# Calculations: Example

Let's see how we can predict the maximum speed of a rider using the information from Figure 5. For this example, we will assume a rider weight (including trolley and harness) of 578 N (or equivalently, 130 pounds).

Figure 5. Measurements for the last zip line on the WVU Canopy Tour. The drawing is not shown to scale.

### Step 1: Determine the Distance Traveled

$$Distance (d) = {\sqrt{horizontal~distance^2 + vertical~drop^2}}$$

Please find the horizontal distance to the lowest point and vertical drop values from Figure 5 and insert them into the equation below. Click on the "Calculate" button to see the result.

 Distance(d) = 2 +  2
Sorry, the values you entered are not correct. Please try again.

$$Distance~(d)= \sqrt{214^2 + 16^2}$$ $$d = \sqrt{45796 + 256}$$ $$d = \sqrt{46052}$$ $$d = 214.597 m$$ $$d \approx 215 m$$

### Step 2: Approximate Acceleration

$$Acceleration~(a) = g \times sin(\theta) - loss$$ $$remember: g = 9.81\frac{m}{s^2}$$

First let’s find sin(θ). It can be calculated by dividing the opposite leg (vertical drop) by the hypotenuse (distance (d)). Enter the two values into the equation below to calculate sin(θ). For simplicity, enter whole numbers only.

sin(θ) = vertical drop / distance(d)

= /

Sorry, the values you entered are not correct. Please try again.

$$sin(\theta) = \frac{16m}{215m} = 0.0746$$

Next, we’ll use the chart below to determine loss. The loss values have been derived through experimentation on the WVU Canopy Tour. In this example the rider is 130 pounds. Find the approximate value of loss in the chart and use it in the equation below.

So, for a rider of 130 pounds, acceleration can then be calculated as: (enter the values of sin(θ) and loss below)

$$Acceleration~(a) = g \times sin(\theta) - loss$$ Acceleration (a) = 9.81 X -
Sorry, the values you entered are not correct. Please try again.

a = 9.81 X 0.0746 - 0.45

a = 0.732 - 0.45

a = 0.282 m/s2

### Step 3: Calculate maximum velocity

$$Maximum~Velocity(V_{max}) = \sqrt{2\times a\times d}$$

Since the values of all variables are now known, the maximum velocity (speed) of a 130 pound rider can be determined. Enter the values into the equation below to find out.

 Vmax = 2 X X
Sorry, the values you entered are not correct. Please try again.

$$V_{max} = \sqrt{2\times(0.282)\times(215)} = 11.0~m/s$$

Thus, knowing the distance of the zip line to the lowest point in the catenary curve, the vertical drop, and the weight of the rider, you can predict the maximum speed on the zip line.

In this example, a 130 pound rider will have a maximum acceleration of 11.0 m/s or 24.7 mph (see unit conversion chart at the right side of this page).

Go to Summary

#### Standard International (SI) to English Conversions

To Convert: Multiply By: To Get:
Length
meters (m)3.28feet (ft)
kilometers (km)0.621miles (mi)
Mass
grams (g)0.0022pound mass (lb)
kilograms (kg)2.202pound mass (lb)
Velocity/Speed
meters per second (m/s)3.28feet per second (ft/s)
meters per second (m/s)2.24miles per hour (mph)
kilometers per hour (km/hr)0.621miles per hour (mph)
Force
newtons (N)0.225pound force (lb)