Tests Performed on Gearbox




We are pleased to present a new type of gearbox comprising a unique gear-set especially designed to achieve a superior efficiency over the common gear trains. Its application can be widely used, either to reduce energy consumption in systems using gear trains or to increase electricity-generating capacity in any types of power plants. When the gearbox operates at a speed ratio of 1: 1, the power put into the unit is amplify by 25% when measured at output.

It might seem to contradict the law of conservation of energy which states that the power put into a system must equal the power coming out, yet the same outcome is prevailed when the gearbox' operation is inverted.

Either reversing the rotation of the applied power or when interchanging the gearbox's supplied power from the Drive side to the Driven side results in a 25% power loss. Therefore, the overall power put into  equals the overall power coming out of the system. 

The formulated combination of the angles at which the gear-sets operates provides an unbalanced distribution of the exerted force put into the unit. When applying a clockwise force at the Drive side of the gear-set, the special combination of the operating angles between the two meshing gears amplifies the exerted power by 25%. However, when reversing the rotation  to a counter clock-wise supplied force, the operating angle now have the reverse effect and a 25% power loss occurs.  These angles are obtain by using different tooth shapes and different circular pitch on each gear of the gear-set. Unlike any common gear-sets where they use the same pitch and tooth shape  on each gears thus resulting in a balance or an even distribution of the applied force regardless of the rotation it is used (explained in details bellow). To explain how it works we employ real life tests and compare the Common to the Custom gearbox. Engineers have already approved one of the prototypes shown bellow, contact us to view the certificates. 

The models discussed here are patent pending and cannot be mass-produced or replicated without our  consent.

 


                                                                                           CHAPTER 1 - TERMINOLOGY                  


 Before understanding how it operates, a few terms have to be reviewed.   

1- What is MECHANICAL ADVANTAGE (MA)?   

Mechanical advantage is a measure of the force amplification achieved by using a tool, mechanical device or machine system. The device preserves the input power and simply trades o forces against movement to obtain a desired amplification in the output force. The model for this is the law of the lever. Machine components designed to manage forces and movement in this way are called mechanisms. An ideal mechanism transmits power without adding to or subtracting from it. This means the ideal mechanism does not include a power source, is friction less, and is constructed from rigid bodies that do not deflect or wear. The performance of a real system relative to this ideal is expressed in terms of efficiency factors that take into account departures from the ideal.  


2- THE LAW OF THE LEVER THE LEVER MODEL DRAWING;


THE FORMULA USED TO CONFIGURE THE MA OF A SYSTEM;


The lever is a movable bar that pivots on a fulcrum attached to or positioned on or across a fixed point. The lever operates by applying forces at different distances from the fulcrum, or pivot. As the lever pivots on the fulcrum, points farther from this pivot move faster than points closer to the pivot. The power into and out of the lever is the same, so must come out the same when calculations are being done. Power is the product of force and velocity, so forces applied to points farther from the pivot must be less than when applied to points closer in. If a and b are distances from the fulcrum to points A and B and if force FA applied to A is the input force and FB exerted at B is the output, the ratio of the velocities of points A and B is given by a/b, so the ratio of the output force to the input force, or mechanical advantage can be known.   

For the common gear-set, the MA can be simply configured by knowing the speed ratio of the gearbox (the number of the Drive shaft revolution VS the number of the Driven shaft revolution). -If the ratio is 1: 1, Ma = 1                                                                                                                                                         

-If the ratio is 2: 1;

MA = 2                                                                                                                                                        

-If the ratio is 1: 2 (the Drive shaft = 1 turns and the Driven shaft = 2 turns);

MA = 0.5

 Another method to configure the system’s MA is using the gear’s radius (as per lever law). This is the method used to calculate the MA of the our Modified gear trains since we need to consider all the functions that the system provides such as forward or reversed rotation and the point where the power is supplied into the system The reason the use of the speed ratio method to find the system’s MA is not accurate in the Modified gearboxes is that the Customs systems are using a power imbalanced method when considering only one of the four ways the system can be used. However, when considering every ways, the power put into and coming out of the system returns to a balance state (overall efficiency of 100%). Unlike the Common gear-sets where all the functions are balanced thus the resulting in the same performance regardless of the rotation or the point where the power is supplied into the system.  


3- The CIRCULAR PITCH of a gear is the distance in between the teeth. 



To have proper meshing between two common gears, the distance between the teeth (circular pitch) has to be the same on both gears (as document bellow states).



4- To calculate the efficiency;


The theoretic efficiency of any power transmitting systems such as a lever system, a tackle and block, a hydraulic jack or a gearbox system must equal 100% without including any friction generated within the unit. Other values than joules can used in the formula such as Watts or Horse Power (HP).


On some of the tests performed below, we will use the Watts and the Horse Power value to calculate the efficiency due to the values used in the various tests performed on the gear-trains.
To calculate the Horse power; 



The speed ratios of the gear trains are using the RPM values of the Drive and the Driven shafts;since a 1:1 speed ratio gear-set provides the Drive and Driven shafts to rotate at the same speed.  In all the examples below,  all the systems are suing a 1: 1 speed ratio which makes the systems easier to be compared.   To calculate Watts; 

                   Watts = Volts x Amps. We use the Watts to know how much power the electric Motor draws VS how much power is produced by the generator.  The result is the real system’s Efficiency of the gear-train.


5- Note that in this document;

- C.P refers to the Circular pitch.

 -V.D refers to the Vertical distance measured between two teeth on a gear.

