Density and Scale Models
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Learning goals
- Calculate and compare average area densities and explain when they support a local estimate
- Use mass density to find mass or volume and distinguish capacity from mass
- Convert lengths, areas, and volumes between a scale model and what it represents
- Combine scale and density in a prediction and check its units and size
How much per square unit?
Two districts can have the same area and very different populations. If one has 72 residents and the other has 108, each across 6 square kilometers, dividing each count by 6 puts the comparison on a common basis. The averages are 12 and 18 residents per square kilometer.
Area density is the quantity per unit of area. A population density uses a population as the quantity. To calculate an average over a region of positive area, divide its total quantity by its area.
That division works even if residents live in clusters. Using the average as a rate for a smaller region assumes that its average density matches the larger region’s. A uniform model treats the density as the same throughout the area. That may be a useful approximation for some subregions. A park surrounded by apartments gives you a reason to question it.
Worked example 1 A whole district and one part of it
Problem. A coastal district has 1,260 residents and covers 7 square kilometers, including a 2-square-kilometer bird sanctuary with no permanent residents. Find the district’s average population density. What would a uniform model predict for the sanctuary, and how does that prediction compare with the given count?
Solution. Divide the district’s population by its full area, including the sanctuary.
The district averages 180 residents per square kilometer.
For the sanctuary, the uniform model applies the same rate to its smaller area.
The model predicts 360 residents. The given count is zero. The district average summarizes the district, but the uniform model gives a poor estimate for the sanctuary.
Check your understanding
A 9-square-kilometer district has 630 residents. Which claim follows from those two numbers alone?
Dividing the total population by the total area gives 70 residents per square kilometer. Where people live within the district is still unknown.
How much mass per cubic unit?
Two solid cubes can occupy the same amount of space and have different masses. If both occupy 125 cubic centimeters, a 250-gram cube averages 2 grams in each cubic centimeter, while a 400-gram cube averages 3.2 grams per cubic centimeter.
Mass density measures mass per unit of volume. For an object of mass and positive volume , the average mass density (the Greek letter rho) is
To find mass from volume, multiply by the applicable density. To find volume from mass, divide by that density, when it is positive. The units show which way to go. For example, grams divided by grams per cubic centimeter gives cubic centimeters. This same rate can apply to another piece of the same homogeneous material, meaning its composition and density are uniform throughout under the conditions of the model.
Worked example 2 Find how much space a batch of resin will fill
Problem. A 96-gram sample of casting resin occupies 80 cubic centimeters. A second batch has a mass of 360 grams and the same mass density. Find that density and the volume of the second batch.
Solution. Divide the first mass by its volume.
The resin has a mass density of 1.2 grams per cubic centimeter.
Divide the second mass by that density.
The second batch occupies 300 cubic centimeters. Multiplying that volume by 1.2 grams per cubic centimeter returns 360 grams.
Check your understanding
A foam sample has mass density 0.3 grams per cubic centimeter. A 60-cubic-centimeter piece of the same foam has that density. What is the piece's mass?
Multiply 0.3 grams per cubic centimeter by 60 cubic centimeters. The cubic centimeters cancel, leaving 18 grams.
Read the scale in the direction you need
On a printed plan, suppose 1 centimeter represents 5 meters. A 2-centimeter side represents 10 meters. A square 2 centimeters on each side therefore represents a 10-meter-by-10-meter square. Its area is 100 square meters. Each of its two dimensions gets the same conversion.
A scale matches a model length to the corresponding actual length. In a ratio such as in model-to-actual order, the lengths are expressed in the same unit. To go from model length to actual length, multiply by 500. To go from actual length to model length, divide by 500. The scale above is because 5 meters is 500 centimeters.
For similar figures, if is the actual-to-model length factor in common units, multiply model lengths by , areas by , and volumes by to get actual measurements in those same units. Reverse the operations to go back. If you instead use a conversion with units, such as 5 meters per model centimeter, square or cube the units along with the number. A scale applies in the directions the model actually represents. A top view gives horizontal dimensions, and a depth needs its own measurement or scale.
Worked example 3 Read a map area from its scale bar
Problem. A rectangular marsh measures 3 cm by 2.5 cm on a paper map. On that copy, a 2 cm scale bar represents 600 m on the ground. Find the actual area in square meters. A straight boardwalk elsewhere on the map is 750 m long in reality. Find its length on the same paper map.
