Generating 3d Terrain

Posted June 29, 2017

by Lensyl Urbano

3d model of the Hawaiian island chain.

3d model of the Hawaiian island chain, rendered in OpenSCAD.

After a lot of hours of experimentation I’ve finally settled on a workable method for generating large-scale 3d terrain.

Data from the NGDC’s Grid Extraction tool. The ETOPO1 (bedrock) option gives topography and bathymetry. You can select a rectangle from a map, but it can’t be too big and, which is quite annoying, you can’t cross the antimeridian.

The ETOPO1 data is downloaded as a GeoTIFF, which can be easily converted to a png (I use ImageMagick convert).

The Hawaiian data with the downloaded grayscale.

The Hawaiian data with the downloaded grayscale.

Adjusting the color scale. One interesting property of the data is that it uses a grayscale to represent the elevations that tops out at white at sea-level, then switches to black and starts from there for land (see the above image). While this makes it easy to see the land in the image, it needs to be adjusted to get a good heightmap for the 3d model. So I wrote a python script that uses matplotlib to read in the png image as an array and then I modify the values. I use it to output two images: one of the topography and one of just land and water that I’ll use as a mask later on.

Hawaiian Islands with adjusted topography and ocean-land.

Hawaiian Islands with adjusted topography and ocean-land.

The images I export using matplotlib as grayscale png’s, which can be opened in OpenSCAD using the surface command, and then saved as an stl file. Bigger image files are take longer. A 1000×1000 image will take a few minutes on my computer to save, however the stl file can be imported into 3d software to do with as you will.

Note: H.G. Deitz has a good summary of free tools for Converting Images Into OpenSCAD Models

Citing this post: Urbano, L., 2017. Generating 3d Terrain, Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Bike Silhouette

Posted April 30, 2017

by Lensyl Urbano

TD cleaning up his new bike.

T. cleaning up his new bike.

One of my students with a TechShop membership wanted a bike silhouette for a wall hanging. He wanted it to be bigger than he could fit on the laser cutter, so I tried doing it on the CNC router. The problem was that to get the maximum detail we needed to use the smaller drill bits (0.125 inches in diameter), however, after breaking three bits (cheap ones from Harbor Freight) and trying both plywood and MDF, we gave up and just used the larger (0.25 inch) bit. Since the silhouette was fairly large (about 45 inches long), it worked out quite well.

Citing this post: Urbano, L., 2017. Bike Silhouette, Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

How to Build an 8-bit Computer on a Breadboard

Posted April 29, 2017

by Lensyl Urbano

Ben Eater’s excellent series on building a computer from some basic components. It goes how things work from the transistors to latches and flip-flops to the architecture of the main circuits (clock, registers etc). The full playlist.

Other good resources include:

Citing this post: Urbano, L., 2017. How to Build an 8-bit Computer on a Breadboard, Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Cell Phone Shelf

Posted March 30, 2017

by Lensyl Urbano

Cell phone rack in use.

Cell phone rack in use.

Managing cell phone usage at school is a tricky topic. We have some teachers who’d like to ban them outright, but we also have a growing number of parents who are expecting to be able to communicate with their kids–to organize pickups and carpooling during the day for example. The phones can be great for data-collection and documentation in classes, and a lot of my upper level math students prefer the Desmos app to using their graphical calculators.

Our current compromise is that middle schoolers have to leave their phones in the front office, where they can check them at lunch time or check them out if a teacher wants them to use them.

The high schoolers are allowed to keep their phones with them, but have to put them in a basket at the front of the classroom. Since they don’t like piling them into the basket, I experimented with the CNC machine to cut some plywood into a cell-phone shelf.

The shelf can hold about 30 phones, and I can easily see how many phones are on there from across the room, so I’d say this one worked out pretty well.

Citing this post: Urbano, L., 2017. Cell Phone Shelf, Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Go Board

Posted March 29, 2017

by Lensyl Urbano

Students playing Go.

Students playing Go.

I recently discovered that, although they may look it, Go boards are not necessarily square. They’re slightly longer in one dimension so that the board looks more square to the players on both sides.

A student asked me to make one for him–he’d ordered a set recently and didn’t like the board it came with–so, I wrote a small python program to generate the Go grid, then lasered it onto a nice piece of sanded plywood.

It worked out quite well. Apparently the plywood makes just the right “thunk” sound when you put down the pieces.

Go board in use.

Go board in use.

