Posts Tagged ‘speed’

Tangential Velocity Dangers

Monday, August 28th, 2017

    We’ve been discussing tangential velocity within the context of a pulley and belt assembly in recent blogs, and you may have wondered whether you encounter this phenomenon in your everyday life.  Undoubtedly you have. Have you ever driven a long stretch of highway at a fast clip and suddenly come upon a curve in the road posted at a lower speed limit?  If you happened not to notice the speed reduction, you may have found yourself slamming on the brakes to regain control of your car.  You’ve been caught in a tangential velocity danger zone.

Tangential Velocity Dangers

Tangential Velocity Dangers


    As this road sign indicates, cyclists must also beware of potentially dangerous circumstances involving tangential velocity.   It warns of an upcoming drop in the road, which, depending on their speed, has the potential to catapult them into the air.

    Next time we’ll resume our discussion of tangential velocity and other factors within the context of our pulley-belt assembly.

Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



Tangential Velocity

Monday, August 14th, 2017

   Last time we introduced the Mechanical Power Formula, which is used to compute power in pulley-belt assemblies, and we got as far as introducing the term tangential velocity, V, a key variable within the Formula.   Today we’ll devise a new formula to compute this tangential velocity.

   Our starting point is the formula introduced last week to compute the amount of power, P, in our pulley-belt example is, again,

P = (T1 – T2) × V                                         (1)

   We already know that P is equal to 4 horsepower, we have yet to determine the belt’s tight side tension, T1, and loose side tension, T2, and of course V, the formula for which we’ll develop today.


Tangential Velocity

Tangential Velocity


   Tangential velocity is dependent on both the circumference, c2, and rotational speed, N2, of Pulley 2.  The circumference represents the length of Pulley 2’s curved surface.   The belt travels part of this distance as it makes its way from Pulley 2 back to Pulley 1. The rotational speed, N2, represents the rate that it takes for Pulley 2’s curved surface to make one revolution while propelling the belt.   This time period is known as the period of revolution, t2, and is related to N2 by this equation,

N2 = 1 ÷ t2                                                                         (2)

   The rotational speed of Pulley 2 is specified in our example as 300 RPMs, or revolutions per minute, and we’ll denote that speed as N2 in light of the fact it’s referring to speed present at the location of Pulley 2.   As we build the formula, we’ll be converting N2 into velocity, specifically tangential velocity, V, which is the velocity at which the belt operates, this in turn will enable us to solve equation (1).

   Why speak in terms of tangential velocity rather than plain old ordinary velocity?  Because the moving belt’s orientation to the surface of the pulley lies at a tangent in relation to the pulley’s circumference, c2, as shown in the above illustration.   Put another way, the belt enters and leaves the curved surface of the pulley in a straight line.

   Generally speaking, velocity is distance traveled over a period of time, and tangential velocity is no different.  So taking time into account we arrive at this formula,

V = c2 ÷ t2                                                                          (3)

   Since the surface of Pulley 2 is a circle, its circumference can be computed using a formula developed thousands of years ago by the Greek engineer and mathematician Archimedes.   It is,

c2 = π × D2                                                            (4)

where D2 is the diameter of the pulley and π represents the constant 3.1416.

   We now arrive at the formula for tangential velocity by combining equations (3) and (4),

V = π × D2 ÷ t2                                                    (5)

   Next time we’ll plug numbers into equation (5) and solve for V.

Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



Work and Energy, Speed, and Calories

Wednesday, January 20th, 2016

    In my work as an engineering expert I’ve never had to convert Joules of work-energy into calories, but that’s exactly what we’ll be doing together today.   Last time we applied the Work-Energy Theorem to the progress of Santa’s sleigh and found that an opposing wind force of 3848.7 Newtons  –or 865.2 pounds for those of us who are American–  slowed his team from an initial velocity of 90 meters per second to a final velocity of 40 meters per second and that it happened over a distance of 760 meters.   Today we’ll see how many calories the reindeer need to expend to get them back up to full delivery speed.

 Work Energy Theorem and Reindeer

Prancer Loves Oats

    Now we know that Santa successfully made all his deliveries on time this past Christmas, so that means that at some point his reindeer team was able to get back up to full sleigh-flying speed.   They did it by expending extra energy.   We’ll use the Work-Energy Theorem to find out how much energy that equates to.  Here’s the Theorem again,

W = ½ × m × [v22v12]

where W is the work/energy required to speed up the sleigh team’s mass, m, from an initial velocity v1 to a final velocity v2.   For a refresher on the special relationship between  work and energy, see our past blog on the subject.

    If Santa’s sleigh has a mass of 900 kilograms and its speed must increase from 40 to 90 meters per second, then the work required to do so is calculated as,

W = ½ × (900 kilograms) × [(90 meters/second)2 – (40 meters/second)2]

W = ½ × (900 kilograms) × (6,500 meters2/second2)

W =  2,925,000 kilogram2 · meters2 per second2 = 2,925,000 Joules

    So Rudolph and his buddies had to expend 2,925,000 Joules of energy to perform 2,925,000 Joules of work.   To understand where Rudolph and his team got that energy, we must state things in terms of nutritional value, that is, units of calories.

