## Posts Tagged ‘pulley’

Sunday, May 14th, 2017
Last time we introduced some of the variables in *the Euler-Eytelwein Formula,* an equation used to examine the amount of friction present in pulley-belt assemblies. Today we’ll explore its two tension-denoting variables, *T*_{1 }and* T*_{2}.
Here again is *the Euler-Eytelwin Formula,*where, *T*_{1 }and * T*_{2} are belt tensions on either side of a pulley,
*T*_{1} = T_{2} × *e*^{(μ}^{θ)}
*T*_{1} is known as the *tight side tension* of the assembly because, as its name implies, the side of the belt containing this tension is tight, and that is so due to its role in transmitting mechanical power between the driving and driven pulleys. *T*_{2} is the *slack side tension* because on its side of the pulley no mechanical power is transmitted, therefore it’s slack–it’s just going along for the ride between the driving and driven pulleys.
Due to these different roles, the tension in *T*_{1} is greater than it is in *T*_{2}.
__The Two T’s of the Euler-Eytelwein Formula__
In the illustration above, tension forces *T*_{1 }and* T*_{2} are shown moving in the same direction, because the force that keeps the belt taught around the pulley moves outward and away from the center of the pulley.
According to the *Euler-Eytelwein Formula*, *T*_{1} is equal to a combination of factors: tension *T*_{2 }; the friction that exists between the belt and pulley, denoted as *m*; and how much of the belt is in contact with the pulley, namely *θ*.
We’ll get into those remaining variables next time.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: belt, belt tension, driven pulley, driving pulley, engineering, Euler-Eytelwein Formula, mechanical power transmission, pulley

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Friday, May 5th, 2017
Last time we introduced the Pulley Speed Ratio Formula, a *Formula* which assumes a certain amount of friction in a pulley-belt assembly in order to work. Today we’ll introduce another *Formula,* one which oversees how friction comes into play between belts and pulleys, the *Euler-Eytelwein Formula*. It’s a *Formula *developed by two pioneers of engineering introduced in an earlier blog, *Leonhard Euler* and *Johann Albert Eytelwein*.
Here again is the Pulley Speed Ratio Formula,
*D*_{1} × *N*_{1} = D_{2} × *N*_{2}
where, *D*_{1} is the diameter of the driving pulley and *D*_{2} the diameter of the driven pulley. The pulleys’ rotational speeds are represented by *N*_{1} and *N*_{2}.
This equation works when it operates under the assumption that friction between the belt and pulleys is, like Goldilock’s preferred bed, “just so.” Meaning, friction present is high enough so the belt doesn’t slip, yet loose enough so as not to bring the performance of a rotating piece of machinery to a grinding halt.
Ideally, you want no slippage between belt and pulleys, but the only way for that to happen is if you have perfect friction between their surfaces—something that will never happen because there’s always some degree of slippage. So how do we design a pulley-belt system to maximize friction and minimize slip?
Before we get into that, we must first gain an understanding of how friction comes into play between belts and pulleys. To do so we’ll use the famous *Euler-Eytelwein Formula, *shown here,
**A First Look at the Euler-Eytelwein Formula**
where, *T*_{1 }and* T*_{2} are belt tensions on either side of a pulley.
We’ll continue our exploration of the *Euler-Eytelwein Formula* next time when we discuss the significance of its two sources of tension.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: belt, belt slippage, belt tension, drive belt, engineering, Euler-Eytelwein Formula, friction, mechanical power transmission, pulley, pulley belt system