 - MA refers the Mechanical Advantage value.



                                                                              CHAPTER 2 - THE COMMON GEAR-TRAIN


 To fully understand how the Modified gearbox operates, we first need to understand the basic principle of operation of the common gearbox. It is well known that a gear-set having a ratio of 1 to 1 has an MA of 1 and the theoretic efficiency equals 100% in any ways it can operate. For example if we apply 100 Ft/lb on the Drive (Input) shaft of a common gear-set having a ratio of 1: 1, a 100 Ft/lb (less friction) will be transferred to the Driven (Output) shaft and results in a 100% efficiency (as per formulas above).

When the rotation of the shafts are reversed, or if the power put into the gearbox is now applied on Driven shaft instead of the Drive shaft, the efficiency remains 100%, Therefore the overall efficiency of the whole gearbox is 100% (as demonstrates the next test).   

Although when using our the Modified gear-trains a change in efficiency results when those aspects are considered. Therefore we need to consider all of the functions (rotation or point where the power is supplied) when configuring the overall efficiency which also always results to 100% theoretically.

 According to encyclopedias the different functions that a system provides are calculated as separate systems and since the common gear-trains (and any other mechanisms) have the same efficiency regardless of the rotation or where the power is applied into the unit, the theory is accurate. Per example if we use a lever having a 10" length on one side and 20" on the other side of the pivot point (fulcrum), and when utilizing one side of the lever as the Input (the point where the force is applied) is considered as one system and when utilizing the other side of the same lever, it is considered as a completely different system since the Mechanical Advantage is not the same, therefore the efficiency is also calculated separately. Although it is obvious that it is the same lever that is used and that the efficiency should be calculated together as one unit, but they calculate them like this; 

 1- If the weakest side of the lever is used (using the shortest lever side as the point where the power is supplied);

the Ma = 0.5 and the efficiency = 100% since the Input point travels half the distance of the Output point.                                        


2- If the strongest side of the lever is used (the longest lever side used as the point where the power is applied);

the MA = 2 and the efficiency = 100% since the Input point travels twice the distance of the Output point.


The lever is only one physical unit and the two functions provided should be considered as one system and not as two different systems. Note that the same applies for gear-trains.   

To demonstrate that the Custom gearbox has a better efficiency than the Common gearbox, we will first start by performing a test on the common gearbox and then we will perform the same test on the Modified gearbox and compare the results. The next video depicts two gear-sets coupled by means of a rigid coupling. To obtain the theoretic efficiency of the system, the torque wrench test method is performed. It consists of installing a torque wrench at the Drive (Input) shaft and installing another torque wrench at the Driven (Output) shaft. Then a force (in Ft/lb) is applied to the system by adjusting the nuts on the threaded rods which are connecting the end of the torque wrenches to the frame of the gearbox. Readings of both torque wrench gauges are recorded to configure the Mechanical Advantage (MA) of the system.  

 

Please watch the video #5 on top of the page.



The readings of the video torque wrench's gauges are; 

- The torque wrench on the Drive shaft reads 10 Ft/lb.

-The torque wrench on the Driven shaft reads 10 Ft/lb. To calculate the HP supplied to the Input; = 10 Ft/lb x 1 RPM / 5252 (as per formula in Chapter 1)

                             = 0.002 HP (RPM is 1 since gear-sets ratio is 1:1) To calculate the HP coming out of the Output; = 10 Ft/lb x 1 RPM / 5252

                             = 0.002 HP To calculate efficiency; = Work Out / 

Work IN x 100 (as per formula in Chapter 1)

                             = 0.002 HP / 0.002 HP x 100

                             = 100%  

Note that since the gears are not turning, no friction is generated thus the result is a theoretical or an ideal value.  A dynamo-meter test needs to be performed in order to configure the efficiency considering the friction generated within the system and is the next test performed.     

If the operation of the gearbox above is reversed by interchanging the point where the power is supplied from the Input (Drive) shaft to the Output (Driven) shaft, the test demonstrate that a balanced force in between the Input and Output shafts is achieved. The efficiency will remain the same and the overall performance of the unit is 100%. The results might seemed obvious but as you will see it is different when using the modified gear-trains.   


Next is a demonstration of the Common gearbox dynamo-meter test; 


The Dynamo-meter test consists of measuring the power put into and coming out of a system at a given system RPM. The system comprises an electric motor coupled to a conventional gearbox's Input shaft and its Output shaft connects to a generator. Then a Volt meter and an Amp meter are connected to the Motor and a separate meter set is connected to the generator. They are used to measure how much power (Watts) the motor draws VS how much power (Watts) the generator produces at a given system's RPM. The RPM monitors measures the speed of the generator. The results are configured and provides the real efficiency of the system using the common gearbox. In Chapter's 3 tests we will use the same Motor and Generator but the Common gearbox is replaced with the Modified gearbox. Then we will compare both results to calculate the efficiency of the Modified gearbox.   

The gearbox has a ratio of 1:1 and consists of an Input gear (Drive), and an Output gear (Driven) (see drawing below). For that reason we connected two common gearbox together since the assembly is to be compared with the Modified gearbox which consists of 2 sets of gears. 

The Common Gearbox internal drawing;



The procedure utilized to perform the dynamo-meter test is fairly simple. We only want to know how much power the motor draws vs how much power the generator supplies at a given RPM. We use the Volts and the Amps that the Motor draws and compare them with the Volts and Amps produced by the generator. With the Volts and Amps values, we can configure the Watts which is the measure for power.

