I am trying to generate a mockup for t-shirts.
The result I am trying to achieve is the following: https://prnt.sc/kzhjk7
Using cloudinary, this is the closest result I have been able to produce:
https://res.cloudinary.com/worldwide-buy-llc/image/upload/c_scale,o_0,w_380/a_0,c_scale,l_TemplateSquare,r_0,w_380,x_900,y_190/c_scale,l_TemplateSquare,w_380,x_-310,y_240/c_scale,u_Mockups:Kids_White,w_3623,x_0,y_0/c_scale,l_Mockups:Kids_Whiteover,o_100,w_3623,x_0/v1538036215/TemplateSquare.png
However, it still looks different from the image that I would like to achieve.
I read that I could apply displacement. For this reason, I do have a displacement map stored at Mockups:Kids_WhiteOver
Do you know how can I apply it? Also the colors of the layers TemplateSquare appear weak in comparison to the target result ( https://prnt.sc/kzhjk7 ).
Any suggestion is very much appreciated since I am literally stuck to achieve that result. Many thanks in advance!
You can try removing the opacity from the transformation and use the multiplying effect.
How about this one: https://res.cloudinary.com/shirly/image/upload/o_0/l_TemplateSquare,w_380,y_300,x_-450/l_TemplateSquare,w_380,y_100,x_650/l_Kids_Whiteover,e_displace,x_10,y_10/u_kids_white,e_multiply/TemplateSquare.png
Let me know if that result can work for you.
Related
T(1)=5,
and for all n>=2: T(n)=2T(n-1)+(3*n+1).
I tried to solve this problem, but I have a problem with 3*n+1. When I put n-1, n-2,..., I don't know how to determine the formula for this problem.
Since there is only (3*n+1) as term and not T(3*n+1) this is solvable. First impression: you have 2T(n-1) as subterm, so the solution is something like 2^n.
Through a simple Excel data analysis I found the solution T(n)=-7-3n+15 * 2^(n-1), I will try to solve it per hand and will update my answer if I found the right path.
Edit: This was more difficult than expected...
Explanation:
first step is to get a sum formula for n. You can derive this pattern from the first few T(n).
Once you get the pattern, try to get rid of the sum.
To solve sums, try to get them in a similar format as sum_(i=1)^n (1) = n or sum_(i=0)^n (2^i) = 2^(n+1)-1
to do this you can manipulate the index such as sum_(i=2)^n (n-i) = sum_(i=0)^(n-2) (i) or to include/exclude elements from the sum.
the trickiest part was to solve sum_(i=0)^n ((n-i)*(2^i)). The idea here is to convert the multiplication (depending on i) to a sum (also depending on i).
please note the changing indice-numbers. sum_(i=0)^n (2^i) is not the same as sum_(i=1)^n (2^i)
The path is not the most efficient one, simplify as you wish.
I defined the following equation to calculate the sum of total power consumed by the system:
TotalPower.. systemPower =e= sum(J,P(J)) + sum(I,CP(I));
However, the variable systemPower gets only the result of the second sum and not both!. The declaration of P(J) is as following:
P.LO(I)=0;
P.up(I)=100;
P.l('i1')=2;
P.l('i2')=3;
Please, Can any one explain why I get the result of a single sum? How I can do to get both?. I tried also to separate them in different values but yet I get the same result.
Thank you in advance.
I though it is a good idea to share this it might help someone else. I used a variable directly instead of an equation and I put it in the following form and it worked.
systemPower.l = sum(I,P.l(I))+sum(I,CP(I));
I'm using Skrollr and trying to set multiple relative keyframes. This doesn't work for me at the moment.
data-bottom-top="opacity:0;transform:translate3d(100%,0,0)" data-top-top="opacity:1;transform:translate3d(0,0,0)" data-top-botttom="transform:translate3d(0,50%,0) data-100p-top-botttom="transform:translate3d(0,50%,0)"
Have been cracking my head over this for the past few days. Any help would be greatly appreciated!
Just to answer the question fully, you must use the same units for each argument:
data-bottom-top="opacity:0;transform:translate3d(100%,0%,0)"
data-top-top="opacity:1;transform:translate3d(0%,0%,0)"
data-top-botttom="transform:translate3d(0%,50%,0)
data-100p-top-botttom="transform:translate3d(0%,50%,0)"
See these:
https://github.com/Prinzhorn/skrollr/issues/134
https://github.com/Prinzhorn/skrollr/issues/315
I'm trying to set a specific property to a non exact value, for example say that I want to define the height of a pine tree to usually between 3-80 m (according to wikipedia). Then I would like to set something like [[Has height::3-80]] (of course this doesn't work) and defining the unit to meters with "custom units". Then I would like to be able to query for example "trees that can reach the height of 70 meters" and the pine tree would be included. I've been searching and trying different angles for hours now and I can't figure it out. Tried with #set_recurring_event but that seems to be only for dates/time. Also understood how to set multiple values for a property with #arraymap but this doesn't seem to help me here. Really would appreciate help with this (it's probably very easy and right in front of me) Thx! COG
There's no such things. But you able to create template, with parameters you want. The you just use code kinda {{range|min|max|units}}. For example your range of heights looks like {{range|3|80|m}}.
I am using iText and am very new to it. There have been several situations where I think I could have figured out the problem with my code if I knew what I was doing - I use examples without knowing the workings behind the code, and even as I look at the source I can't figure out what the programmer was thinking.
What does setTextMatrix of contentByteArray in iText do? And how do I figure out the parameter values I need?
For example:
cb.setTextMatrix(1, 0);
The input parameters are x,y coordinates in points, unless CTM scaling was defined.
0,0 would be the bottom left of the template you are referencing.
The position is the 'baseline' of the text, rather than the top or the bottom.
Transcribed from this source:
https://sourceforge.net/p/itext/mailman/message/12855218/
first parameter sets left margin, second parameter sets bottom margin.