August 25, 2013

Skating Race Comes Down to Inches

Filed under: Complex Unit Converter — MikeVV @ 10:10 am

I recently watched the World Cup short-track ice skating races at the Olympic Oval in Kerns, Utah, USA.  Two skaters crossed the finish line within 0.01 seconds of each other.  How far apart were they?  The complex converter web application enabled me to determine this distance.  First, distance = rate x time.  I know that the winning race time was 43.00 seconds and the race distance was 500 meters.  The race speed (or rate) was = 500 / 43.00 [m/s] = 26.0108350951374 [mi/hr] .  At that rate, one travels a distance in 0.01 seconds = 26.0108350951374 * .01 [(mi/hr)*s] = 4.58 [in].  The placement of the skaters was within 5 inches of each other!

May 18, 2013

Utah Professional Engineer Conference 2013

I had a great time speaking at the Utah Professional Engineer’s Conference in Sandy, Utah on 5/18/2013. I was able to demonstrate FreeStyle workbook features such as interpolation, VBA, graphic maps and data digitization.  Mr. Brad Allen did a great job of coordinating activities and several interesting talks were given by other Utah PE’s including Jerry Simpson (Life Science Test Facility overview), David Horne (Gas Engine Mileage Improvements), George Adamson Jr (Founding an Engineering Firm), Jason Dahl (Sensors and Data Acquisition) and Daniel Donahoe (Economic Changes in Engineering Profession).

FS-4Bar.xls workbook to analyze linkages

4-Bar Linkage Modeler at

My presentation, Engineering Workbooks, is available for download.

Several attendees were interested in the 4-bar linkage workbook that allows users to plot the resulting curve of a point on any 4-bar linkage.  The workbook is available for download here.  The FreeStyle.xls workbook describing digitizing and including graphical slide rules was also interesting to those in attendance.


July 18, 2012

Calories Burned on the Crag Crest trail hike

Filed under: Complex Unit Converter — MikeVV @ 12:05 am

How many “hamburgers” did I “burn” while hiking up the Crag Crest trail at the Grand Mesa in Colorado?  The human body consumes energy that it gets from food.  The “calorie,” or more accurately, “KCAL,” is used to define this food energy content.  Approximately 20% of the food we eat actually goes into useful work (like hiking) according to Stephen Seiler (see  We can relate that to the energy required to lift a mass to a given height to determine how much caloric intake is needed.  Dividing the work produced by the efficiency to obtain the intake is easy for the – Complex Unit Converter.  Calculations like this demonstrate its ability to relate known quantities in their given units directly which simplifies the analysis process which improves accuracy.   Consulting the internet, we determine that the elevation change up the trail is 900 ft.

Energy Consumed = (my mass * trail elevation gain) / efficiency
my mass = 185 lbm (that’s what I “weigh” in mass terms),
elevation gain = 11,220 ft – 10,320 ft = 900 ft
(see for trail-head elevations)
efficiency for human body energy consumption = .20

Crag Trail - Grand Mesa, CO

A “kcal” is used to define food caloric quantity.  It relates to the mass * height as an energy quantity.  Substituting into the AnalysisChamp Complex Converter:

Input Values = (185*1*900)/.2
Input Units = (lbm*g*ft)/item
Output Unit = KCAL
   gives: 270 KCal <==

which relates to approximately one single patty hamburger (according to .  I think a good hamburger is a nice reward for hiking to the top of Crag Crest! Please note that we needed to include “1-g” in the analysis.  Mass is not the same as force.  Energy is determined from force multiplied by distance.  The English measurement system measures mass and force using the “pound,” but they are distinguished as “LBM” (pound-mass) and “LBF” (pound-force) in calculations.  It is important to keep them properly assigned.

July 1, 2012

Kinetic Energy: How Much does Earth Have?

Often we try to relate physical processes to human technology.  As an example, how does the inherent kinetic energy of the Earth in its orbit around the sun compare to something man-made, like an atomic bomb?  The Earth orbits around the sun because it has enough kinetic energy to do so.  How much is that?  How do we use the website to perform this calculation?

The kinetic energy of any mass = 1/2 * mass * velocity^2.  What is the mass of the Earth?  It is already available as “EARTHMASS” in the Complex Converter (many useful values are found in the “Unit String” webpage link).

EARTHMASS Constant lookup in unit name table

EARTHMASS Constant lookup in unit name table

The Earth’s orbital velocity = 1 REV/YR * 1 Astronomical Unit (AU) (revolution, year and astronomical unit are also in the “Unit String” webpage).  Although we typically define velocity in terms of “distance / time,” we can express it using the terms that we have.  The Complex Converter webpage has an “Expression Builder” to assist in entering the “input value,” “input unit string” and “output unit string” for us.  It is particularly useful when using a small screen and keypad like those on iPhones, Android smart-phones and other hand-held internet devices.

View of Expression Builder in Complex Unit Converter

Expression Builder in Complex Converter webpage

The expression builder allows you to select unit strings and paste them into a text editing field that also has “calculator” functions.  When the string is acceptable, you then paste it into the Complex Converter’s “input number,” “input unit,” or “output unit” fields.  Unit strings like “AU” (Astronomical unit = distance from Earth to Sun), which are infrequently used by many, can be identified and used this way.