    Did you know that 1 calorie is equal to 4,184.43 Joules?   Applying that equivalency to our situation we get,

Nutritional Energy Required = (2,925,000 Joules) × (4,184.43 Joules/calorie)

= 699.02 calories

    The net result is Santa’s team expended a total of 699.02 calories for all the reindeer to regain their full speed of 90 meters per second.   That’s the nutritional energy found in slightly more than one cup of oats.   Now everybody knows that Santa takes good care of his reindeer, so undoubtedly they were fed plenty of oats and hay before takeoff.   This was stored in their body fat for future, on-demand use.

    Sadly, Christmas is over, and it’s time to get back to the more mundane aspects of life.   Next time we’ll apply the principles behind the Work-Energy Theorem to calculate the braking force required to stop a car in motion.

Copyright 2015 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog



Earth’s High Speed Race Around the Sun

Tuesday, July 28th, 2015

      Last time we began our discussion on velocity by focusing on one of its aspects, distance, and we calculated Earth’s orbital distance around the sun to be 5,816,023,200 miles.   Today we’ll focus on velocity’s other aspect, time.   Together, these aspects will allow us to solve for Earth’s orbital velocity, aka speed.

      When early astronomers monitored Earth’s journeys, they found it took exactly one year for it to complete its orbit around the sun.   They combined this fact with Earth’s orbital travel distance of 5,816,023,200, or 9.36 × 1011 meters — meters being the unit of measurement most often used in scientific computations — and solved for Earth’s orbital velocity as follows,

v = 9.36 × 1011 meters ÷ 1 year = 9.36 × 1011 meters per year

      The scientific notation of 9.36 × 1011 equates to 936,000,000,000 meters, a large and unruly number to work with.   We can simplify things further by breaking this number down into units of meters per second, which will then allow us to arrive at Earth’s velocity in terms of miles per hour, something most Earthlings can relate to.

      One meter per second is equal to 2.237 miles per hour, and there are 31,536,000 seconds in one year, so breaking Earth’s orbital velocity down into meters per second we arrive at,

v = 9.36 × 1011 meters ÷ 31,536,000 seconds = 29,680 meters per second

v = (29,680 meters per second) × (2.237 mph/m/sec) = 66,394 miles per hour

      Yes, it’s true, Earth whips around the sun at warp speed.  Our fastest man made rockets only achieve speeds of about 25,000 miles per hour.   Earth beats them two-to-one!

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      Now that we know Earth’s orbital velocity, we have everything we need to calculate the gravitational force exerted on Earth by the sun.   We’ll do that next time.


Earth’s Orbital Velocity

Sunday, July 19th, 2015

      Last time we introduced Newton’s equation to calculate the sun’s gravitational force acting upon Earth, and today we’ll begin solving for the last remaining unsolved variable within that equation, v, Earth’s orbital velocity.

      Here again is Newton’s equation,

Fg = [m × v2] ÷ r

      For a refresher on how we solved for m, Earth’s mass, and r, the distance between Earth and the sun, follow these links to past blogs in this series, What is Earth’s Mass and Calculating the Distance to the Sun.

      Velocity, or speed, as it’s most commonly referred to, is based on both time and distance.   To bear this out we’ll use an object and situation familiar to all of us, traveling in a car.   The car’s velocity is a factor of both the distance traveled and the time it takes to get there.  A car traveling at a velocity of 30 miles per hour will cover a distance of 30 miles in one hour’s time.   This relationship is borne out by the formula,

vCar = distance traveled ÷  travel time

vCar = 30 miles ÷ 1 hour = 30 miles per hour

      Similarly, v is the distance Earth travels during its orbital journey around the sun within a specified period of time.   It had been observed since ancient times that it takes Earth one year to complete one orbit, so all that remained to be done was calculate the distance Earth traveled during that time.   Vital to calculations was the fact that Earth’s orbit is a circle, which allows geometry to be employed and calculations to be thereby simplified.   Refer to Figure l.

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Figure 1

      From geometry we know that the circumference of a circle, C, is calculated by,

C = 2× π × r

where π is a constant, the well known mathematical term pi, which is equal to 3.1416, and r is the radius of Earth’s circular orbit, determined, courtesy of the work of Johannes Kepler and Edmund Halley, to be approximately 93,000,000 miles.   Stated in metric units, the unit of measurement most often employed in science, that comes to 149,000,000,000 meters.

      Inserting these numerical values for π and r into the circumference formula, scientists calculated the distance Earth travels in one orbit around the sun to be,

C = 2 ×  π × 149,000,000,000 meters = 9.36 x 1011 meters

      Next time we’ll introduce the time element into our equations and solve for v, and from there we’ll go on and finally solve for Fg, the sun’s gravitational force acting upon Earth.