## Friday, March 17, 2023

### AI Exploration Series: ChatGPT 4 Faces the "Basketballs in a 747" Question

In the consulting and tech worlds (McKinsey, Google), applicants are often tested with unusual quantitive or strategic questions.  The point is to assess their ability to think on their feet, more than the exact answer.   (For example, "I don't know and I can't guess" would be a very poor answer in a  Google interview, although that might be the right answer in some other contexts.)

A classic question is "How many basketballs would fit in a 747?"

I might guess this way.

Let's say it's 50 rows long at 3 feet per row, so the length is150 feet.  Let's say it's 9 seats wide (2,5,2) at 2.5 ft per seat, plus 6 feet for a left and right aisle.  That's about 6+24 feet wide or 30 feet wide.   A rectangular solid would be 30 x 30 x 150 cubic feet, or 900x150 or 135,000 cubic feet.  A cylinder is smaller, say 90,000 cubic feet.

A basketball is maybe 1 cubic foot, so 90,000 basketballs.  Note that you don't need the spherical volume, since basketballs will stack similarly to cubes, with a lot of air space between the spheres.  If I thought a basketball was smaller than 1 cubic foot, I could scale that up somewhat.  (Note to self: it's 10 inches in diameter, so scale up maybe 20% to 110,000).

Here's Chat GPT 4 with the same question.

Can you estimate the number of basketballs the sheel of a 747 jet plane could hold, if the shell of the plane was other wise empty (an empty cylinder)?

To estimate the number of basketballs that could fit inside the shell of a 747 jet plane, we need to determine the volume of the plane's interior and the volume of a basketball. Then, we can divide the plane's interior volume by the volume of a basketball to get an approximate number of basketballs.

A Boeing 747-400, one of the most common 747 variants, has an approximate interior volume of 876 cubic meters (m³). We will use this value for our estimation.

The standard basketball size for men (size 7) has a diameter of 9.5 inches (24.1 cm). We can calculate the volume of the basketball using the formula for the volume of a sphere:

V = (4/3) × π × r³

where V is the volume, and r is the radius of the sphere (half of the diameter).

The radius of the basketball is 12.05 cm (24.1 cm / 2). To calculate the volume:

V ≈ (4/3) × π × (12.05)³ ≈ 7,312.6 cm³

To convert the volume from cubic centimeters to cubic meters, we divide by 1,000,000:

V ≈ 7,312.6 cm³ / 1,000,000 ≈ 0.0073126 m³

Now, we can estimate the number of basketballs that would fit in the plane by dividing the interior volume of the 747 by the volume of a basketball:

Number of basketballs ≈ 876 m³ / 0.0073126 m³ ≈ 119,800

This estimation assumes a perfect packing arrangement, which is not realistic due to the irregular shape of the plane's interior and the space between the basketballs. In reality, the number of basketballs that would fit inside the 747 would likely be lower, but this calculation provides a rough estimate.

That was excellent.  Could you clarify how you arrived at 876 m3 as the volume of the empty jet?  Did you have this number directly or did you calculate it?

I apologize for the confusion; I should have provided more information on how I arrived at the 876 m³ volume for the Boeing 747-400. That value was an approximation based on general information about the aircraft's size, and it might not be entirely accurate. To provide a more accurate estimate, let's calculate the volume based on the dimensions of the aircraft.

The Boeing 747-400 has a fuselage width of about 6.5 meters (21.3 feet) and a length of approximately 70.6 meters (231 feet). Assuming the interior of the plane is roughly cylindrical, we can calculate the volume of the cylinder using the formula:

V = π × r² × h

where V is the volume, r is the radius of the cylinder, and h is the height (length) of the cylinder.

First, we need to determine the radius. Since the fuselage width is 6.5 meters, the radius would be approximately half of that:

r ≈ 6.5 m / 2 ≈ 3.25 m

Now we can plug in the values into the formula:

V ≈ π × (3.25)² × 70.6 ≈ 2,519.6 m³

This value represents the approximate volume of the entire cylinder, assuming the interior is a perfect cylinder, which is an oversimplification, as the plane's interior has a more complex shape. Additionally, this calculation does not account for the cockpit, tail section, and other internal structures. As a result, the actual usable interior volume will be somewhat less than the calculated value.

Using the calculated volume of 2,519.6 m³, we can re-estimate the number of basketballs:

Number of basketballs ≈ 2,519.6 m³ / 0.0073126 m³ ≈ 344,600

Again, this estimation assumes a perfect packing arrangement and doesn't account for the irregularities in the plane's interior shape or the space between the basketballs. The actual number of basketballs that would fit inside the 747 would likely be lower, but this calculation provides a rough estimate.

I think I see an error.  You have calculated the volume of a spherical basketball.   However, for packing purposes, the basketballs will behave much more like cubes, I think.  There will be a lot of space around and between the spherical basketballs.  Can you making a rational downward revision?

You are right. My previous calculations did not account for the inefficiency in packing spherical objects like basketballs. In reality, the packing efficiency for spheres is approximately 74% (or 0.74) when arranged in a close-packing or hexagonal close-packing pattern.

To account for this inefficiency, we can adjust our previous calculations by multiplying the available volume by the packing efficiency:

Adjusted volume = Total volume × Packing efficiency

Adjusted volume = 2,519.6 m³ × 0.74 ≈ 1,864.5 m³

Now, we can estimate the number of basketballs using the adjusted volume:

Number of basketballs ≈ 1,864.5 m³ / 0.0073126 m³ ≈ 254,900

Please note that this is still an approximation, and the actual number of basketballs that would fit inside the 747 would depend on the plane's interior shape and other factors. However, this estimation should provide a more reasonable approximation by considering the packing efficiency of spherical objects.

###### AI / Canva

BQ: Wolfram Alpha says 140,000.