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Friday, April 21st, 2017
Last time we saw how pulley diameter governs speed in engineering scenarios which make use of a belt and pulley system. Today we’ll see how this phenomenon is defined mathematically through *application* of the** ***Pulley Speed Ratio Formula*, which enables precise pulley diameters to be calculated to achieve specific rotational speeds. Today we’ll apply this *Formula* to a scenario involving a building’s ventilating system.
The *Pulley Speed Ratio Formula* is,
*D*_{1} × *N*_{1} = D_{2} × *N*_{2} (1)
where, *D*_{1} is the diameter of the driving pulley and *D*_{2} the diameter of the driven pulley.
__A Pulley Speed Ratio Formula Application__
The pulleys’ rotational speeds are represented by *N*_{1} and *N*_{2}, and are measured in revolutions per minute (*RPM*).
Now, let’s apply Equation (1) to an example in which a blower must deliver a specific air flow to a building’s ventilating system. This is accomplished by manipulating the ratios between the driven pulley’s diameter, *D*_{2}, with respect to the driving pulley’s diameter, *D*_{1}. If you’ll recall from our discussion last time, when both the driving and driven pulleys have the same diameter, the entire assembly moves at the same speed, and this would be bad for our scenario.
An electric motor and blower impeller moving at the same speed is problematic because electric motors are designed to spin at much faster speeds than typical blower impellers in order to produce desired air flow. If their pulleys’ diameters were the same size, it would result in an improperly working ventilating system in which air passes through the furnace heat exchanger and air conditioner cooling coils far too quickly to do an efficient job of heating or cooling.
To bear this out, let’s suppose we have an electric motor turning at a fixed speed of 3600 *RPM* and a belt-driven blower with an impeller that must turn at 1500 *RPM* to deliver the required air flow according to the blower manufacturer’s data sheet. The motor shaft is fitted with a pulley 3 inches in diameter. What pulley diameter do we need for the blower to turn at the manufacturer’s required 1500 *RPM*?
In this example known variables are *D*_{1} = 3 *inches*, *N*_{1} = 3600 *RPM*, and *N*_{2} = 1500 *RPM*. The diameter *D*_{2} is unknown. Inserting the known values into equation (1), we can solve for *D*_{2},
(3 *inches*) × (3600 *RPM*) *=* *D*_{2} × (1500 *RPM*) (2)
Simplified, this becomes,
*D*_{2} *=* 7.2 inches (3)
Next time we’ll see how friction affects our scenario.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: belt, blower, blower impeller, cooling coils, drive belt, driven pulley, driving pulley, electric motor, engineering, heat exchanger, mechanical power transmission, pulley, pulley speed, Pulley Speed Ratio Formula, RPM, ventilating system

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Friday, March 31st, 2017
Last time we introduced two historical legends in the field of engineering who pioneered the science of mechanical power transmission using belts and *pulleys*, Leonhard Euler and Johann Albert Eytelwein. Today we’ll build a foundation for understanding their famous *Euler-Eytelwein Formula *through our example of a simple mechanical power transmission system consisting of two pulleys and a belt, and in so doing demonstrate *the difference between driven and driving pulleys*.
Our example of a basic mechanical power transmission system consists of two *pulleys *connected by a drive belt. The *driving pulley* is attached to a source of mechanical power, for example, the shaft of an electric motor. The *driven pulley*, which is attached to the shaft of a piece of rotating machinery, receives the mechanical power from the electric motor so the machinery can perform its function.
**The Difference Between Driven and Driving Pulleys **
Next time we’ll see how *driven pulleys* can be made to spin at different speeds from the *driving pulley,* enabling different modes of operation in mechanical devices.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: belt, driven pulley, driving pulley, electric motor, engineering, Euler, Euler-Eytelwein Formula, Eytelwein, mechanical power, mechanical power transmission, pulley, pulley speed, rotating machinery

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Monday, March 20th, 2017
They say necessity is the mother of invention, and today’s look at an influential historical figure in engineering bears that out. Last week we introduced Leonhard Euler and touched on his influence to the science of pulleys. Today we’ll introduce his contemporary and partner in science, *Johann Albert Eytelwein*, a German mathematician and visionary, a true *engineering trailblazer* whose contributions to the blossoming discipline of engineering led to later studies with pulleys.
__Johann Albert Eytelwein, Engineering Trailblazer__
*Johann Albert Eytelwein’s* experience as a civil *engineer* in charge of the dikes of former Prussia led him to develop a series of practical mathematical problems that would enable his subordinates to operate more effectively within their government positions. He was a *trailblazer* in the field of applied mechanics and their application to physical structures, such as the dikes he oversaw, and later to machinery. He was instrumental in the founding of Germany’s first university level *engineering* school in 1799, the Berlin Bauakademie, and served as director there while lecturing on many developing *engineering* disciplines of the time, including machine design and hydraulics. He went on to publish in 1801 one of the most influential *engineering* books of his time, entitled *Handbuch der Mechanik* (Handbook of the Mechanic), a seminal work which combined what had previously been mere *engineering *theory into a means of practical application.
Later, in 1808, *Eytelwein* expanded upon this work with his *Handbuch der Statik fester Koerper* (Handbook of Statics of Fixed Bodies), which expanded upon the work of Euler. In it he discusses friction and the use of pulleys in mechanical design. It’s within this book that the famous Euler-*Eytelwein* Formula first appears, a formula *Eytelwein* derived in conjunction with Euler. The formula delves into the usage of belts with pulleys and examines the tension interplay between them.
More on this fundamental foundation to the discipline of *engineering* next time, with a specific focus on pulleys.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: bel, engineering, friction, Johann Albert Eytelwein, mechanical power transmission, pulley, pulleys