Solution. Divide the matched lengths by 2 to get 300 actual meters per map centimeter.
Marsh area. The two side lengths become 900 m and 750 m. Multiply those actual lengths.
The marsh covers 675,000 square meters. Each map square centimeter represents square meters, so multiplying 7.5 map square centimeters by that conversion gives the same area.
Boardwalk length. The actual length is known, so divide by 300 actual meters per map centimeter.
The boardwalk measures 2.5 cm on the paper map. This is longer than the 2 cm bar, which agrees with 750 m being greater than 600 m.
Check your understanding
On a printed map, 1 cm represents 40 m. A region covers 5 square centimeters on the map. What actual area does it represent?
Each map square centimeter represents a square 40 m on a side, or 1,600 square meters. Five of them represent 8,000 square meters.
Scale the space inside a container
A container’s capacity is the volume it can hold. Model that space using inside dimensions and the intended fill level. For a rectangular interior, use the inside floor area times the fill depth. The rim gives one possible limit. A lower fill line gives a smaller volume.
A cubic decimeter is exactly 1 liter. Since a decimeter is 10 centimeters, 1 liter is 1,000 cubic centimeters. The capacity in liters tells you the available volume. To predict the mass of the contents, you also need their density.
Worked example 4 Compare a rim estimate with a fill-line estimate
Problem. A tank model has a rectangular interior 12 cm long and 5 cm wide. Its inside depth to the rim is 5 cm. A fill line is 4 cm above the inside floor. The corresponding actual inside lengths all use model-to-actual scale . Find the modeled actual capacities at the rim and at the fill line, in liters.
Solution. The actual-to-model length factor is 12, so the volume factor is , or 1,728.
At the rim. The model interior volume is 300 cubic centimeters. The actual estimate is
That gives 518.4 L.
At the fill line. The model volume below the line is 240 cubic centimeters. The actual estimate is
That gives 414.72 L. The floor area is fixed, so four fifths of the depth holds four fifths of the rim volume. The two results satisfy that check. Both describe the rectangular model. An actual tank with rounded interior corners or fittings that take up space may hold less.
Check your understanding
A container and its model have similar interiors at a model-to-actual length scale of 1:4. The model holds 250 cubic centimeters at the rim. What actual rim capacity does the model predict?
Multiply 250 cubic centimeters by the volume factor of 64 to get 16,000 cubic centimeters. There are 1,000 cubic centimeters in each liter.
Connect scale to density
Changing the size of a drawing leaves the number of people it represents fixed. If you double its lengths, the paper area quadruples. The number of residents per square centimeter of paper falls to one fourth, while the actual district’s population density stays the same.
Compare that with a solid sample and a larger, similar solid made of the same homogeneous material under the same conditions. If the lengths double, there is eight times as much volume and eight times as much mass. Dividing mass by volume gives the same density. A miniature made of different material can have a different density, so use the actual material’s density for the actual mass estimate.
Worked example 5 Estimate the mass of soil for the garden bed
Problem. A plan shows a rectangular bed’s inside floor as 4.5 cm by 2.5 cm. Each plan centimeter represents 0.8 m. The planned fill depth is 0.25 m in the real bed, and the soil is expected to have a bulk mass density of 720 kilograms per cubic meter. Bulk density here includes the space between soil grains. Model the fill as a rectangular prism. Estimate the mass of soil needed. Compare it with the mass of 2 cubic meters of the same soil as a check.
Solution. Convert the plan dimensions to actual lengths. They become 3.6 m and 2 m.
Volume. The fill depth is already an actual measurement. Multiply the three actual dimensions.
The modeled fill volume is 1.8 cubic meters.
Mass. Use the expected bulk density for the soil in the bed.
The estimate is 1,296 kg. The volume is just below 2 cubic meters, so the mass should be just below 1,440 kg at this density. The estimate fits that check. How tightly the soil is packed can change the bulk density. Building and Checking a Geometric Model explains how to check which inputs a prediction depends on.
Check your understanding
A miniature and a larger solid are similar. Each length in the larger solid is three times its matching miniature length. Both are solid pieces of the same homogeneous material under the same conditions. What changes from miniature to larger solid?
Volume has factor 27. At the same mass per unit of volume, mass has factor 27 as well. The common factor cancels when you divide mass by volume.