It can be seen in the video that the electric meters are individually tagged to show the value that they are measuring. Per example one meter measures the Volts of the Generator. The RPM meter is held by hand on the generator's and motor's shaft. We utilize a 120 Volts ac / 10 Amps electric motor, and a 120 Volts ac / 30 Amps generator. Note that no meter is connected to the motor to measure the Volts as it is connected to a household plug supplying a constant 120 Volts. 

 Please watch the Video #2 on top of the page


Table's Parameters explanation; 

- "NO LOAD" = There is no load on the generator (the light bulbs are turned "OFF"). 

- “6 Bulbs" = There is a load on the generator, the light switch is turned "ON" and the six 100 Watts (ea) light bulbs are lit.

 - "M" = Motor 2 HP @ 1750 RPM

 - "G" = Generator 120 VAC\ 30 Amps 

- "RPM" = Speed

 - "Gratio" = The gearbox's speed ratio.




To calculate the Efficiency, we need to calculate how many Watts the Motor is drawing and compare it to the generator supplied Watts when they both are under load. We also have to consider the RPM of the motor and the generator in order to fairly compare this version with the Modified gearbox (Next Chapter) since varying the speed ratio between the motor and the generator acts the Mechanical advantage of the system.     

-The Motor draws; 120 Vac x 13.95 Amps = 1674 Watts 

-The generator produces; 63.5 VAC x 3.1 Amps = 200.2 Watts

- The System's Efficiency = Power Out \ Power In x 100

= 200.2 \ 1674 = 11.9 %  

The same test is performed on the Modified gearbox, with the results we will be able to determine its efficiency since we know the one of this Common gearbox. Note that the 11.9 % is the whole system efficiency, the gearbox’s efficiency is stated in the manufactures specs and is discussed in the next chapter.




 CHAPTER  3

 THE MODIFIED SPUR TYPE  GEAR-TRAIN



The types of gear-sets that we offer transfers the power supplied to the unit and coming out differently than the common gear-sets. Instead of having the exerted force distributed evenly through one operable function, the applied force is distributed in a non-balanced manner between two operable functions of the unit, such as in the next prototype example;

 

Applying 100 Ft/lb on the Drive (Input) shaft of the modified gearbox (speed ratio of 1 to 1) results in 125 Ft/lb transferred to the Driven (Output) shaft. Although when we apply 100 Ft/lb to the Driven (Output) shaft instead of the Drive (Input) shaft, only 80 Ft/lb is transferred to the Drive (Input) shaft resulting in 100% overall efficiency.  The "meshing stroke" is the amount of degrees rotated by the gears occurring between the points where the Drive gear's tooth engages the Driven gear's tooth, to the point where the next Drive gear's tooth starts to engage the next Driven gear's tooth (also refers to the lower and upper operating angles). In a gear-set having 10 teeth on each gears, the meshing stroke is 36 degrees (where there is 36 degrees between the lower and upper operating angles). The power applied into the Modified gear-sets doesn't balance out within the meshing stroke (like the common gear-sets), rather the power balances when its operable function is reversed. Although its disadvantage over the common gear-set is that when its operation is reversed, the efficiency is proportionally less The next video demonstrates our latest design. The modified gearbox comprises an Input gear where its tooth engages the First gear’s tooth and forms the first gear-set. The First and the Second gears are fabricated as one unit, just like a common double-sided crown gear and it is used as an idler gear. The Second gear’s tooth engages the Output gear’s tooth and forms the second gear-set. The Input and Output gear have 10 tooth each.

 The First and the second gear have 32 tooth each. This combination provides a ratio of 1:1.  

Please watch the video#4 on top of the page and also watch the video #8 (on top of the page) which uses digital torque wrenches and provides more accurate readings.




To calculate MA using the video torque wrench's gauges reading values in the video; At the start of the meshing stroke; 

- The gauge on torque wrench installed on the Input shaft reads = 10 FT/lb                                                                   

 - The gauge on the torque wrench installed on the Output shaft reads = 12.5     

 MA = 12.5 / 10 = 1.25   

When we apply 100 Ft/lb at the Input shaft, an average of 125 lb is transferred to the Output shaft since the ratio of this gearbox is 1: 1 (as shown in the video), the efficiency is 125% We can now compare the torque wrench test of the Common gearbox performed above to the Modified gearbox. We can see that at same speed ratio, the gearboxes have different MA.


Next is the dynamo-meter test performed on the Modified Gearbox;

 



Please watch video #1 at the bottom of the page.

The values were recorded on the table below;



 

To calculate the Efficiency, we need to calculate how many Watts the Motor is drawing and compare it to the generator supplied Watts when they are both under load. We also have to consider the RPM of the motor and the generator in order to fairly compare this version with the Common gearbox since varying the speed ratio between the motor and the generator changes the Mechanical advantage of the system.

 - The Motor draws; 120 Vac x 12.5 Amps = 1500 Watts 

- The generator produces; 64.3 VAC x 3.5 Amps = 225.1 Watts 

- The system's Efficiency = Power Out \ Power In x 100                                         

                                             = 225.1 \ 1500 = 15 %  

The common gearbox is 11.9 % and the Custom gearbox is 15 % Efficient.  

 We also need to consider that the common gearbox isn't 100% efficient. The specifications on the sticker on the common gearbox, if 3.675 Hp is applied on the input, we will get 125 in/lb. @ 1750 RPM. 

