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Zolo company bangaloreThank you for helping keep Eng-Tips Forums free from inappropriate posts. The Eng-Tips staff will check this out and take appropriate action. Click Here to join Eng-Tips and talk with other members! Already a Member? Join your peers on the Internet's largest technical engineering professional community. It's easy to join and it's free.

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Webpack export default imported as was not found inAlready a member? Close this window and log in. Are you an Engineering professional? Join Eng-Tips Forums! Join Us! By joining you are opting in to receive e-mail. Promoting, selling, recruiting, coursework and thesis posting is forbidden. Students Click Here. Related Projects. I'm trying to calculate the total system mass for a friction hoist. My question is in the conversion of the moment of inertia of a hoist drum and pulleys.

If a catalogue gives me the moment of inertia I, then would I be able to calculate the equivalent mass or mass required to turn the drum from rest using the inertia and radius of the drum? But since it has a different shape, how do I calculate this? Can you rephrase what you are trying to do?

It sounds like you want inertia, and they give you inertia. Or do you want mass? In that case, call the manufacturer? Yes I am looking for an equivalent mass. Is the formula I listed valid?

Timer contactor diagramBesides calling the manufacturer is there a way to compute this? The formula you listed being valid will depend on how accurate you need to be. You can get closer by breaking the shape up into segments and using the moment of inertia transfer formulas to get your total inertia.

Diablo 3 macros razerThen solve it for mass. Look up 'Inertia Transfer Formula', or 'Parallel Axis Theorum' An alternative would be to draw the drum in a 3d cad program solidworks, inventor, etc and then tweak the material density until the inertia matches the catalog value. Then read off the mass. Quote: I be able to calculate the equivalent mass or mass required to turn the drum from rest. Thanks, to verify, the mass that is determined is the minimum mass required to change the components velocity?

Or to start turning the drum from rest? That does not makes sense. Inertia cannot be converted to mass. An inertial system can be converted into a simpler equivalent inertial system only.Adding a gear set or gearbox between the motor and the load can improve the load-motor inertia ratio. Image credit: Omron Industrial Automation. In an attempt to control the load, the motor will draw higher current, which decreases efficiency and increases wear on the motor and electrical components.

Note that the inertia of the gear set or gearbox J G is added to the load inertia, but its effect is typically small compared to the reduction provided by the gear ratio. In addition to optimizing the load-motor inertia ratiogearboxes are often used in motion control applications to increase the torque delivered to the load from the motor, but they also decrease the rotational speed delivered to the driven component from the motor, by an amount equal to the gear ratio.

Although it is possible to use a gearbox or gear set that is configured to reduce torque and increase speedin motion control applications, a more typical solution would be to choose a smaller motor. You must be logged in to post a comment. Leave a Reply Cancel reply You must be logged in to post a comment.Mechanical power transmission between shafts can be done in several ways.

In most engineering practice, major power transmission used is gear system, belt drive, chain or rope drive. However, among these, gear system is the most efficient. Gear system is efficient because it can perform high consistency of connection to produce high speed and load transfer with minimal noise of operation. It is analogical with belt drive system because belt drive is a flexible, easy to install way of power transmission mode.

The efficiency can be affected by belt drive slip, centrifugal effect and creep factor. Gears are used in many machines such as metal cutting machine tools, automobiles, hoists, rolling mill and so on. The function of gear is to transmit mechanical power from one shaft to another shaft with a certain speed ratio.

Gear system parts include at least a set of gear that consist of Driver Gear and Driven Gear. Driver Gear is the gear that actuates power while Driven Gear is the gear that receives the power.

A series of gear set is called Gear Train. Gear can be classified according to the relative position of the axes of mating gears. Example of this type of gear is Spur Gears and Helical Gears. Example of gear is bevel gears. Some of the important terminology of gear system is: Pitch Circle : An imaginary circle which by pure rolling action, would produce the same motion as the toothed gear wheel.

Circular Pitch : The distance measured along the circumference of the pitch circle from a point on one tooth to the corresponding point on the adjacent tooth. Addendum Circle : Circle that limits the top of the teeth. Driven gear Driver gear.

## Equivalent Moment of Inertia

Gear ratio n is defined as ratio of speed of driven gear with the speed of driver gear. Combination of gear wheels by means of which motion is transmitted from one shaft to another shaft is called Gear Train.

In simple gear train, each shaft carries one gear only.Different parts of a load may be coupled through different mechanisms, such as gears, V-belts and crankshaft. These parts may have different speeds and different types of Motions such as rotational and translational. If the losses in transmission are neglected, then the kinetic energy due to equivalent inertia must be the same as kinetic energy of various moving parts.

Power at the loads and motor must be the same. If in addition to load directly coupled to the motor with inertia J 0 there are m other loads with moment of inertias J 1J 2. If m loads with torques T l1T l2. If loads are driven through a belt drive instead of gears, then, neglecting slippage, the equivalent inertia and torque can be obtained from Eqs.

Let us consider a motor driving two loads, one coupled directly to its shaft and other through a transmission system converting rotational motion to linear motion Fig. If the transmission losses are neglected, then kinetic energy due to equivalent inertia J must be the same as kinetic energy of various moving parts. Moment of inertia can be calculated if dimensions and weights of various parts of the load and Motor Design Parameters are known.

Frankenstein f series headsIt can also be measured experimentally by retardation test. In retardation test, the drive is run at a speed slightly higher than rated speed and then the supply to it is cut off. Drive continues to run due to kinetic energy stored in it and decelerates due to rotational mechanical losses. Variation of speed with time is recorded.

