Since an actual tyrannosaurus was about 15 meters long, the ratio of the actual
dinosaur to the model is 50 to 1, because 15/0.3 = 50. Use the density ratio of
mass
volume
to determine the mass of the tyrannosaurus. Most animals and reptiles
have a density near
0.95 =
m
v
, so the mass of the tyrannosaurus can be calculated
once the volume is found. The volume of the actual tyrannosaurus can be calcu-
lated by using the cube of the ratio of the lengths of the actual dinosaur to the
model. The cube of the ratio is used, because volume is a measure of three
dimensions. Therefore the volume of the actual Tyrannosaurus will be 50
3
, or
125,000 times the volume of the dinosaur model.
You can measure the volume of an irregular object, such as a dinosaur model,
by submersing it in a bucket of water. Place a bucket of water filled to the brim
(and larger than the dinosaur model) inside a larger empty bucket. Drop the
dinosaur model into the bucket of water, and the excess water will spill over the
sides into the empty bucket. Pour the excess water into a graduated cylinder,
which is a tool to measure the volume of water. This volume should be the same
as the volume of the dinosaur model, because the model replaced the same
amount of space in the bucket as the excess water. Suppose that the volume of
the model is 61 milliliters. This means that the volume of the actual tyranno-
saurus was about 125,000 times 61, or 7,625,000 milliliters, or 7,625 liters. Since
density equals mass divided by volume, the equation
0.95 =
m
7,625
can be used
to predict the mass,
m, of the tyrannosaurus. Note that the units of density are
kilograms per liter, so volume units are in liters and calculated mass units are in
kilograms. The solution to the equation predicts the tyrannosaurus’s mass to
equal approximately 7,243 kilograms, which is about 16,000 pounds. That is the
same as 100 people that have an average mass of 160 pounds. Most football
coaches would like to recruit a tyrannosaurus for their teams!
Similarity is sometimes not used in models, which as a result can cause mis-
conceptions about length and size. Most models of the solar system are inaccu-
rately proportioned so that they can be easily stored, carried, and viewed within
a reasonable amount of space. If a teacher wants to illustrate planetary motion
on a solar-system model, he or she needs to be able to move the planets around
fairly easily, and students need to see all of them. Realistically, however, this
type of model is inaccurate, because the planet sizes vary tremendously and are
spread apart by vastly different distances. For example, if an accurate scale
model of the planets in the solar system were used in a classroom with the sun
at the center of the room, then the first four planets would be within 227 cm of
the center, and the remaining planets would be stretched out to almost 6 meters
away! The large variability in distances among the planets would make it diffi-
cult to build a movable model that illustrates rotation around the sun. Further-
more, the volumes of the planets vary considerably. Large planets, like Jupiter
and Saturn, have diameters that are about ten times larger than the earth. If the
planets were built to scale, these giant planets would have to be a thousand times
larger than the earth, because the ratio of volumes between similar figures is
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