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Same answer as before David - sorry, it has to come down to the quality of your connections.
If you are melting the fuse enclosure before the fuse itself melts, then you do not have an over-current situation but a voltage drop situation (the combination of which, with current - which is a given - is power dissipated as heat) - that can only be caused by resistance which can only be caused by the quality of wire & the connection.
That's the whole story, nothing else.
P= V*I or P=R*I^2
I is a fixed requirement of the system so the only thing that can increase P is R (will increase V but that is consequence of R, not the other way around)
Doesn't take much since the power increases exponentially with the current.
Maybe you could include some picture of how you did the mod ...
Take your voltmeter and stick one of the probes through the insulation at the output of the R/R itself (close up to where it exits the unit) - use a needle or something if you prefer not to make a hole hole in the insulation; put your other probe on the battery positive terminal. Start bike, have high beams etc on and any other load you would typically have on your bike.
What is the voltage you are measuring?
Check at idle and also at 5K or so
p.s. of course if current demand is higher than normal that would also have an effect - but with quality fuse holder and 12ga wire, you should still not be melting anything - and obviously not drawing in excess of 30A or fuse filament itself would blow.
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