 


-Torque = 5252 x HP / RPM

= 5252 x 3.675 HP / 1750                                                                                                                 

 = 11 Ft/lb. x 12  = 132 in/lb. 


To determine the efficiency using" Force out / Force in"

  = 125 / 132 = 94.6 % efficient 

On the common gearbox test we coupled two common gearbox together, therefore;

= 94.6 + 94.6 / 2

= 94.6 and 100 - 94.6

= 6.4.

To find out the real efficiency of the Modified gearbox;

= 126.1% - 6.4%

= 119.7 % efficient.   



We can even compare the Modified gearbox's dynamo-meter test with the Direct version's dynamo-meter test which simply consists of using the same motor connected to the same generator by means of a belt (without including a gearbox in the system). 


 Please watch video#3 at the bottom of the page.   



To calculate the Efficiency, we need to calculate how many watts the Motor is drawing and compare it to the generator supplied Watts while they are under load. 

-The Motor draws; 120 Vac x 11.55 Amps

= 1386 Watts.


- The generator produces; 63 .2 VAC x 2.75 Amps

= 138.6 Watts.


- The Efficiency = Power Out \ Power In x 100

= 136.8 \ 1386

= 12.5 %  

To compare with the Modified gearbox;

= 15% / 12.5%

= 120% efficient.

We can see that by comparing with either version (Common gearbox or Direct version) that the Modified gearbox is around 120% in both version.     


  CHAPTER 4 - THE CAGE STYLE GEAR BOX    


 The next video demonstrates the first prototype gearbox that we designed and is 107% efficient including the system’s friction. Continued researches and developments provided us with much more efficient models. We apologize for the poor model and video quality but it still demonstrates the results obtained when using our gearbox.

 The first part of the next video demonstrates the Direct version which consists of an electric motor coupled to a generator by means of a belt. This part is to configure how much power is taken by the motor VS how much power the generator will produce when the motor is coupled directly to the generator. Note that the efficiency is rather low; the motor takes 8.6 Amps @ 120 V/AC and the generator supplies 2.9 Amps @ 42 V/AC. The efficiency is then around 11.8%. This is normal and it is mainly due that we utilize 120 volts 1 phase motor and generator which are known to be around 30% efficient each therefore = 15% if coupled together plus the friction of the belt drive.

  Since we now know the efficiency when using the motor directly coupled to the generator "The direct-version", we can compare it with "The gearbox-version". Normally including a gearbox into the system will only create more load and friction and will result in a decrease of efficiency in The system. The difference in efficiency is what we want to measure in order to know the real efficiency (including friction) of the cage style custom gearbox; we simply subtract the efficiency of the "Gearbox" version to the efficiency of the "Direct" version. 


The second part of the video depicts the Gearbox version which consists of an electric motor, a gearbox, a generator. The same electric motor is coupled to the drive (Input gear's) shaft of the gearbox by means of belts. The Driven (Output gear's) shaft is connected to a generator also by means of belts. Note that the same motor, generator and the same length of belts are utilized in both versions. The gearbox has a ratio of 1:1 which is used to fairly compare to the "Direct" version since it doesn't change the speed (RPM) of the generator. The gearbox is an open-gearbox model that consists of an Input gear (Drive), a First gear (idler gear 1), a Second gear (idler #2) and an Output gear (Driven).

It utilizes cage style gears that are no longer used in the industry but are the easiest types of gears to fabricate. The cage style gears are also very efficient ranging from 98 to 99% efficient compared to the spur gear types that are from 95 to 98% efficient. Note that all the gears have different circular pitch unlike any other gearbox available (view drawing bellow for more info).

 Please watch the video #7 at the bottom of the page.   


The video demonstrates the procedure utilized to perform the dynamo-meter test where the goal is to determine and compare the real efficiency between the "direct" and the "gearbox" version including the system friction. 

- The yellow volt meter gauge is connected to the AC input of the motor and is set on AC amps.

- The green and yellow volt meter gauge is connected to the AC output of the generator and is also set on AC amps.

- The blue volt meter gauge is connected to output of the generator and is set on AC volts.

- The RPM meter is held by hand on the output generator shaft. We utilize a 120 Vac / 10 Amps electric motor, and a 120 Vac / 30 Amps generator.

Note that no gauge is connected to the motor to measure the Volts as it is connected to a household plug supplying a constant 120 Volts. 

In the first part of the video "The direct version"; with the motor and the generator running with no load ("NO LOAD" section on parameters chart).

- The motor draws 6.1 Amps and the generator turns at 1507 RPM.

When the switch is turned "ON", the six light bulbs are lit and the motor takes 8.6 amps;

- The generator turns at 1254 RPM and delivers 42 volts AC and 2.9 amps.

 In the second part of the video "The gearbox version"; With the motor, the gearbox, and the generator running with no load;

- The motor draws 6.1 amps and the generator turns at 1530 RPM (difference with the direct version is explained below).

When the switch is turned "ON", the six 100-watt light bulbs are lit;

- The motor draws 8.3 amps.

- The generator produces 42 volts and 3.0 amps and turns at 1254 RPM. 

The values are recorded and put on a parameter sheet (below); 

The efficiency table of various gear types;


Parameters  explanation;

 - Direct version = The electric motor connected directly by means of a belt to the generator (Alternator). 

- Gearbox version = The electric motor is connected by means of a belt to the drive shaft of the gearbox.