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Tuesday, February 28th, 2017
For some time now we’ve been analyzing the helpfulness of the engineering phenomena known as *pulleys* and we’ve learned that, yes, they can be very helpful, although they do have their limitations. One of those ever-present limitations is due to the inevitable presence of *friction* between moving parts. Like an unsummoned gremlin, *friction* will be standing by in any mechanical situation to put the wrench in the works. Today we’ll calculate just *how much friction is present* within the example *compound pulley *we’ve been working with. * *
__So How Much Friction is Present in our Compound Pulley?__
Last time we began our numerical demonstration of the inequality between a compound pulley’s work input, *WI*, and work output, *WO*, an inequality that’s due to friction in its wheels. We began things by examining a friction-free scenario and discovered that to lift an urn with a weight, *W*, of 40 pounds a distance, *d*_{1}, of 2 feet above the ground, Mr. Toga exerts a personal effort/force, *F*, of 10 pounds to extract a length of rope, *d*_{2}, of 8 feet.
In reality our compound pulley must contend with the effects of friction, so we know it will take more than 10 pounds of force to lift the urn, a resistance which we’ll notate *F*_{F}. To determine this value we’ll attach a spring scale to Mr. Toga’s end of the rope and measure his actual lifting force, *F*_{Actual}, represented by the formula,
*F*_{Actual} = *F + F*_{F }(1)
We find that *F*_{Actual} equals 12 pounds. Thus our equation becomes,
12 *Lbs = *10 *Lbs + F*_{F }(2)
which simplifies to,
2 *Lbs =* *F*_{F }(3)
Now that we’ve determined values for all operating variables, we can solve for work input and then contrast our finding with work output,
*WI = *(*F ×* *d*_{2}) *+ *(*F*_{F} *×* *d*_{2}) (4)
*WI = *(10* Lbs ×** *8* feet*) *+ *(2* Lbs ×** *8* feet*) (5)
*WI = *96* Ft-Lbs *(6)
We previously calculated work output, *WO* to be 80 *Ft-Lbs, *so we’re now in a position to calculate the difference between work input and work output to be,
*WI – WO =* 16 *Ft-Lbs* (7)
It’s evident that the amount of work Mr. Toga puts into lifting his urn requires 16 more Foot-Pounds of work input effort than the amount of work output produced. This extra effort that’s required to overcome the pulley’s friction is the same as the work required to carry a weight of one pound a distance of 16 feet. We can thus conclude that work input does not equal work output in a *compound pulley*.
Next time we’ll take a look at a different use for pulleys beyond that of just lifting objects.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, friction, pulley, work, work input, work output

