Your own neighborhood mini roller coaster can be a bicycle
ride down a slope near your home.

For instance, we used a shallow hill
of less than 10 feet rise over about a 200 foot run to get data for this
project. It is best to find about a two hundred foot distance of
pavement that is sloped overall but flat like an incline board. We
placed orange soccor cones every twenty feet and used a video tape
recorder to collect data. Video tapes can be played back in slow motion
so one can count the number of frames per second then use that information
to mark the time of an event. Be sure to keep the clock timer running
on your video camara while you film the bicycle trial runs.

We are interested in a math model of our mini bicycle roller coaster from the point of view of slope. We define slope to be the the value, call it m, computed by taking the rise of a hill divided by the run along the horizontal or level underground base of the hill. Consider the rise as an upright height or leg of the right triangle associated with the level run leg of the triangle then you are traveling on the longer third, road, side of the triangle called the hypotenuse of the triangle. In our first graph the bicycle moves down the hypotenuse road side of the triangle. The height along the y axis of about six large tick marks would be the rise for the straight line hypotenuse while the length along the x axis of about 29 large tick marks would be the run. So the slope is m = ___________

As we show in this next graph, the slope, m, of the straight line hypotenuse on which the bicycle is rolling may also stand for any short run length. So here we have changed the units and consider a smaller triangle but the same slope number, m, works for the shorter run and rise. The reason the same number works is due to having similar triangles. The large triangle is similar to the small triangle. Their corresponding sides are in proportion.

Then the ratio rise/run is equal to m which is also the ratio m/1. So for a run of one unit, the road distance is (1 + m² )^0.5 which is the square root of (1 + m² ). This relationship is a result of using the Pythagorean Theorm. Also note that the rise is the product of the run and the slope, (run value)×m, whenever you want to use a longer run and calculate the corresponding rise.

We did the hill in the above color picture represented by the following graph and data.

The following points represent twenty foot intervals covered in the corresponding experimental times:

A, 0 feet in 0 seconds

B, 20 feet in 2.87 seconds

C, 40 feet in 5.17 seconds

D, 60 feet in 6.94 seconds

E, 80 feet in 8.50 seconds

F, 100 feet in 9.90 seconds

G, 120 feet in 11.14 seconds

H, 140 feet in 12.40 seconds

I, 160 feet in 13.92 seconds

J, 180 feet in 15.30 seconds

The hill we used was not a flat slope, in fact the last two points, I and J, did not have near the fall (negative rise) that the other points had. So we used points B, D, and H to get a second degree equation

s = .506 t² + 4.852 t + 1.876 where s is the road distance in feet.

This equation seems to be a close fit to most of the points A through H but not I and J. If you use your calculator with time t = 8 seconds, you can find from the equation that s about equal to 73 feet. In other words this equation will tell you how far you traveled down the road for time t seconds.

Graph the results to get a picture of the distance, s, versus the corresponding time, t, with t plotted on the horizontal axis.

Now locate the data points: A, B, ..., J on your graph.

Perhaps you can find another equation relating s and t. Hint: Try using three other points.

Which equation best fits the data.

y = 16 t² + ? t + ?? with a constant 16 for the t squared's coefficient,

we have:

s = .506 t² + 4.852 t + 1.876 with a constant .506 for the same term.

Realizing that the constant 16 acts straight down and our coefficient .506 acts in the direction of the road or hypotenuse gives use the following picture correspondence.

The two triangles within the large triangle have different scales but they are similar due to having a right angle, a common side and sides perpendicular. That means:

.506/16 = m/((1 + m² )^0.5).

Then by squaring both sides, multiplying both sides by 16 and (1+m²), we get:

.256036(1 + m²) = 256 m²

.256036 + .256036 m² = 256 m²

.256036 = (256 - .256036) m²

m² = .001001141

m = .03164

So in a run of around 140 feet we have a rise given by:

140×.03164 = 4.4277 feet.

We checked this calculation with some surveying data and calculations and found 4.4 feet in the ball park.

Now we can go back and draw the slope of the hill with a 4.4 foot rise, the road, hypotenouse, around 140 plus feet and the run around 140 feet.

To learn more about the physics of this situation you may enjoy reading David A. Sandborg's Roller Coaster Physics, © 1996.

Frank Rowland Whitt and David Gordon Wilson's book "Bicycling Science", 2nd ed. printed by MIT Press in 1995 is full of lots of interesting studies.