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 T_{1}, and the temperature at the outer edge of the wall is the 5th node (T_{5}) so:

- T
_{1}= 1560 K - T
_{5}= 300 K

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

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:

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:

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

HEAT OUT: Heat flows out to Node 3 so:

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

so:

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:

And now we solve for temperature at T_{2} to get:

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.

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

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.

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

- T
_{1}= 1560 K - T
_{2}= 0.5 T_{1}+ 0.5 T_{3} - T
_{3}= 0.5 T_{2}+ 0.5 T_{4} - T
_{4}= 0.5 T_{3}+ 0.5 T_{5} - T
_{5}= 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.

- T
_{1}= 1560 K - -0.5 T
_{1}+ T_{2}– 0.5 T_{3}= 0 - -0.5 T
_{2}+ T_{3}– 0.5 T_{4}= 0 - -0.5 T
_{3}+ T_{4}– 0.5 T_{5}= 0 - T
_{5}= 300 K

Now to matrixize:

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.

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 (T_{i}) 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.

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.

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 March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

I used the computer controlled (CNC) Shopbot machine at the Techshop to drill out 64 square pockets in the shape of a chessboard. One of my students (Kathryn) designed and printed the pieces as part of an extra credit project for her Geometry class.

The pockets were then filled with a clear eqoxy to give a liquid effect. However, I mixed in two colors of pigmented powder to make the checkerboard. The powder was uv reactive so it fluoresces under black (ultra-violet) light.

The powder also glows in the dark.

**Citing this post**: **Urbano**, L., 2017. Liquid Chessboard, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

If the Earth behaved as a perfect black body and absorbed all incoming solar radiation (and radiated with 100% emissivity) the we calculated that the average surface temperature would be about 7 degrees Celsius above freezing (279 K). Keeping with this simplification we can think about how the Earth’s temperature could change with time if it was not at equilibrium.

If the Earth started off at the universe’s background temperature of about 3K, how long would it take to get up to the equilibrium temperature?

Using the same equations for incoming solar radiation (E_{in}) and energy radiated from the Earth (E_{out}):

Symbols and constants are defined here except:

- r
_{E}= 6.371 x 10^{6}m

At equilibrium the energy in is equal to the energy out, but if the temperature is 3K instead of 279K the outgoing radiation is going to be a lot less than at equilibrium. This means that there will be more incoming energy than outgoing energy and that energy imbalance will raise the temperature of the Earth. The energy imbalance (ΔE) would be:

All these energies are in Watts, which as we’ll recall are equivalent to Joules/second. In order to change the temperature of the Earth, we’ll need to know the **specific heat capacity** (c_{E}) of the planet (how much heat is required to raise the temperature by one Kelvin per unit mass) and the mass of the planet. We’ll approximate the entire planet’s heat capacity with that of one of the most common rocks, granite. The mass of the Earth (m_{E}) we can get from NASA:

- c
_{E = 800 J/kg/K} - m
_{E}= 5.9723×10^{24}kg

So looking at the units we can figure out the the change in temperature (ΔT) is:

Where Δt is the time step we’re considering.

Now we can write a little program to model the change in temperature over time:

*EnergyBalance.py*

from visual import * from visual.graph import * I = 1367. r_E = 6.371E6 c_E = 800. m_E = 5.9723E24 sigma = 5.67E-8 T = 3 # initial temperature yr = 60*60*24*365.25 dt = yr * 100 end_time = yr * 1000000 nsteps = int(end_time/dt) Tgraph = gcurve() for i in range(nsteps): t = i*dt E_in = I * pi * r_E**2 E_out = sigma * (T**4) * 4 * pi * r_E**2 dE = E_in - E_out dT = dE * dt / (c_E * m_E) T += dT Tgraph.plot(pos=(t/yr/1000,T)) if i%10 == 0: print t/yr, T rate(60)

The results of this simulation are shown at the top of this post.

What if we changed the initial temperature from really cold to really hot? When the Earth formed from the accretionary disk of the solar nebula the surface was initially molten. Let’s assume the temperature was that of molten granite (about 1500K).

**Citing this post**: **Urbano**, L., 2017. Modeling Earth's Energy Balance (Zero-D) (Transient), Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

Energy and matter can’t just disappear. Energy can change from one form to another. As a thrown ball moves upwards, its kinetic energy of motion is converted to potential energy due to gravity. So we can better understand systems by studying how energy (and matter) are conserved.