Now drive is reconnected to the supply and run at rated speed and rotational mechanical power input to the drive is measured. This is approximately equal to P. Now J can be calculated from Eq. Main problem in this method is that rotational mechanical losses cannot be measured accurately because core losses and rotational mechanical losses cannot be separated. In view of this, retardation test on a dc separately excited motor or a synchronous motor is carried out with field on.

Now core loss is included in the rotational loss, which is now obtained as a difference of armature power input and armature copper loss. In case of a wound rotor induction motor, retardation test can be carried out by keeping the stator supply and opening the rotor winding connection.

J can be determined more accurately by obtaining speed time curve from the retardation test as above and also rotational losses vs speed plot as shown in Fig. Using these two plots, rotational losses vs time plot can be obtained, e. Then for this speed rotational loss P 1 is obtained from the plot of rotational loss vs speed and plotted against t 1.

### Inertia Balance in a Gearbox

Area A enclosed between the rotational loss vs t plot and the time axis shaded areais the kinetic energy dissipated during retardation test. December 22, Paragraph Text.Update Reset Print.

Enter number data in appropriate fields - calculated results in RED. Motor torque and speed are NOT at absolute max values, but rather at max efficiency. Nm Motors' rotary inertia, Jmotor kg-m 2g-mm 2.

If known, enter motor inertia else, enter "0"Jm kg-m 2g-mm 2. Enter motor inertia in g-mm 2 if different than above. Max motor torquegammax m-N-m, N-m. Transmission Planetary or known transmisssion ratio and inertia. Planet carrier inner diameter, Dpid mm, m.

Planetary total rotary inertia, Jplanets kg-m 2. Output shaft rotary inertia, Jouts kg-m 2. Total Nm planetary transmissions' rotary inertia, Jtrans kg-m 2. Transmisssion efficiency includes car wheelsetatrans. If known, enter transmission inertia else, enter "0"Jt kg-m 2g-mm 2.

If known, enter transmission efficiency, etat else enter "0". Coefficient of friction wheel-to-ground, mu static friction coefficient??

Optimal Transmission ratio by Matched Inertia Doctrine. Actual equivelent linear inertia of motor and tranny, mtrans kg. Total actual sysetm equivelent inertia, Mtotal kg. Triangular Velocity Profile Motion Results. The wheels will slip if the static traction force is less than the maximum motor traction force. Time to accelerate to speed at max motor h theoreticaltaccel seconds.

## Gears and Systems with both Rotation and Translation

Distance travelled during acceleration to vmaxeff, Xaccel m. Time to travel the desired travel distance full speed at end of move.Hot Threads. Featured Threads. Log in Register. Search titles only. Search Advanced searchâ€¦. Log in. Support PF! Buy your school textbooks, materials and every day products Here! JavaScript is disabled.

For a better experience, please enable JavaScript in your browser before proceeding. Finding equivalent mass moment of inertia for gear system. Thread starter immel Start date Jan 22, Homework Statement Two masses, having mass moments of inertia J1 and J2, are placed on rotating rigid shafts that are connected by gears. Homework Helper. Thanks for the help tiny-tim. I am confused about what variable I am solving for. In the problem statement " I've come up with a solution, but using the equivalent kinetic energy method.

Thanks tiny-tim! Much appreciated. I know this is about 2 years later but in case anyone comes across this looking for help I wanted to correct an error I noticed in the calculations.

If you think about it less teeth equals smaller diameter and when a constant torque is applied the smaller gear and shaft coupled to it will spin faster than the gear with more teeth.Gears perform many functions, in this section we look at gears that increase or reduce angular velocity while simultaneously decreasing or increasing torque, such that energy is conserved. In many ways gears act in rotating systems as do levers in translating systems.

### Finding equivalent mass moment of inertia for gear system

A picture of gears is shown below, along with a schematic representation. We will specify gears by their radii. When you buy gears you normally specify the number of teeth and the number of teeth per inch or centimeter. These two measures are equivalent: the number of teeth and teeth per inch give the circumference, which in turn determines the radius. If the number of teeth are doubled, the radius is doubled.

If we consider two gears in equilibrium and in contact with each other, we can two very useful relationships. First we note the geometric relationship that results from the path that the arc lengths along their circumference must be equal as the gears turn.

We derive a second relationship from a torque balance. Before we can do so we must define a force between the gears termed a "contact force. We start by drawing free body diagrams with a contact force where the gears meet. The contact force is tangent to both gears and so produces a torque that is equal to the radius times the force. We are not usually interested in f cso we remove from the equations and we get. It is easy to show that the same relationship exists even if the contact forces are defined in the opposite direction up on gear 1, down on gear 2.

It, in turn, is connected to gear 2 with moment of inertia J 2 and a rotational friction B r. We start by drawing free body diagrams, including a contact force that we will arbitrarily choose to be down on J 1 and up on J 2. The directions of the reaction forces due to inertia and friction are chosen, as always, opposite to the defined positive direction. We convert the force to a torque note: we could have skipped the previous step and done this directly.

When you use the arc length expression you must be careful of signs. The final result from the example is important deserves some notice. The multiplication of J 2 and B r by the square of the ratio of the radii is a well known, and general phenomenon. If a load is attached to a system through gears, the perceived effect is multiplied by the square of the ratio of the radii.

We can use this to our advantage to make a load appear smaller if we have a weak driving torque - for example, a small motor. You will see on the next web page that the product of angular velocity and torque is power. So the last equation simply states that power is preserved across a set of gears.

**Reduced Mass Moment of Inertia (Gears) Excel Sheet - Spreadsheet Download**

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