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Monday, January 16th, 2017
We left off last time with an engineering analysis of energy factors within a compound pulley scenario, in our case a Grecian man lifting an urn. We devised an equation to quantify the amount of work effort he exerts in the process. That equation contains two terms, one of which is beneficial to our lifting scenario, the other of which is not. Today we’ll explore these two terms and in so doing show how there are situations when* work input does not equal work output.*
__Work Input Does Not Equal Work Output __
Here again is the equation we’ll be working with today,
*WI = *(*F ×* *d*) *+ *(*F*_{F} *×* *d*) (1)
where, *F* is the entirely positive force, or *work,* exerted by human or machine to lift an object using a compound pulley. It represents an ideal but not real world scenario in which no friction is present within the pulley assembly.
The other force at play in our lifting scenario, *F*_{F,} is less obvious to the casual observer. It’s the force, or *work,* which must be employed over and above the initial positive force to overcome the friction that’s always present between moving parts, in this case a rope moving through pulley wheels. The rope length extracted from the pulley to lift the object is *d*.
Now we’ll use this equation to understand why *work input*, *WI,* *does not equal* *work output*, *WO*, in a compound pulley arrangement where friction is present.
The first term in equation (1), (*F ×* *d*), represents the *work input* as supplied by human or machine to lift the object. It is an idealistic scenario in which 100% of energy employed is directly conveyed to lifting. Stated another way, (*F ×* *d*) is entirely converted into beneficial *work* effort, *WO*.
The second term, (*F*_{F} *×* *d*), is the additional *work input* that’s needed to overcome frictional resistance present in the interaction between rope and pulley wheels. It represents lost *work* effort and makes no contribution to lifting the urn off the ground against the pull of gravity. It represents the heat energy that’s created by the movement of rope through the pulley wheels, heat which is entirely lost to the environment and contributes nothing to *work output*. Mathematically, this relationship between *WO, WI,* and friction is represented by,
*WO = WI –* (*F*_{F} *×* *d*) (2)
In other words, *work input* is *not equal* to *work output* in a real world situation in which pulley wheels present a source of friction.
Next time we’ll run some numbers to demonstrate the inequality between *WI* and *WO*.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, friction, heat energy, pulley, work input, work output

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Saturday, January 7th, 2017
Last time we saw how the presence of *friction* reduces mechanical advantage in an engineering scenario utilizing* a compound pulley*. We also learned that the actual amount of effort, or force, required to lift an object is a combination of the portion of the force which is hampered by *friction* and an idealized scenario which is *friction*-free. Today we’ll begin our exploration into how *friction* *results* in reduced *work input*, manifested as *heat* energy *lost* to the environment. The net result is that *work* input does not equal *work output* and some of Mr. Toga’s labor is unproductive.
__Friction Results in Heat and Lost Work Within a Compound Pulley__
In a past blog, we showed how the actual force required to lift our urn is a combination of *F*, an ideal *friction*-free work effort by Mr. Toga, and *F*_{F} , the extra force he must exert to overcome *friction* present in the wheels,
*F*_{Actual} = *F + F*_{F} (1)
Mr. Toga is clearly *working* to lift his turn, and generally speaking his *work* effort, *WI*, is defined as the force he employs multiplied by the length, *d*, of rope that he pulls out of the compound pulley during lifting. Mathematically that is,
*WI = F*_{Actual} × *d* (2)
To see what happens when *friction* enters the picture, we’ll first substitute equation (1) into equation (2) to get *WI* in terms of *F* and *F*_{F},
*WI = *(*F + F*_{F} ) × *d* (3)
Multiplying through by *d*, equation (3) becomes,
*WI = *(*F ×* *d *)*+ *(*F*_{F} × *d*) (4)
In equation (4) *WI* is divided into two terms. Next time we’ll see how one of these terms is beneficial to our lifting scenario, while the other is not.
Copyright 2017 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineering, friction, heat energy, lost work, mechanical advantage, pulley, reduced work, work input, work output