The driven shaft of the gearbox is connected by means of a belt to the generator. (Note that the same motor and generator are used in both scenarios and we utilize a diameter adjustable sheave on the electric motor. Also note that the gearbox has a speed ratio of 1 to 1) -The section "NO LOAD" = There is no load on the generator (the light bulbs are turned "OFF"). It demonstrate that in both the direct and the gearbox version we starts of at the same motor amperage. 

- The section "WITH LOAD" = There is a load on the generator, the light switch is turned "ON" and the six 100 Watts light bulbs are lit.   

Explanation of why we do not have the same RPM's on both versions in the "NO LOAD" section; In the "NO LOAD" section on the parameter chart above, the generator's RPM is higher on the "Gearbox" version than on the "Direct" version when the light bulbs are "OFF". This is due to the fact that we need to compare both version’s efficiency when the light bulbs are lit ("WITH LOAD" section) and NOT when they are turned "OFF" since efficiency equals power IN / power OUT and we do not have any power OUT when the light bulbs are not lit, we can't make the calculation without any power value. It is by using those measurements that we calculate and compare the efficiency of both systems and figure out if it is advantageous to use the gearbox over the Direct way. 


To calculate the efficiency of the custom spur type gearbox using the dynamo-meter test; 

 -Power IN "Direct" version; Current draw from the motor: 8.6 Amps x 120 Volts/AC motor voltage = 1032 Watts  

-Power OUT "Direct" version;  2.9 Amps is drawn from the generator to supply the light the bulbs. 42 Volts/AC x 2.9 Amps = 121.8 Watts.  

 -Power IN "Gearbox" version; Current draw from the motor = 8.3 Amps x 120 Volts/AC = 996 Watts

-Power OUT "Gearbox" version; 42 Volts/AC of output from the generator x 3.0 Amps that supplies the light bulbs = 126 Watts. 

To calculate the final efficiency; For the motor point of view;

Divide 1032 by 996 = 1.036 x 100 = 103.6%. We gain 3.6% on the power usage of the motor when using the "Gearbox" version over the "Direct" version.

 

For the generator point of view;

Divide 126 watts by 121.8 = 103.5%.

We gain 3.5% more power when using the “Gearbox" over the "Direct" version.


 For the total Power gain;

Add 103.6 + 103.5 = 107.1% or 7.1% more efficient than the direct version.  

This demonstrates that the gearbox is 107 % efficient. Note that the motor still takes more power than the generator supplies and that the efficiency of the motor and generator is not 107.1%. This also means that if we were to add this gearbox in any power plants using a shaft to connect the turbine to the generator, we would increase their power generating capacity. Or if we were to replace the common gearbox with this gearbox in conveyor system or in a drive terrain of a vehicle, a saving in power consumption will result.   

The efficiency can be geometrically calculated  (please contact us for more details).

  

 

Now we will view the torque wrench test on the Custom Cage type gearbox.

Note that the gearbox has been modified from the model disused due to development of the model.    

Please watch the video #6 at the bottom of the page  

This method makes it very easy to calculate the efficiency since we know the ratio is 1 to 1; simply divide 38 FT/LB for the Output reading by 30 FT/LB for the Input reading

= 1.2666 x 100 = 126 % efficient.

Note that if we use the Output' shaft as the Drive shaft where the force is applied to the system, then the efficiency will only be 73.33%. Therefore the overall efficiency is 100% geometrically calculated.

We can use the same coefficient as the above dynamo meter to figure out the real efficiency since it is the exact same gearbox. Only the angle of operation and the circular pitch of the input gear has been modified;

= 126.666 - 5.34 = 121.32% real efficiency.

 


 


The efficiency table of various gear types;

This is just to demonstrate that gears are very efficient, the worst type is the worm gear set. As you can see in the drawing bellow the drive gear has the shape of a screw resulting in a lower efficiency due to that the power applied to drive gear is transmitted to the drive gear in a 100% sliding manner. The modified gear set gear-set has a sliding effect percentage proportional to the common spur gear set. Therefore the loss due to friction will be similar to a spur gear type since the gear's shaft are installed parallel to each other. 





 CHAPTER 5 - MODELS TESTED USING DIGITAL GAUGES FOR EVERY VALUES MEASURED

Still not convinced, please watch the next two videos, the same dynamo-meter tests are performed on the same gearboxes (the Motion X and the Common gearbox), but to be more precise, we use all digital meters. Note that the generator RPM is lower than in the tests performed above and results in more Amps being drawn by the motor.

 

Please watch video #9 above, the Motion X gearbox dynamo with digital gauges.


The table bellow shows the values of each measurement values taken from the video.

                                                     

  


Now we can calculate how much Watts the motor draws and the generator produces.

Motor Watts = 13.5 Amps x 120 Volts

                      = 1620 Watts


Generator Watts = 55.2 Amps x 3.6 Volts

                              = 198.72 Watts


The whole system  Efficiency = 198.72 / 1620

                                                    = 12.3%



In order to confirm the gearboxes efficiency we need to compare it to the Common gearbox at the same RPM,

Please watch video # 10 above.

The table bellow shows the value of each measurement values taken from the video


We can now calculate the whole system efficiency;

The motor Watts drawn = 16.7 Amps x 120 Volts

                                            = 2004 Watts


The generator Watts produced = 57.2 Amps x 3.7 Volts

                                                         = 211.1 Watts


The whole system Efficiency = 211.1 / 2004

                                                    = 10.4 %


To know the Motion X gearbox efficiency, we need to compare it with the Common gearbox;

Efficiency = 12.35% / 10.4%

                   = 119%

By using all digital meters, it definitely confirms that the Motion X gearbox is over efficient.



