Prototyping a Soup Ladle Bowl for 3D Printing using TINKERCAD

We just discussed a math practice solution for soup ladles in kitchen math.

We focused on the relationship between the diameter of a soup ladle bowl and the standard sizes used in recipes and food service plans.

Now, we will look at how mathematics goes into 3D modeling of soup ladles and similar objects. TINKERCAD is the main tool we will use to visualize some important math concepts.

This activity looks closely at the geometry related to soup ladle bowls.

Positive and negative volumes

TINKERCAD provides a wonderful platform for applying math knowledge from the classroom. For example, the solid objects represent “positive” volumes (or volume values with positive signs). In contrast, the hole objects represent “negative” volumes (or volume values with negative signs).

During this activity, let us also keep track of the total amount of volume on the workplane. Right now, there are no objects on the workplane. The total volume is zero.

V_TOTAL = 0

Let’s dive right into Part 1.

Part 1

Okay, there is a lot here.

Before we began 3D modeling in Part 1, we chose a system of units. This project used inches.

The hemisphere was placed on the “work plane” with the flat side down. We wanted the flat side up, so we rotated the hemisphere 180 degrees about a single axis.

CCSS.MATH.CONTENT.HSG.GMD.A.3: Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

The shape that was added from the library was a hemisphere (half-sphere). The length and the width (the diameters) of the hemisphere were equal, but the height was half the diameters.

These rules were applied when the length and width were changed. To maintain the regular hemisphere shape, the height scaled (“proportionally”) at the same rate as the length and width.

Also, when we duplicated the solid hemisphere object, we doubled the total volume of the objects sitting on the “work plane”..

Let us continue with Part 2.

Part 2

In Part 2, we reduced the length and the width of the second object to 1.8 inches. The height was proportionally reduced to be half that measurement, which was 0.9 inches.

Also, something significant happened.

There was a sign change. The positive volume of the second object was converted into a negative volume.

Now, the total volume on the “work plane” comprises the positive volume of the first object (represented by the solid) and the negative volume of the second object (represented by the hole).

V_TOTAL = (+)V_SOLID + (-)V_HOLE

V_TOTAL = V_SOLID — V_HOLE

The next step is to combine both volumes. Perhaps adding a negative volume to a positive volume will result in less volume. Let us find out in Part 3.

Part 3

As we saw in Part 3, when both objects occupy the same space, we can apply additive properties. The (negative) hole object removed material from the (solid) positive object.

The result is a new shape that represents the positive (solid) object minus the negative (hole) object.

This is the same as digging a hole. Whenever we dig a hole, we are adding a negative volume to dirt.

Bigger holes, thinner walls

Let us suppose we want to decrease the wall thickness of the soup bowl ladle. One solution would be to increase the size of the (negative) hole object. Let us see what happens when the (negative) hole object’s volume is increased while the (positive) solid object’s volume remains unchanged.

Part 4

The hole is bigger.

The size of the length and width were increased to 1.95 inches — and the diameter (0.975 inches) was changed to proportionally match the new value.

Now, we must add the volumes again to form the new shape.

Part 5

Conclusion

This activity required a special type of computational thinking. This is visuospatial computational thinking, which is important in manufacturing, logistics, fashion, and the performing arts.

There are also some important takeaways. Volume and density are related to weight (and mass).

Weight = Density * Volume

If density stays constant, then a decrease in wall thickness (an increase in the hole size) would cause the soup ladle bowl weight to decrease. Less material is less expensive to produce and transport. However, minimal material may affect product quality and user experience.

These are design issues to think about in cooperative and collaborative projects. Math is no longer about the “answer in the back of the book” in the workplace. Instead, math is now about creating new solutions and using critical thinking to continuously improve products and create new opportunities.

MathForWork delivers distributed learning systems, instruction support, and test preparation for all learners. Learn more at MATHFORWORK.COM. MathForWork is a Bitwise Thermodynamics project.

Creating a Frustum for 3D Printing using TINKERCAD

TINKERCAD is web-based 3D modeling software that allows you to create objects for 3D printing. In TINKERCAD, you can build a frustum by starting with a solid pyramid and subtracting a “hole” from the top of that pyramid.

The following video provides a demonstration.

Volumes can be added (like when we finished the pyramid).

In this video demonstration, a cylinder and a hemisphere (half-sphere) are combined to form a new volume.

FUTURE WORK: What equation do we get when we add a hemisphere to cylinder as we did in the video?

CCSS.MATH.CONTENT.HSG.GMD.A.3: Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

MathForWork delivers distributed learning systems, instruction support, and test preparation for all learners. Learn more at MATHFORWORK.COM. MathForWork is a Bitwise Thermodynamics project.

Shape Algebra

Add two shapes in TINKERCAD and you get a new shape. The sphere and cylinder combine here to form a half-capsule. What is the new equation that best describes this new shape??

CCSS.MATH.CONTENT.8.G.C.9: Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.

Finishing the Pyramid in TINKERCAD with Math in Wolfram Alpha

The frustum can be seen as a Rorschach test. Some people see a gold bar, and some people see an unfinished pyramid.

Frustum in TINKERCAD

One way to create a frustum in TINKERCAD is to start with a solid pyramid basic shape. The top part of the pyramid is then removed by a “hole” that adds negative volume to the shape. The result is a frustum. The following video demonstrates this method:

Before removing the top part of the pyramid, we need to know the dimensions of the solid pyramid’s base. We also need to know the height of the pyramid.