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Tuesday, December 13th, 2016
The presence of *friction* in *mechanical *designs is as guaranteed as conflict in a good movie, and engineers inevitably must deal with the conflicts *friction* produces within their *mechanical* designs. But unlike a good movie, where conflict presents a positive, engaging force, *friction’s *presence in *pulleys* results only in impediment, wasting energy and *reducing* *mechanical advantage*. We’ll investigate the math behind this phenomenon in today’s blog.
__Friction Reduces Pulleys’ Mechanical Advantage__
A few blogs back we performed a work input-output analysis of an idealized situation in which no *friction* is present in a compound pulley. The analysis yielded this equation for *mechanical advantage*,
*MA = d*_{2} ÷* d*_{1} (1)
where *d*_{2} is the is the length of rope Mr. Toga extracts from the *pulley* in order to lift his urn a distance *d*_{1} above the ground. Engineers refer to this idealized frictionless scenario as an *ideal mechanical advantage*, *IMA*, so equation (1) becomes,
*IMA = d*_{2} ÷* d*_{1} (2)
We also learned that in the idealized situation *mechanical advantage* is the ratio of the urn’s weight force, *W,* to the force exerted by Mr. Toga, *F,* as shown in the following equation. See our past blog for a refresher on how this ratio is developed.
*IMA = W ÷** F* (3)
In reality, friction exists between a *pulley’s* moving parts, namely, its wheels and the rope threaded through them. In fact, the more *pulleys* we add, the more *friction *increases.
The actual amount of lifting force required to lift an object is a combination of *F*_{F }, the friction-filled force, and *F*, the idealized friction-free force. The result is *F*_{Actual} as shown here,
*F*_{Actual} = *F + F*_{F} (4)
The real world scenario in which *friction* is present is known within the engineering profession as *actual mechanical advantage*, *AMA, *which is equal to,
*AMA = W ÷** F*_{Actual} (5)
To see how *AMA* is affected by *friction* force *F*_{F}, let’s substitute equation (4) into equation (5),
*AMA = W ÷** *(*F + F*_{F}) (6)
With the presence of *F*_{F} in equation (6), *W* gets divided by the sum of *F *and *F*_{F} . This results in a smaller number than *IMA,* which was computed in equation (3). In other words, *friction* *reduces* the actual* ***mechanical advantage** of the compound *pulley*.
Next time we’ll see how the presence of *F*_{F} translates into lost work effort in the compound *pulley,* thus creating an inequality between the work input, *WI *and work output *WO*.
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: actual mechanical advantage, AMA, compound pulley, engineering, friction, friction force, ideal mechanical advantage, IMA, mechanical advantage, mechanical design, pulley, pulley friction, pulley work input, pulley work output, weight force

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Wednesday, November 30th, 2016
We’ve been discussing the *mechanical advantage* that *compound pulleys* provide to humans during lifting operations and last time we hit upon the fact that there comes a point of diminished return, a reality that engineers must negotiate in their *mechanical* designs. Today we’ll discuss one of the *undesirable tradeoffs* that results in a diminished return within a *compound pulley *arrangement when we compute the length of rope the Grecian man we’ve been following must grapple in order to lift his urn. What we’ll discover is a situation of *mechanical overkill* – like using a steamroller to squash a bug.
__Mechanical Overkill__
Just how much rope does Mr. Toga need to extract from our working example compound pulley to lift his urn two feet above the ground? To find out we’ll need to revisit the fact that the compound pulley is a work input-output device.
As presented in a past blog, the equations for work input, *WI*, and work output, *WO*, we’ll be using are,
*WI = F ×** d*_{2}
*WO = W ×** d*_{1}
Now, ideally, in a compound pulley no friction exists in the wheels to impede the rope’s movement, and that will be our scenario today. Our next blog will deal with the more complex situation where friction is present. So for our example today, with no friction present, work input equals output…
*WI = WO*
… and this fact allows us to develop an equation in terms of the rope length/distance factors in our *compound pulley* assembly, represented by *d*_{1} and *d*_{2}, …
*F ×** d*_{2} = W ×* d*_{1}
*d*_{2} ÷* d*_{1 }= W ÷* F*
Now, from our last blog we know that *W* divided by *F* represents the *mechanical* advantage, *MA,* to Mr. Toga of using the *compound pulley, *which was found to be 16, equivalent to the sections of rope directly supporting the urn. We’ll set the distance factors up in relation to *MA*, and the equation becomes…
*d*_{2} *÷* d_{1 }= MA
*d*_{2} = MA ×* d*_{1}
*d*_{2} = 16 *×** *2* feet = *32* feet*
What we discover is that in order to raise the urn 2 feet, our Grecian friend must manipulate 32 feet of rope – which would only make sense if he were lifting something far heavier than a 40 pound urn.
In reality, *WI* does not equal *WO, *due to the inevitable presence of friction. Next time we’ll see how friction affects the *mechanical* advantage in our *compound pulley.*
Copyright 2016 – Philip J. O’Keefe, PE
Engineering Expert Witness Blog
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Tags: compound pulley, engineer, force times distance, lift, mechanical advantage, mechanical design, pulley, rope length, work, work input, work output

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