CHAPTER 6 -TO CALCULATE THE CUSTOM GEARBOX USING A FORMULA

We first need to review some aspects which are used in the formula;

1A- The push force (line) angle effect on the Load of the system (can also be called the Resting angle and relates to the pressure angle);

The example bellow utilizes a lever system to explain the effect of the push force angle on the Load (weight attached on the end of the lever). 

As per test performed with the help of a lever, a 103 grams weight and a scale, the Resting angle (or push angle effect on the Output weight) was configured performing live tests which results are as per drawing bellow.

Drawing Resting angle effect on a weight.

 

The drawing depicts a lever system where the fulcrum point is depicted by the red circle in the center, a lever is represented by the different colors line (to show different lever orientation) and a 103 grams weight shown by the circles at the lever end where their color matches the lever color.

To perform the test of the Resting angle effect on the weight, a scale is held underneath the weight (attached to the lever end) at various lever angles, then the scale reading is recorded.

As it can be seen on the drawing the closer to the vertical the lever and weight are positioned, the less force is needed to lift the weight (103 grams). At vertical, the weight is completely resting on the fulcrum and the scale reads 0 grams. At horizontal, the whole weight isn’t resting on the fulcrum at all and the scale reads 103 grams.

The Sin of the lever angle (measured from the vertical axis) has the same result than in the test performed. When the weight and lever are at vertical the scale reading = 0 grams and the sin of 0 degrees = 0. At horizontal the scale reads 103 grams, and the Sin of 90 degrees = 1. Therefore the Sine of the lever angle x 100 = the % of the weight that needs to be supported.

Per example, when the lever and the weight are at 30 degrees (from vertical), the scale reading = 51.5 grams. The Sin of 30 degrees = 0.5 x 100 = 50% of the weight is required to balance with 103 grams weight.

As it can be seen, a mechanical advantage occurs when this aspect is present in the system as the force needed to counteract (hold the weight in place) is only a % of the weight where it is less than the weight itself.

This principle also applies to gears but only on the Drive gear of the gear-set. Imagine that the Drive gear isn’t turning another gear but instead that it is lifting a weight. Or imagine that a weight is attach on the Driven gear's tooth, the closer the operating angle is from the Drive gear's vertical axis the less force is required by the Drive gear in order to lift it (or to be in a balance state).

Note that this is only accurate if the force pushing down on the weight s performed in a vertically manner such as the gravity force of the weight. Therefore the same applies if instead of a weight, a vertical force such as pushing with your hand (in a vertical manner) on the lever at various angles provides the same results. The same principle then applies when considering this aspect in a gear set, the drive gear’s tooth pushing angle must be equal to vertical when pushing on the driven gear’s tooth.

This Step needs to be considered on the Drive gear in every Customized gear-sets and results in a gain on the system's MA.



1B- The effect of the pushing force line on the applied force (can also be called the Effort angle and relates to the pressure angle);

Again a lever system is used to explain the effect of the push force angle on the Input (or the force applied into the system).

The drawing bellow simply demonstrates a pushing force (the red arrows) performed vertically and its effect on the lower quadrant. The principle is the same as in the above explanation 1A, If a push force is performed in a upwards (in a vertical manner) direction against the lever end, the force acting to rotate the lever varies when orienting the lever at different angles. Per example, if the lever is at 30 degrees from vertical, only 50% of the force is acting to rotate the lever. Sin of 30 degrees x 100 = 50%. of the force put into the system is used to rotate the lever (or the gear)

As it can be seen, a mechanical disadvantage occurs when this aspect is present in the system as only a % of the force applied is acting to rotate the lever (gear).




In every Custom gear-set, this effect needs to be considered and result in a loss in the system's MA.


2- The next aspect to review is the method used to measure the radius of the gears when the pushing force is performed vertically as in the drawing bellow;
 


 

As can be seen on the top right picture, the distance from the fulcrum to the force point (arrow in red) needs to be taken horizontally(we don’t use whole length of the wrench). This is due that the pushing force is applied vertically. This method also applies to gears as will be explained in details bellow.





To calculate the efficiency of the Spur Custom gearbox demo;

For ease of understanding, the gearbox has been separated into the Input/First gear-set and the Second/Output gear-set.


The Input/First gear-set:




The Second/ Output gear-set:


The drawings above depicts the actual gearbox arrangement and its dimensions.


As explained above, this method used to calculate the gear-sets MA only applies when the pushing force is performed vertically. In order to provide this aspect, we need to rotate the gear-sets until their average pushing force (line) angle is equal to the vertical axis. This provides the appropriate angle values and radius distances to be used. It also provides the true working distance occuring in the gear set (0.610" in the embodiment bellow).


The next drawing is the Input/First gear-set where it is rotated until its average pushing force angle (blue lines) are averaging an equal to the vertical axis push [demonstrated by the blue lines (where teeth contact occurs) at the lower and upper operating angles].

-The Input gear’s radius = 1.1”, its C.P = 0.6912”. It has 10 teeth, the radius measured horizontally = 1.041” average.

- Its average operating angle is 14.745 degrees above horizontal.


-The First Gear’s radius equals 3.356”, its C.P = 0.6589”. It has 32 teeth, the radius measured horizontally = 3.325” average.


-The First gear’s average operating angle = 2.797 degrees above horizontal .


-The Input gear is the Drive gear and rotates C.W.


