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Segnala un problema nella traduzione
Thank you
You correctly identified this as the critical number. The first parameter is "OpCodes.Ldc_R4", which tells the computer to push the operand (1.33f) onto the stack. On the next line, you can see it yields "OpCodes.Div", which is the division operator. To see this as a percent, assume you have 1 already on the stack, as the amount of energy in the battery. So the VM does 1/1.33, which leaves 0.75 on the stack. (Obviously, that number gets scaled by whatever number the battery returned).
Bottom line: If you want to have it only take 25% of your stored energy, change that 1.33f to 4.0f.
I saw 1.33f in the code.
What does this mean?
That's better
You must have Harmony mod for this to work.