# Illustrator

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## "Blend" Tool Problems

### Nov 19, 2012 9:37 AM

Hi, I am using CS and the “blend” tool to create a train track but I cannot seem to get the spacing to look right. I tried adding a 3rd railroad tie in the middle but that didn’t work either. Any ideas how I can get more spacing in the foreground and less in the distance?

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Nov 19, 2012 9:56 AM   in reply to Comp. 792

You need to create handles on that spine path and adjust their length:

There must be another thread from this year, but I couldn't find it

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Nov 19, 2012 11:59 AM   in reply to Comp. 792

Maybe you'd like a completely different approach.

You could create the crossties like they look from above. Then either use the perspective grid or apply the 3D-effect Rotate

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Nov 19, 2012 3:35 PM   in reply to Comp. 792

Comp,

You may also do this in a simple (or maybe not so simple) 2D way.

Your first and topmost/smallest sleeper/crosstie shaped as a trapezium has a certain height H (basic width of sleeper as seen at the viewing angle) and two horizontal segment lengths, the top length W (basic width/length of sleeper) and the larger bottom length WB; you may wish to adjust the length of the bottom segment to get the desired growth in width; you can see the misfit between the angle and the growth in your image(s) yielding a ragged appearance.

You have to decide the relation between the height and distance between the sleepers, and have the angle at the ends fit the desired growth in width.

If you settle for an approximate solution (which is probably only insignificantly inaccurate) where each distance between sleepers has a certain ratio R to the height (width on the ground) of the previous one, you may:

1) If needed, adjust the length of the bottom segment (see below);

2) Effect>Distort & Transform>Transform with Horizontal Scale = Vertical Scale = 100+(1+R)*(WB-W)*100/W%,  Vertical Move = -(1+R)*H (negative value), and the desired number of Copies.

If you dislike the growth of the width, you may repeat 1) and 2), starting with an adjustment of the length of the bottom segment, then recalculating.

If you dislike the ratio of distance to height, you may just repeat 2).

To illustrate, if you start with a first sleeper having H = 2, W = 20, and WB = 21, and a distance ratio  R = 1 (same distance as width), you will get the following values in 2):

Horizontal = Vertical Scale = 100+(1+1)*(21-20)*100/20 = 100+2*1*100/20 = 110%

Vertical Move = -(1+1)*2 = -4.

Number of Copies as desired (19 for 20 sleepers in total or similar).

Obviously, the greater the relative difference between WB and W (the further from a rectangle), the lower the viewing angle and the faster growth in the size of the sleepers.

And obviously, the greater the ratio R, the less dense appearance.

Here are images based upon H = 2 and:

A) W = 20/WB = 21/R = 1 (Horizontal = Vertical Scale = 110, Vertical Move = -4)

B) W = 40/WB = 24/R = 1(Horizontal = Vertical Scale = 105, Vertical Move = -4), and

C) W = 20/WB = 21/R = 2 (Horizontal = Vertical Scale = 115, Vertical Move = -6):

You (probably) do not want to know how to do it for full (insignificantly improved) accuracy.

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Nov 19, 2012 7:06 PM   in reply to Comp. 792

I made this with envelope distort mesh with one column and row. On the envelope object shown in the second image on the top two points of the mesh I reduced the length of the horizontal and vertical handles to zero by dragging them to their corner points. I pulled the vertical handles of the bottom points up the same distance with the help of a guide. On the third image I select the two points on each left and right side and used the Shear tool holding Shift while dragging it horizontally with its center of shear set at the lower point to make the shape narrow at the top

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Nov 20, 2012 4:12 AM   in reply to Comp. 792

For my part you are welcome, Comp.

I believe we all look forward to our seeing your findings.

It seems that any handle dragging solution leads to a distribution of distances and heights that corresponds to a top view of something following the curved surface of a(n elliptic) cylinder rather than a perspective view of something that is straight.

I am afraid this is an inherent limitation to an otherwise simple and obvious solution.

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