To calculate the force put into the system at the Input gear’s shaft and coming out at the First gear’s shaft:

 

A-  We will use 100 in/lb as the force put into the system at the Input gear’s shaft, this way it is easily converted to a percentage. We then have to configure the effect of the Input gear’s radius by using the force lost due to the lever law, in this case creates a mechanical disadvantage.

 = 100 in/lb / 1.041” (average radius)

 = 96.1 in/lbs present at the Input gear’s tooth.


B- As explained in the #1A and 1B explanations, we need to consider the Input gear average operating angle (lever angle) effect. Note that in any common gear-set, this calculation is not necessary since the operating angle occurs evenly on each sides of the gear-set’s horizontal axis (after it has been rotated until the pushing force angle is equal to the vertical axis). The same also occurs on the Custom Cage type Input and Output gears since they also operate evenly under and above their horizontal axis.

 To use the correct effect (1A Resting angle or 1B Effort ngle), simply remember that the advantage occurs on the Drive gear and the disadvantage occurs to the Driven gear. 

the Input gear is the Drive; simply imagine that it isn’t moving the First gear but instead that it is lifting a weight attached (to its tooth) to a lever as in explanation 1A, where the operating angle affects the force required to lift the weight where it is pushing down on the Drive gear’s tooth, therefore the higher the operating angle (from the horizon axis) occurs, the less force is required by the Drive gear in order to lift the weight.

 We then have to configure the average operating angle of the Input gear (the resting angle effect as per explanation 1A):

= -0.05 degrees (minus because it occurs bellow the horizontal axis) - 31.25 degrees / 2 - 1.76. Note that 0.05 degrees is not demonstrated in the drawing due to the limiting space available.

= 15.6 degrees from horizontal, therefore, 90 - 15.6

= 74.4 from vertical axis average Resting angle as per 1A explanation..

The Sin of 74.4 degrees = 0.963, since it is an advantage, we need to use: 0.963 x 100 = 96.3;

= 100 - 96.3 = 3.7. Therefore, a gain of 3.7% occurs due to the Resting angle.  

= 96.1 (as per step A) + 3.7

= 99.8 In/lbs present at the First gear’s tooth

 Note that a gain occurs due to the pushing force effect of Load reduces the force required to lift it since the average (lever) operating angle is 15.6 degrees and as in 1A explanation, a gain occurs in the system. If we were to use a Common gear-set, we wouldn't use this calculation as it operates evenly under and above their horizontal axis (when the pushing force line is rotated so that the push is performed vertically). We would use 100% of the force is required to lift the load and not 96.3%. 


C- Next we have to use the First gear’s radius which provides a mechanical advantage due to the lever law.

= 99.8 (as per step 1B above) x 3.325” (average First gear’s radius)

= 332.83 in/lbs present at the First gear's shaft (not including the Effort angle as per 1A explanation).

 

D- Finally we have to use the effect explained in 1B, the First gear’s operating angles effect the disadvantage occurring when pushing on a lever which is oriented at an angle which doesn’t occur evenly above and below its horizontal axis. It has to be viewed as the pushing force supplied from the Input gear’s tooth where it is partially using its force to rotate the First gear..

 We need to configure the First gear’s average operating angle;

= 2.797 degrees from the horizontal axis. Therefore 90 - 2.797 = 87.203 degrees from the vertical axis.

 The Sin of 87.203 degrees = 0.999, since it is a disadvantage, we need to use;

= 332.83 (as per step C above) x 0.999

= 332.8 in/lb is present at the First gear’s shaft. The First and Second gear’s shaft are using the same shaft. Therefore the force is present at the Second gear’s shaft. The value is then used to configure the force from the second gear’s shaft to the Output


To calculate the force transferred from the Second gear to the Output gear's shaft;


We then use the values in the next drawing where it depicts the Second/Output gear-set rotated so that the average pushing force line (at the meshing lower and upper points) are equal to the gears' vertical axis.


- The Second gear’s radius = 4.919” x 32 tooth = 0.966” C.P. It has 32 teeth, the radius measured horizontally = 2.9535” average

 - Resting angle average = 42.96 to 33.38 degrees operating angles, average = 38.17 degrees from vertical axis. NOTE THAT THE DIFFERENCE BETWEEN THE OPERATING ANGLEES SHOULD BE = 11.25 DEGREES (360 DEGREES / 32 TEETH)  BUT IT DOESN’T SINCE THE OUTPUT TOOTH MOVES BACKWARDS AT A CERTAIN POINT IN THE STROKE DUE TO THE TOOTH SHAPE, THIS IS WHY WE NEED TO USE THOSE ANGLES

 

-The Output gear’s circular pitch = radius of 1.29” x 10 tooth = 0.8105” C.P It has 10 teeth, the radius measured horizontally = 0.9195” average

 - Output gear’s Effort angle = 64.36  to 31.95 degrees operating angles, average = 48.15 degree from horizontal axis. NOTE THAT THE DIFFERENCE BETWEEN THE OPERATING ANGLES SHOULD = 36 DEGREES (360 DEGREES / 10 TEETH)  BUT IT DOESN’T SINCE THE OUTPUT TOOTH MOVES BACKWARDS AT A CERTAIN POINT IN THE STROKE DUE TO THE TOOTH SHAPE, THIS IS WHY WE NEED TO USE THOSE ANGLES

 

The Second gear is the Drive gear and rotates C.C.W. Note that the Input / First gear set above couples with this gear set and forms the whole gearbox. Therefore, the supplied power at the Input gear is transfers to the First gear. Then it is transferred to the Second gear and finally the power comes out of the Output gear’s shaft.