Let’s start by considering the Earth as a simple system, a sphere that takes energy in from the Sun and radiates energy off into space.

At the Earth’s distance from the Sun, the incoming radiation, called insolation, is 1367 W/m^{2}. The total energy (wattage) that hits the Earth (E_{in}) is the insolation (I) times the area the solar radiation hits, which is the area a cross section of the Earth (A_{cx}).

Given the Earth’s radius (r_{E}) and the area of a circle, this becomes:

The energy radiated from the Earth is can be calculated if we assume that the Earth is a perfect black body–a perfect absorber and radiatior of Energy (we’ve already been making this assumption with the incoming energy calculation). In this case the energy radiated from the planet (E_{out}) is proportional to the fourth power of the temperature (T) and the surface area that is radiated, which in this case is the total surface area of the Earth (A_{surface}):

The proportionality constant (σ) is: σ = 5.67 x 10^{-8} W m^{-2} K^{-4}

Note that since σ has units of Kelvin then your temperature needs to be in Kelvin as well.

Putting in the area of a sphere we get:

Now, if the energy in balances with the energy out we are at equilibrium. So we put the equations together:

cancelling terms on both sides of the equation gives:

and solving for the temperature produces:

Plugging in the numbers gives an equilibrium temperature for the Earth as:

T = 278.6 K

Since the freezing point of water is 273K, this temperature is a bit cold (and we haven’t even considered the fact that the Earth reflects about 30% of the incoming solar radiation back into space). But that’s the topic of another post.

**Citing this post**: **Urbano**, L., 2017. Modeling Earth's Energy Balance (Zero-D) (Equilibrium), Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

This little embeddable, interactive app uses *n*th order polynomials to approximate a few curves to demonstrate the Taylor Series.

**Citing this post**: **Urbano**, L., 2017. Demonstrating Taylor Series Approximations with Graphs, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

Hat tip to Ms. Douglass for this link.

**Citing this post**: **Urbano**, L., 2017. Elements by Their Uses, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

**Citing this post**: **Urbano**, L., 2017. The Essential Biopolymers, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

I had my Numerical Methods student calculate the angle that would give a ballistic projectile its maximum range, then I had them write a program that did the the same by just trying a bunch of different angles. The diagram above is what they came up with.

It made an interesting pattern that I converted into a face-plate cover for a light switch that I made using the laser at the TechShop.

**Citing this post**: **Urbano**, L., 2017. Projectile Paths, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

One of the things my students had to figure out was what angle would give the maximum range of the projectile? You can figure this out analytically by finding the function for how the horizontal distance (x) changes as the angle (theta) changes (i.e. x(theta)) and then finding the maximum of the function.

In a nutshell, to find the distance traveled by the potato we break its initial velocity into its x and y components (v_{x} and v_{y}), use the y component to find the flight time of the projectile (t_{f}), and then use the v_{x} component to find the distance traveled over the flight time.

Starting with the diagram above we can separate the initial velocity of the potato into its two components using basic trigonometry:

,

so,

,

Now we know that the height of a projectile (y) is given by the function:

(you can figure this out by assuming that the acceleration due to gravity (a) is constant and acceleration is the second differential of position with respect to time.)

To find the flight time we assume we’re starting with an initial height of zero (y_{0} = 0), and that the flight ends when the potato hits the ground which is also at zero ((y_{t} = 0), so:

Factoring out *t* gives:

Looking at the two factors, we can now see that there are two solutions to this problem, which should not be too much of a surprise since the height equation is parabolic (a second order polynomial). The solutions are when:

The first solution is obviously the initial launch time, while the second is going to be the flight time (t_{f}).

You might think it’s odd to have a negative in the equation, but remember, the acceleration is negative so it’ll cancel out.

Now since we’re working with the y component of the velocity vector, the initial velocity in this equation (v_{0}) is really just v_{y}:

so we can substitute in the trig function for v_{y} to get:

Our horizontal distance is simply given by the velocity in the x direction (v_{x}) times the flight time:

which becomes:

and substituting in the trig function for v_{x} (just to make things look more complicated):

and factoring out some of the constants gives:

Now we have distance as a function of the launch angle.