The output units are chosen to be in KILO*TONTNT which is related to atomic bombs.  This unit is often used to express the energy released by our most powerful weapons.  How does the natural energy contained by the Earth in its orbit compare to these man-made devices?  The AnalysisChamp – Kinetic Energy of Earth Orbit calculation is expressed as:

  • Input Number: 1/2*1*(1*1/1)^2
  • Input Unit: EARTHMASS*(PI*AU/YR)^2
  • Output Unit: kilo*TONTNT


The result = 1.585E+20 KiloTonTNT! This is a huge value and shows how massive the Earth-Sun system is when compared to something on the scale of a man-made system.  This article also shows that the Complex Converter can be applied to problems consisting of constants, like the Earth’s mass and distance from sun, using “unit names” instead of numerical values.


June 27, 2012

How Much Do I Weigh If the Earth Stops Spinning?

Filed under: Complex Unit Converter — MikeVV @ 12:02 pm

A recent book, “The Age of Miracles,” by Karen Thompson Walker, was reviewed on National Public Radio.  In it, she describes the effects on humanity of a world that stops rotating.  One of the many postulated effects is that we become heavier because the centrifugal / centripetal force caused by the Earth’s rotation ceases to exist.

How much would our weight change really change?  The Wikipedia provides an article applicable to this question, a discussion of “Centripetal Force.” (A discussion of the “Earth’s Gravity” force is also available for those interested in the general topic.)  What is the magnitude of the centripetal force on a person who’s mass is 185 lbm when rotating about the Earth center at 1.0 revolutions per day?  What is that magnitude when the revolution speed is 0.0 revolutions per day?

The Complex Converter at is used to calculate this force.  Enter “185*1*(1/1)^2″ into the number field, “LBM*EARTHRM*(REV/DAY)^2″ into the Input Unit field and “LBF” in the Output Unit field.  Press “Calculate” and find the force to be 0.636 lbf.  If the rotation speed is zero the centripetal force will also be zero so this represents the effect postulated in the book.  Hmm, doesn’t seem like much for a 185 lb person.

I expect the book to be interesting and entertaining but do not want to delve into it too deeply for fear of spoiling the adventure.  On another note, did you notice that the Eartrh’s radius is already built into the converter?  There was no need to look it up.  Also, the equation for calculating the centripetal force is provided in Example 19 on the webpage.  You can find it by using your browser’s “Search” function.   Be sure to ponder the difference between a “pound-mass” (LBM) and a “pound-force” (LBF).  The English system requires one to distinguish between these units for proper calculations.  The SI system uses “KG” for mass and “N” for force.

September 3, 2011

Where are Cubic Feet?

Filed under: Complex Unit Converter — MikeVV @ 11:13 am

A new Complex Converter user recently tried to convert using “cubic feet” and couldn’t find them in the list of units.  Are they there?  What are they called?  The power of the complex converter lies in its ability to handle any kind of unit expression.  Cubic Feet are expressed as “ft^3” or “ft*ft*ft” or “ft^2*ft” or “in^2*ft” or any unit of length taken in three multiples.

Example Conversion ScreenShot

Example Conversion ScreenShot

This technique can be combined with other base units to evaluate complex conversions.  An example is converting “gallons per second” (GPS) into “cubic feet perminute” (CFM).

The input string, “gal/s” and output string, “ft^3/min” are entered into the Complex Converter and evaluated using the common Windows syntax of Visual Basic.  That’s why units like “GPS” or “CFM” are not found in the list of unit names.  They are not needed because the Complex Converter uses expressions instead.  (Most online converters are hindered by the limits of unit names available, The Complex Converter has no such limit.)


Try it out and experiment with your own daily calculation examples!

August 29, 2011

How Much Dry Ice?

Filed under: Complex Unit Converter — MikeVV @ 10:59 pm

I came home to a bad surprise tonight; my refrigerator door was left open all day and the food inside was warmed up! I figure there is 50 lbm of food (with properties of water) affected. Most of the food in the freezer is still frozen and the food in the refridgerator is still chilled but needs to cool down 40 degrees F as soon as possible. I can get some ice but how much? Perhaps dry ice is better (it’s colder, doesn’t melt and bathes the food in carbon dioxide).

Heat to draw from food, Q = mass * specific heat * temperature change; gives property for water as 4.1813 J/(gm*degK) so Q = 50 * 4.1813 * 40 [lbm * J/(gm * degK) * degR] = 1997 BTU (degR = DegF difference). How much dry ice is needed to extract this much energy when sublimating into a gas?

The latent heat of sublimation for dry ice is given at as 246 BTU/lbm. 1997 BTU / 246 (BTU/lbm) = 8.12 lbm dry ice needed to chill things down. I bought 10 lbm and put 2/3 in the refrigerator and 1/3 in the freezer. After 2 hours the freezer was at 10 degF and the refrigerator was a 45 degF with half the dry ice remaining.

The calculations are combined at follows: Note: When working with temperatures: Use absolute whenever performing calculations and when working with temperature changes – degF changes are = degR and degC changes are = degK.

Reply with any questions or comments and keep on crunching those numbers!

August 27, 2011

Filed under: Uncategorized — MikeVV @ 1:42 am

All things for students, educators and technical professionals involved with STEM and applied problem solving activities

Powered by WordPress – Hosted by <>