Example 1 - Calculate Tank capacity

Assume we are trying to determine the capacity of a rectangular water-tank. Further, we know the length/width/depth measurements, but they are all in different units. With other programs, we would have to convert each measurement before performing the simple capacity calculation. Worse still, most programs would force us to navigate a series of cascading menu-selections before even stumbling across the units you want. With Univerter however, just type what what you have in one simple equation...

Since inches, centimeters, and feet are all predefined units, Univerter automatically understands your equation and does conversions on the fly. Using Univerter we're done in a fraction of the time that we would have spent searching for just the right units-combination with those other programs.

Example 2 - Complex Pressure Calculation

In this next example, we're trying to calculate pressure. Although we want the answer to be in inches of Mercury, the measurements were made in a variety of different units.

With Univerter, this isn't a problem. If we had made a mistake and the input equation didn't equate to pressure, Univerter would alert us to the problem.

Notice also that the output was formatted for three decimal places -- Univerter will format your results almost any way you want.

Example 3 - Dimensional Analysis

In this example, we want to calculate how much power would be required to lift a 125 pound mass at a constant rate of 4 feet per second. Univerter will not only let you type variables in whatever units you choose, it will also let you know what you’ve typed. Let the cursor hover for a moment over either the input or output field, and a small window will pop up to identify the units group your equation describes.

In this figure, the input is being identified as "Power" (or "Radiant Flux"). The window also displays the base units (metric) that your equation could be ‘distilled’ into.

Example 4 - Motor Torque

Given a 2.5 horsepower air motor, what's that most torque you could expect if the motor runs at a constant 2250 RPM?

Let's perform this solution step by step...

Start with the knowledge that force times distance is energy, and that energy per unit of time is power.
As shown in the diagram to the left, a force "F" applied tangentially to a rotational shaft at a distance "r" from the center of rotation acts over a distance of 2 × pi × r (the circumference) for each revolution. Thus the energy spent per revolution is (F×2×pi×r). Since we have the "Energy" part of the equation, all we have left is to divide by time to get power.

Taking RPM and dividing by revolutions leaves "per minute". If we multiply the energy part of the equation by RPM/Rev we now have a result that is units of power.

Now that we've derived the basic equation, we can simply rearrange the variables so that each side of the equation is force times distance (i.e. torque):
F×r = HP / (RPM/Rev) / (2×pi)

Enter the new equation -- with supplied values -- as shown above and request the answer in foot × pounds. Of course you need to use "lbf" (pounds-force) rather than "lb" which is a unit of mass, not force. If you change the Output Units to "Newton×meter" you'll have your answer in metric units.