We can simplify this a little by using the double-angle formula:

to get:

How do we find the maxima for this function. Sketching the curve should be easy enough, but because we know a little calculus we know that the maximum will occur when the first differential is equal to zero. So we differentiate with respect to the angle to get:

and set the differential equal to zero:

and solve to get:

Since we remember that the arccosine of 0 is 90 degrees:

And thus we’ve found the angle that gives the maximum launch distance for a potato gun.

**Citing this post**: **Urbano**, L., 2016. Maximum Range of a Potato Gun, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

I painted the wall on my new space in the basement to make it a dry-erase surface. Unfortunately, I did not have an eraser to use on it, so, I decided to make my own down at the TechShop. And what started as a simple project turned into a bit of a rabbit hole.

The Shopbot CNC router is great for cutting shapes out of wood. I started with simple rectangular 2 inch by 4 inch blanks with designs and patterns, but that truly does not take advantage of the technological possibilities. Map projections can have some interesting shapes, so I tried a few that I could find black and white vector-graphic maps for on the Wikimedia commons (Mollweide and Sinusoidal projections).

After a little sanding (of the edges and sides in particular) I put the wooden blanks on the laser. It helped to cut out a template for the wooden blanks to sit in so I could do multiple blocks at the same time.

I put on a few coats of polyurethane to protect the wood surface (I also tried a spray on sealer I had sitting around–we’ll see which one works better) and then attached velcro strips to the bottom.

One of my old sweatshirts served as material for the erasing.

**Citing this post**: **Urbano**, L., 2016. Making Dry-Erase Erasers, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

This September a TechShop branch opened up in St. Louis. I’ve been aware of these neat Makerspaces for a few years now, so it was a pleasant surprise when one turned up in town. Even more surprising (and just as pleasant) was that a parent at our school, who was so excited by the opportunities that a place like the TechShop would offer to a school that tries to emphasize hands-on, experiential education, donated four memberships to the school–one for a faculty member and three for students.

Since there are some age restrictions on which machines minors can use–a lot of the woodshop is off limits until they’re 16 and even then adult supervision is required, I arranged a small application for the student memberships that was only open to middle and high-schoolers. Based on the response I got back, we split the annual memberships by semester, so we have three students using it this fall and three more will have access to them in the spring.

The way the TechShop works is that they have a wide range of equipment under one roof and once you take a safety and basic usage (SBU) class on the particular machine you want to use you can reserve time on the machines. There’s a wood shop with saws, sanders, a lathe, and a CNC machine; a metal shop with the same; a set of 3d printers; a set of laser cutters/etchers; a fabric shop with some serious sowing machines, including one that is computer controlled; an electronics shop; a plastics work area with vacuum forming and injection molding machines. They also do a set of interesting classes on using the design software and some interesting projects that can take advantage of the tools available–I have my eye on the Coptic Bookbinding, and the Wooden Bowl making classes. Finally, they’re set up with classrooms where you can bring students in for small STEAM classes, which includes things like using Arduinos.

So far, we’ve all taken the Laser class, and there’s just so much that you can do with the laser that we’ve been spending a lot of time experimenting. The students have been etching signs–including a grave marker for our goat MJ who recently passed away–as well as pictures, luggage tags, and making presents. Since this is a machine that the older students can use independently I’ve lost track of everything they’ve been doing.

I’ve also taken the woodshop wood CNC class, so my own experiments have been a bit more expansive, including making dry-erase erasers, floor-holders for quivers for the archery program, simple chemistry molecular model sets (just 2d), boxes for Ms. Fu’s math cards, and I’m trying to figure out how to make a clock.

**Citing this post**: **Urbano**, L., 2016. TechShop, Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.

This TedEd video explains a few common sorting methods used in computer science. Sorting can be a challenging computational problem because of the enormous number of comparisons between items that can be involved, so computer scientists has spent a lot of time looking into it.

The video below shows the Quick Sort method using Hungarian folk dance.

**Citing this post**: **Urbano**, L., 2016. Sorting algorithms (with Hungarian Dance), Retrieved March 25th, 2017, from *Montessori Muddle*: http://MontessoriMuddle.org/ .**Attribution (Curator's Code )**: Via: ᔥ Montessori Muddle; Hat tip: ↬ Montessori Muddle.