We can do this by using our prior knowledge and best practices. One of the most basic uses for algebra is providing a simple way to express how two things add up to one thing.

We know that a frustum is the “bottom section” (in blue) of a pyramid. This means that if we add the missing “top section” (in pink) to the frustum, we will have a complete pyramid.

That relationship can be described by the following equation:

We know the formula for the volume of the complete (big) pyramid,

with the base area B, the height of the frustum h, and the unknown height of the top pyramid H.

We also know the volume of the top section of the complete pyramid (which can also be described as a mini pyramid),

with the base area A (the same area as the top area of the frustum), and the unknown height of the top pyramid H.

The volume of the bottom section of the complete pyramid can be described as a frustum.

With a base area B of 6″ x 6″. let us suppose that the height of the frustum h is 10 inches, and the top area of the frustum A is 2″ x 2″. This means that the total height of the complete pyramid H + h provides us clues to the information that we need finish the pyramid.

To help finish the pyramid, we now have a useful equation that can be re-used and re-purposed in different workplace contexts:

In Wolfram Alpha, type this:
h=10;A=2*2;B=6*6;(1/3)(H+h)*B=(1/3)A*H + (h/3)(A+B+SQRT(A*B))
(view query)

Notice that we never actually calculated the volumes in the equations. Instead, we used the additive relationships between the volumes, areas and heights to complete a specific workplace task.

The 3D models related to the “finish the pyramid” math practice activity are available on GitHub. Also, the 3D models can be directly imported into TINKERCAD. NOTE: Model units are in inches.

MathForWork delivers distributed learning systems, instruction support, and test preparation for all learners. Learn more at MATHFORWORK.COM. MathForWork is a Bitwise Thermodynamics project.

CCSS.MATH.CONTENT.5.MD.A.1: Coffee Jar Flower Vase Activity

The following activity is an opportunity for math practice in situations when measurement requires math-based design solutions. If possible, read the problem statement aloud. After reading the problem statement, we will have a brief discussion.

Problem Statement

An empty coffee jar has been converted into a flower vase. The 3D modeling of this coffee jar flower vase is described as two frusta stacked end-to-end to form an hourglass container. The top areas of both frusta share a common square cross-section, which forms a bottleneck at 3.5″ above the bottom of the jar (sitting on a table). The capacity of the container is 800 mL of water. The total height of the container is 5.5″. The base area for the lower frustum and upper frustum are 3.5″ x 3.5″ and 3″ x 3″, respectively. What are the dimensions of the cross-sectional area at the bottleneck?

Coffee Jar

Discussion

Wow! 😂😂😂 Lots of words. That is the point.

This activity is designed to help you practice reading a lot of words in test items. You will gain experience organizing these words and ideas using computational thinking.

Let us take a moment to visualize the solid model described above with 3D file viewer on GitHub. Or you can simply view the video below:

Before we begin this problem, we need to recall some prior knowledge and best practices. First, we choose and confirm the units of the final answer. In this case, the units are [inches x inches] or [in. x in.].

Note that the question did not ask for the side length or the actual area. The call to action asked for the dimensions. In the workplace, it is very important to pay attention to what the request for proposal (RFP) needs.

Second, we convert all quantities into compatible units. We convert 800 millimeters into cubic inches. The result is about 49 cubic inches.

In Wolfram Alpha, type this:
800 ml in cu in (view)

For voice assistants, try this command:
800 milliliters in cubic inches

CCSS.MATH.CONTENT.5.MD.A.1: Convert among different-sized standard measurement units within a given measurement system (e.g., convert 5 cm to 0.05 m), and use these conversions in solving multi-step, real world problems.

Next, we look at the formula for the volume of a frustum.

V = π/3 h (A + B + sqrt(A B))

There are two frusta in this problem, so we will be using the same formula twice to help solve for a system of equations.

Let us place the container on the table. The table will be our reference point. The total height of the container is the distance from the table to the top of the container (see Figure 1). The total height (in inches) of the container is H = 5.5.

Figure 1: Upper frustum and lower frustum modeled in TINKERCAD

This measurement excludes the the jar mouth section, where the screw top lid has been removed.

For the lower frustum volume Y, we suppose that the base area (in inches squared) is A = 3.5 * 3.5 and the height (in inches) from the table is h = 3.5.

The units of area are square inches. We will let Wolfram Alpha make the calculations for us by leaving the dimensions as multiplicands. This methods lets us make design changes to the dimensions as needed without re-building our solution.

Figure 2: Lower frustum dimensions in TINKERCAD

For the upper frustum volume Z, we conclude that the base area is B = 3 * 3, with a height of H – h.

We know the exact height of the upper frustum, but we will still use variables to express the difference between the total height and the height from the table.

Both frusta share the common bottleneck area X. Their combined volume is 49 cubic inches; that is, Y + Z = 49.

Our analysis is complete. We now know enough information to solve for the unknown bottleneck area X.

In Wolfram Alpha, type this:
Y + Z = 49; H = 5.5; A = 3.5 * 3.5; B = 3 * 3; h = 3.5; Y = (h/3) * (A + X + SQRT(A*X)); Z = ((H-h)/3) * (B + X + SQRT(B*X)); J = SQRT(X);
(view)

The dimensions of the bottleneck cross-section are 2.63″ x 2.63″.

Key vocabulary: request for proposal, units, unit conversion

MathForWork delivers distributed learning systems, instruction support, and test preparation for all learners. Learn more at MATHFORWORK.COM. MathForWork is a Bitwise Thermodynamics project.