MSefi.com read only forum (Jan.15,2018)

I'm about to modify a 6-0 trigger wheel by removing one of the teeth to make it a 6-1 trigger wheel.

Here's a picture of what the trigger wheel looks like:


I'm going to remove the "long tooth" because it will keep the overall situation a simple as possible, i.e. if I where to choose to remove any of the other teeth I'd still have to come back to the long tooth and cut it to the exact size of the other teeth. So removing the long tooth seems to be the logical choice.

This trigger wheel is from a Suzuki motorcycle and it will be spinning at up to 14,000 RPM so having the wheel perfectly balanced is important. It seems wise to only cut away the minimum amount of metal from the long tooth that would prevent the VR sensor from "seeing" the tooth as it passes by. Removing as little metal as possible means that I will have to do less work (cutting on the other side of the wheel) to rebalance the wheel.

I'm posting here to ask: How much metal would I need to remove from this trigger wheel's long tooth to keep the sensor from "seeing" it at all as it passes by? Or, to put it another way, How big would I have to make the air-gap between the bike's VR sensor and the cut-down tooth to make it fail to register as it passes the VR sensor?.
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I've seen comments about tooth designs to the effect that the missing tooth should be ~ 1/2 height vs fully-removed, as it has some benefit to the signal form. The discussion was about high-count wheels, 36-1 and 60-2, but it would certainly suggest that a half-tooth height on a *60* tooth wheel - far smaller than your big clunky 6-teeth - is sufficient to show as 'missing'.