The script to generate the grid.
go_board_2.py

from visual import *
from svgInator_3 import *

length = 424.2  #mm
width = 454.5   #mm
nLines = 19
dx = length/(nLines-1)
dy = width/(nLines-1)

print "Lenght = ", length
print "dx = ", dx

f = svgInator("go_board.svg")

lineStyle = {"stroke": "#000", "stroke-width": "2pt",}

#lines
for i in range(nLines):
    x = i * dx
    y = i * dy
    #vertical
    f.line(pos=[vector(x,0), vector(x,width)], style=lineStyle)
    #horizontal
    f.line(pos=[vector(0,y), vector(length,y)], style=lineStyle)

#circles
grid_pos = [(3,3), (3,9), (3,15),
            (9,3), (9,9), (9,15),
            (15,3), (15,9), (15,15)]

for i in grid_pos:
    (x, y) = (i[0]*dx, i[1]*dy)
    f.circle(pos=vector(x,y), radius=2.0,
             style={"stroke": "#000", "fill":"#000"})

#bounding box
f.rect(dim=vector(length,width), style=lineStyle)

f.close()

Now I just have to learn to play.

Citing this post: Urbano, L., 2017. Go Board, Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Our Natural Bridge

Posted March 28, 2017

by Lensyl Urbano

Crossing the bridge.

Crossing the bridge.

Inspired by a video of a temporary bridge built out in the woods for mountain biking, my students wanted to try building a “natural” bridge with no fasteners–no screws, no nails–over a small ravine that feeds into our creek.

The base of the bridge.

The base of the bridge.

We found a couple large fallen logs to cut into two 10 foot lengths for the basic structural support for the bridge. These were dug into the ground to anchor them on either side of the ravine. We then chopped a couple more logs into 2 foot sections to go across the structural logs. The dense mud from the banks of the creek was then packed onto the top to hold it all together.

Packing mud.

Packing mud.

In the end, the bridge turned out to be pretty solid, and definitely usable.

The bridge holds up.

The bridge holds up.

Citing this post: Urbano, L., 2017. Our Natural Bridge, Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Building Bridges (Literally)

Posted March 28, 2017

by Lensyl Urbano

Small, movable bridge.

Small, movable bridge.

My crew from the Gaga Ball pit decided to make a trail through the woods and across the creek. So they built two short (12 ft long) bridges to cross the creek itself, and a third, “natural” log bridge to cross a small ravine that runs into the creek and cuts across the trail.

The short bridges were made of overlapping 2×4’s for structure (held together by 2.75 inch structural screws), with 24 inch long, 1×6 planks across the top.

Short bridge under construction.

Short bridge under construction.

The short bridges needed to be small and light enough to be moved when the creek rises, like it did today. I’ll attest that they can be moved, but not easily. They’re pretty heavy: it took a team of three or four middle schoolers to get it down to the creek, and it was hard going trying to drag it over to the side by myself this afternoon. Note to self: next time make sure the structural cross pieces are not at the very end of the bridge.

The rising creek.

The rising creek.

Citing this post: Urbano, L., 2017. Building Bridges (Literally), Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

Solving 1D Heat Transfer Problems with Matricies

Posted March 5, 2017

by Lensyl Urbano

Say, for example, we have a furnace with an interior temperature of 1560 K (hot enough to melt copper) while the outside temperature is a balmy 300 K (27 Celcius). The wall of the furnace is 1 meter thick (it’s a big furnace). Can we figure out how the temperature changes inside the wall as goes from hot on the inside to cool on the outside?

Figure 1. Five nodes (points of interest) where we will calculate the heat flow through a furnace wall.

Figure 1. Five nodes (points of interest) where we will calculate the heat flow through a furnace wall.

To start with, we’ll look at the temperature at five points (call them nodes): three are in the wall and two at the edges (see Figure 1). We’ll say the temperature at node 1 which is next to the furnace is T1, and the temperature at the outer edge of the wall is the 5th node (T5) so:

  • T1 = 1560 K
  • T5 = 300 K

Now we need to find the temperature at nodes 2, 3, and 4.

Heat Flow at Equilibrium

For each of the interior nodes, we can consider how things are at equilibrium. Heat is always moving through the wall, but as some point in time the heat flowing from the furnace into the wall will be equal to the heat flowing out of the wall, and for each node the heat flowing in will equal the heat flowing out.

