John Rostron Posted January 13, 2018 Share Posted January 13, 2018 @atfitzy posted a thread about re-shaping a text block. I tried various Equations in Filter> Distort and the best I could come up with is: x=x y+(h-y)*(w-x)*x/w/w/a This produces the required arch in the upper margin of the block. The a parameter allows the user to increase the stretching effect. The default of one has no effect; reducing it will increase the effect. However, I find that reducing the parameter has no effect until the value goes below half, after which it has the desired effect. I can get the desired effect by putting a multiplier at the start of the equation: y+2*(h-y)*(w-x)*x/w/w/a Here are a couple of arched images using this formula: Otherwise the formula works as desired. Before I commit this to a macro can anyone explain this unexpected behaviour in the parameter value? John Polygonius 1 Quote Windows 10, Affinity Photo 1.10.5 Designer 1.10.5 and Publisher 1.10.5 (mainly Photo), now ex-Adobe CC CPU: AMD A6-3670. RAM: 16 GB DDR3 @ 666MHz, Graphics: 2047MB NVIDIA GeForce GT 630 Link to comment Share on other sites More sharing options...
Staff stokerg Posted January 16, 2018 Staff Share Posted January 16, 2018 Hi John, This is a bit over my head but hopefully @James Ritson or @Andy Somerfield would be able to clear up why it's acting the way it does John Rostron 1 Quote Link to comment Share on other sites More sharing options...
John Rostron Posted January 16, 2018 Author Share Posted January 16, 2018 1 hour ago, stokerg said: Hi John, This is a bit over my head but hopefully @James Ritson or @Andy Somerfield would be able to clear up why it's acting the way it does Thanks, @stokergI will live in hope. John Quote Windows 10, Affinity Photo 1.10.5 Designer 1.10.5 and Publisher 1.10.5 (mainly Photo), now ex-Adobe CC CPU: AMD A6-3670. RAM: 16 GB DDR3 @ 666MHz, Graphics: 2047MB NVIDIA GeForce GT 630 Link to comment Share on other sites More sharing options...
v_kyr Posted January 16, 2018 Share Posted January 16, 2018 Works for me when applied on a custom SVG grid under OSX. - The default of one has an streching effect on the grid here and also altering the "a" parameter slider continously updates the whole streching effect. John Rostron 1 Quote ☛ Affinity Designer 1.10.8 ◆ Affinity Photo 1.10.8 ◆ Affinity Publisher 1.10.8 ◆ OSX El Capitan ☛ Affinity V2.3 apps ◆ MacOS Sonoma 14.2 ◆ iPad OS 17.2 Link to comment Share on other sites More sharing options...
John Rostron Posted January 16, 2018 Author Share Posted January 16, 2018 The expression (w-x)*x/w/w should be zero at each end, rising to a maximum of 0.25 in the middle. Thus the arch should rise up a quarter of the current position above the x-axis (h-y). The a parameter at the default of 1 would have no effect. As a decreases, the value of the expression should rise in inverse proportion. It should do so strictly monotonically. It is not doing this on my Windows 10 PC. Note the 2* is just a fudge factor to make it behave as expected. (Note that strictly monotonic means that the rise should always be positive and non-zero.) John Quote Windows 10, Affinity Photo 1.10.5 Designer 1.10.5 and Publisher 1.10.5 (mainly Photo), now ex-Adobe CC CPU: AMD A6-3670. RAM: 16 GB DDR3 @ 666MHz, Graphics: 2047MB NVIDIA GeForce GT 630 Link to comment Share on other sites More sharing options...
Polygonius Posted February 2, 2018 Share Posted February 2, 2018 I have modified this equation from top a little bit. Please give a try - it should eliminates this A-jump. y+(h*c-y)*(w-(x*b))*x/w/w/(a*0.5) If B =1 and C = 0.5 than A is just scaling from the middle BUT!!! it works different for different layers (with different sizes). So my question is it possible that this filter is computing the whole canvas, not the size of the selected layer? Quote OSX 12.5 / iMac Retina 27" / Radeon Pro 580X / Metall: on! --- WWG1WGA WW! Link to comment Share on other sites More sharing options...
