Howard, there are hundreds of things that you can calculate utilizing a thermal imaging camera. And what you are asking is one of them.
Unfortunately there are a bucket load of camera owners here at think the expense of equipment and training is a bunch of malarkey.
Did you know that in level II training you walk away with the ability to determine how many watts of electricity is being consumed by a lightbulb hidden underneath a 12 quart aluminum pot!?
Let’s try to wrap our minds around this for a minute and see where it takes us.
#1 we are trying to analyze the effectiveness of a radiant barrier.
The key here is “radiant”.
We’re not measuring how hot the barrier is with relation to other things in the attic we want to know how we can remove the amount of radiant heat that is affecting the total heat in the attic. After all that’s why we’re putting it in there.
So, what is a radiant barrier? Is something that blocks radiant heat (not conduction and not convection). It doesn’t matter if the radiant barrier is 100°. What matters is that what it is constructed of does not “radiate”.
So of you simply take your temperature readings as indicated below you will see that the truss members that are shadowed by the radiant barrier have a significant lower temperature than those exposed to the roof deck.
Divide this into each other and it will give you a percentage.
Note, if you put a substance on the radiant barrier that has a high emissivity (something that you can measure accurately) you are going to find the temperature of that radiant barrier is as close in temperature to that of the roof deck.
Conduction + convection + radiation = total heat load in the attic.
Because there is a huge amount of radiation affect going on in an attic space when you address the radiation amount of heat load significantly drops.
What happens if your roof is covered by a tree shadow?
Answer: you wasted your money on a radiant barrier.