UcLc kerning glyphs set with 'no kerning'

[7roy]

Hi everyone, I'm new to KernOn, so I'm still in the newbie experimenting phase. I’ve kerned my Uc-Uc and Lc-Lc, setting my flat-sided glyphs to 'no kerning.' But now I see this causes problems for my Uc-Lc kerning. For example, my 'Wn' pair does not kern because the left side of my 'n' is set to 'no kerning'. I thought I would set this to Ind. and all would be fixed but when I 'Kern-On' it resets it to 'no kerning'. What am I doing wrong, or what should I be doing for these pairs?

[Tim Ahrens]

You should definitely not set the left side of the /n to “No kerning” as there are many combinations with /n as the second glyph that require kerning.

I’d strongly advise against using “No kerning” for any letters.

[7roy]

Thanks Tim

[7roy]

Hey Tim, another newbie question for you.

I have a string showing HHBBHH. KernOn keeps telling me the right side bearing of my B is too tight so it kerns it away from the H by 7. Visually it’s now too loose to my eye, so I make a model pair and set it to zero. But now that affects my nn pair, which is also a model pair set to zero. So my next step is to set my BH pair as independent, but when I KernOn the BB pair does not pick up the same kern as my BH pair when both left side bearings and forms are identical. What am I supposed to do here?

[7roy]

Another response I had: "Removed model status for ẞH as it is too tight compared to ẞD"

[Tim Ahrens]

KernOn keeps telling me the right side bearing of my B is too tight so it kerns it away from the H by 7. Visually it’s now too loose to my eye, so I make a model pair and set it to zero.

Yes, that’s the way to go!

But now that affects my nn pair, which is also a model pair set to zero.

That’s impossible. If nn is a model, it cannot be affected by other models.

So my next step is to set my BH pair as independent

I would strongly advise against that. Don’t work with independent pairs if they are pairs that should fit into the normal system of spacing and kerning between the letter shapes.

Independent pairs are a “parking” for deactivated models. The only reason why models, when they have to be removed, are not switched to auto pairs is that they would lose their previous kerning value. By switching them to independent pairs this information is retained.

I’d recommend to really try to resolve this without using any independents.

"Removed model status for ẞH as it is too tight compared to ẞD"

It’s hard to say without seeing the font. Assuming H and D are (nearly) identical on the left, why would you want ẞH as well as ẞD as models?

You can send me the font if you want, then I can have a closer look.

[7roy]

Thanks Tim