 A- We first have to consider the Second gear’s radius as per lever law which is a disadvantage;

   = 332.8 (as per step D above) / 2.9535” Second gear’s radius

   = 112.68 in/lb of force acting on the Output gear’s tooth.

 

 B- As explained in the #1A and 1B explanations, we need to consider the Second gear’s average operating angle (lever angle). Another way to explain this method is to take per example the wrenches drawing above in explanation 2 above, when looking at the top right picture, it can be seen that if 50 lbs of force is applied vertically to the end of the wrench, a 75 FT/lb torque is transferred to the nut (fulcrum point). The reverse occurs when we put in 75 FT/lb of torque at the fulcrum (in a CCW manner), then a 50 lb force is transferred to the end of the wrench (pushing vertically up). Since the wrench is oriented around 30 degrees from the vertical. As in explanation 1A, installing the Load (weight) of 100 lb at the end of a lever oriented at 30 degrees from vertical, only 50 lb of force is needed to lift it due to the weight resting on the fulcrum where the Sine of 30 = 0.5 x 100 = 50 % of the Load (weight) is required and result in an advantage. Therefore the 50 lb present at wrench end can lift a 100 lb weight.  

To apply this method to the Second gear;

 We then have to configure the average operating angle of the Second gear (Resting angle Effect as per explanation 1A):

= 42.96 degrees - 33.38 degrees / 2 + 33.38 = 38.17 degrees from vertical.

 The Sin of 38.17 degrees = 0.618, since it is an advantage we then need to;

= 100 x  0.618

= 61.8 in/lb required to counteract the 100 In/lb( if we were to add a force of 100 in/lb at the output side). This is due to the Effort angle as explained above when reversing the wrench calculation.

Therefore we have to use; 112.8 in/lb  (as per step A above) - 61.8 in/lb. We use 112.8 and not 100 because only 61.8 in/lb is required to counteract a 100 in/lb which refers to the common gear-set having a ratio of 1:1 where the force put into the system equals the force coming out of the system where 100 is the value used. While we have 112.8 in/lb at the Second gear's tooth and not 100.  

 = 112.8 - 61.8 = 50.9 in/lb more than needed in order to counterbalance a resistance of 100 in/lb at the Output gear's tooth, Therefore

 = 112.8 + 50.9

= 163.56 in/lb present at the Output gear's tooth.

 

C- Next we have to use the Output gear’s radius which provides a mechanical advantage due to the lever law.

= 163.56 (as per step B above) / 0.9195” (average Output gear’s radius)

= 177.9 in/lb present at the Output gear's shaft (not including the Effort angle factor).

 

D- Finally we have to use the effect explained in 1B, the disadvantage occurring when pushing at an angle while the gear’s operating angle doesn’t occur evenly above and below its horizontal axis (like in common gears). It has to be viewed as the pushing force supplied from the second gear’s tooth is partially using its force to rotate the Output gear (refers to the Effort angle).

 We need to configure the Output gear’s average operating angle to configure the disadvantage (the Effort angle as per explanation 1B):

= 64.36 degrees - 31.95 degrees (Output gear's operating angles) / 2 + 31.95

 = 48.15 degrees from horizontal. Therefore 90 - 48.15

= 41.85 from the vertical axis.

 The Sin of 41.85 degrees = 0.6672, since it is a disadvantage, we need to use;

= 177.9 (as per step C) x 0.6672

= 118.7 % is transferred  at the Output gear’s shaft or isthe power coming out of the system..

 Therefore If a 100 in/lb is applied at the Input shafts of the gearbox, 118.7 in/lb is measured at the Output gear’s shaft. This takes care of the force factor but what about the working distances travelled by the Input and the Output.

 

E- In order to calculate the Efficiency, we need to consider both the Forces and the Distances travelled by the Input and the Output. As can be seen on the Input /First and the Second / Output gear-set drawings above, the Blue dimension represents the working distance travelled. They are both equal to 0.610” while If 100 in/lb is supplied to the Input (power put into the system), 118.7 in/lb will be measured at the Output (power coming out).

Therefore the Efficiency = 118.7 %.

 As it can be all of the tests results are equal to around 118% Efficient. Note that the force needs to be applied to the appropriate shaft and also have the proper rotation or other wise the efficiency will be only 82%. As any scientific principles, the power put into and the power coming out needs to balance within a system;

 - 118% + 82% 

  = 200 / 2

  = 100% overall efficiency.

 

 

 CONCLUSION


Not only are we able to sell gearboxes but we can also license or sell the Motion x gearbox patent itself. Or we can be a part of the team that fabricate and install the right gearbox for your particular application. After installation, the Motion x gearbox will allow your plant’s equipment to operate more efficiently (between 7% and 25% more efficiency, depending on which model is chosen).

 If you have any questions or queries regarding the gearbox, drawings, documents or videos, we will be more than happy to answer. The gearbox needs to be manufactured accordingly to a specific application, every plant uses different size of shaft for generator requirements, so we don't have many sizes of gearboxes already manufactured yet but we are prepared to have a gearbox manufactured to meet all your requirements.

We have exactly the gearbox that will be beneficial for your plant and to be ready in a short period of time. We are also willing to do a live presentation of the tests performed on our latest design at any place ad time so you can confirm for yourself these gearboxes are really over efficient when used in proper manner.

Please contact us for any inquires at: ridoux@yahoo.com

 

 
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