John Rostron Posted February 2, 2018 Author Share Posted February 2, 2018 @Polygonius, I shall have to work this one on both on paper and in Affinity. Thanks for your interest. John Quote Windows 10, Affinity Photo 1.10.5 Designer 1.10.5 and Publisher 1.10.5 (mainly Photo), now ex-Adobe CC CPU: AMD A6-3670. RAM: 16 GB DDR3 @ 666MHz, Graphics: 2047MB NVIDIA GeForce GT 630 Link to comment Share on other sites More sharing options...
John Rostron Posted February 4, 2018 Author Share Posted February 4, 2018 (edited) @Polygonius, I have now tried your equations. Putting /(a*0.5) at the end is algebraically the same as I had with /a at the end and 2* at the beginning. However, I now find that the a parameter still does not cut in at first, but only when it reaches a value of about 0.7-0.8. Curiouser and curiouser! (The link explains the origin of this expression for those unfamiliar with it.) Your use of parameters b and c are interesting. Parameter b just seems to scale along the y-axis, enhancing the value of a - not what I would have expected. Parameter c is more intriguing. It shifts the effective horizontal axis of the effect upwards. If this is used with an enhanced curve-shift (with parameter b), it gives the appearance of a rounded corner. I was applying these parameters to the square grid as in my first posting. Here is the grid with a=0.7, b=0.9 and c=0.3 (all values approximate). I am interested to see these effects on a real image! John Edited February 4, 2018 by John Rostron Coirrected link! Quote Windows 10, Affinity Photo 1.10.5 Designer 1.10.5 and Publisher 1.10.5 (mainly Photo), now ex-Adobe CC CPU: AMD A6-3670. RAM: 16 GB DDR3 @ 666MHz, Graphics: 2047MB NVIDIA GeForce GT 630 Link to comment Share on other sites More sharing options...
Alfred Posted February 4, 2018 Share Posted February 4, 2018 27 minutes ago, John Rostron said: Curiouser and curiouser! (The link explains the origin of this expression for those unfamiliar with it.) Dead link. John Rostron 1 Quote Alfred Affinity Designer/Photo/Publisher 2 for Windows • Windows 10 Home/Pro Affinity Designer/Photo/Publisher 2 for iPad • iPadOS 17.4.1 (iPad 7th gen) Link to comment Share on other sites More sharing options...
R C-R Posted February 4, 2018 Share Posted February 4, 2018 35 minutes ago, Alfred said: Dead link. John Rostron and Alfred 2 Quote All 3 1.10.8, & all 3 V2.4.1 Mac apps; 2020 iMac 27"; 3.8GHz i7, Radeon Pro 5700, 32GB RAM; macOS 10.15.7 Affinity Photo 1.10.8; Affinity Designer 1.108; & all 3 V2 apps for iPad; 6th Generation iPad 32 GB; Apple Pencil; iPadOS 15.7 Link to comment Share on other sites More sharing options...
John Rostron Posted February 4, 2018 Author Share Posted February 4, 2018 36 minutes ago, Alfred said: Dead link. I have corrected this link. John Alfred 1 Quote Windows 10, Affinity Photo 1.10.5 Designer 1.10.5 and Publisher 1.10.5 (mainly Photo), now ex-Adobe CC CPU: AMD A6-3670. RAM: 16 GB DDR3 @ 666MHz, Graphics: 2047MB NVIDIA GeForce GT 630 Link to comment Share on other sites More sharing options...
John Rostron Posted February 4, 2018 Author Share Posted February 4, 2018 16 minutes ago, R C-R said: @R C-R, for a moment I thought that you had applied this filter to Alice and the pig. I must try it! For Alice in Wonderland, a stretch transform seems appropriate. This has a=1, b=0, b=0.9, c=0.35 (all approximate). John Quote Windows 10, Affinity Photo 1.10.5 Designer 1.10.5 and Publisher 1.10.5 (mainly Photo), now ex-Adobe CC CPU: AMD A6-3670. RAM: 16 GB DDR3 @ 666MHz, Graphics: 2047MB NVIDIA GeForce GT 630 Link to comment Share on other sites More sharing options...
Alfred Posted February 4, 2018 Share Posted February 4, 2018 1 hour ago, John Rostron said: For Alice in Wonderland, a stretch transform seems appropriate. Or a shrink transform, depending on whether you’re illustrating “Eat Me” or “Drink Me”. Quote Alfred Affinity Designer/Photo/Publisher 2 for Windows • Windows 10 Home/Pro Affinity Designer/Photo/Publisher 2 for iPad • iPadOS 17.4.1 (iPad 7th gen) Link to comment Share on other sites More sharing options...
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