and tell me what I'm doing wrong

Thanks,

John