The heat flow (Q) from one location to another is driven by the difference in temperature (ΔT): heat flows from high temperature to cooler temperatures (which makes ΔT negative). However, you also need to consider the distance the heat is traveling (Δx) since, given the same temperature drop, heat will flow faster if the distance is short. So, it’s best to consider the temperature gradient, which is how fast the temperature is changing over distance:

\text{Temperature gradient} = \frac{\Delta T}{\Delta x}

How fast heat flows is also mediated by the type of material (different materials have different thermal conductivities (K)), so our heat flow equation is given by the equation:

Q = - K \frac{\Delta T}{\Delta x}

Consider Node 2

HEAT IN: Heat is flowing into Node 2 from Node 1, so:

Q_{in} = - K \frac{T_2 - T_1}{\Delta x}

HEAT OUT: Heat flows out to Node 3 so:

Q_{out} = - K \frac{T_3 - T_2}{\Delta x}

At equilibrium the heat into the node equals the heat out:

Q_{in} = Q_{out}

so:

- K \frac{T_2 - T_1}{\Delta x} =  - K \frac{T_3 - T_2}{\Delta x}

Which we can simplify a lot because we’re assuming the thermal conductivity of the wall material is constant and we’ve (conveniently) made the spacing between our nodes (Δx) the same. Therefore:

T_2 - T_1 =  T_3 - T_2

And now we solve for temperature at T2 to get:

T_2 =  \frac{T_1 +T_3}{2}

Finally, we can break the fraction into separate terms (we need to do this to make it easier to solve using matricies as you’ll see later) and start using decimals.

T_2 =  0.5 T_1 + 0.5 T_2

If we do the same for all the internal nodes we get:

T_3 =  0.5 T_2 + 0.5 T_3

T_4 =  0.5 T_3 + 0.5 T_5

You should be able to see here that the temperature at each node depends on the temperature of the nodes next to it, and we can’t directly solve this because we don’t know the temperatures of the interior nodes.

A System of Equations

Let’s collect all of our information to get a system of equations:

  • T1 = 1560 K
  • T2 = 0.5 T1 + 0.5 T3
  • T3 = 0.5 T2 + 0.5 T4
  • T4 = 0.5 T3 + 0.5 T5
  • T5 = 300 K

Now to rewrite this as a matrix. We start by putting all terms with variables on the left and all the constants on the right.

  • T1 = 1560 K
  • -0.5 T1 + T2 – 0.5 T3 = 0
  • -0.5 T2 + T3 – 0.5 T4 = 0
  • -0.5 T3 + T4 – 0.5 T5 = 0
  • T5 = 300 K

Now to matrixize:

\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ -0.5 & 1 & -0.5 & 0 & 0 \\ 0 & -0.5 & 1 & -0.5 & 0 \\ 0 & 0 & -0.5 & 1 & -0.5 \\ 0 & 0 & 0 & 0 & 1  \end{bmatrix} \begin{bmatrix} T_1 \\ T_2 \\ T_3 \\ T_4 \\ T_5  \end{bmatrix} = \begin{bmatrix} 1560 \\ 0 \\ 0 \\ 0 \\ 300 \end{bmatrix}

This you can solve on paper, and, if you’d like, check your answer using Alex Shine’s Gaussian Elimination solver, which gives step by step output (although it’s a little hard to follow). I used Alex’s solver to get the result:

Solution
Variable 1 = 1560.0
Variable 2 = 1245.0
Variable 3 = 930.0
Variable 4 = 615.0
Variable 5 = 300.0

This problem is set up to be easy to check because the results should be linear (if you plot temperature versus distance through the wall you’ll get a straight line). It will also give you whole number results up to 10 nodes.

Next Steps

Any Number of Nodes

This procedure for setting up the matrix give the same basic equations no matter how many nodes you use, because as long as the distance between the nodes (Δx) and the thermal conductivity (K) are constant, the equation for each internal node (Ti) will be:

[math] T_i = \frac{T_{i-1} + T_{i+1}}{2}

and the matrix will continue to just have three terms along the diagonal.

Non-uniform Wall and Non-uniform Node Spacing

If the wall is not uniform then the thermal conductivity coefficient does not just drop out of the equation, so you’ll have to pay attention to the conductivity going into and out of each node. Same goes with the node spacing.

Alternative Solution Methods

Students could try:

  • writing their own matrix solvers.
  • setting up the system of equations in a spreadsheet program (they’ll need to change your program’s settings so that it uses its iterative solver).

Citing this post: Urbano, L., 2017. Solving 1D Heat Transfer Problems with Matricies, Retrieved July 29th, 2017, from Montessori Muddle: http://MontessoriMuddle.org/ .
Attribution (Curator's Code ): Via: Montessori Muddle; Hat tip: Montessori Muddle.

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