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### Author Topic: Increase efficiency of rotating shaft.  (Read 8305 times)

#### jigar

• Newbie
• Posts: 10
##### Increase efficiency of rotating shaft.
« on: October 09, 2006, 06:30:43 PM »
swamishriji

I am mechanical engineering student . Here I am describing one mechanical engineering situation that is very useful in any kind of mechanical and energy consumed industries. It is not even in any applied mechanics book. I hope you will try to understand it and use it. If you find any difficulty to understand ,feel free to ask me.
we can increase efficiency of rotating shaft. we can lift mass using less energy by using simple mechanism. please take few minute to read .we can use this technology in power plants and transportation vehicles.
The base of this theorey is force or weight is stationary and it can help to rotate pulleys . we can convert gravitational force into energy while the mass or weight is stationary.
Click on Photos to View Figures(Energyefficiency and Energy efficiency1)

The figure(energy efficiency) is in cross section. as shown in figure, there are two cases. in each case there are two pulleys of same diameter. each pulley is of exactly circular shape.In first case 50 k.g weight is fixed with each pulley as shown in figure. The center of each pulley is fixed. Between these two pulleys there is a stationary plate. plate will remain stationary, while rotating the pulleys because centers of pulleys are fixed. 100 k.g weight is put on this plate. The force or weight of plate is applied on these two pulleys in vertical downward direction. now we try to rotate slowly first pulley in clockwise direction and second pulley in anti clockwise direction. lubrication is provided between the contact surfaces of stationary plate and pulleys.
In second case two pulleys of same diameter. 50 k.g weight is also fixed with each pulley, but there is no stationary plate and 100 k.g weight. Now we try to rotate slowly first pulley in clockwise direction and second pulley in anti clockwise direction.
As free body diagram in first case we can rotate pulleys using less moment of force. we can lift mass using less energy.

1. In case 1) the moment necessary to rotate each pulley is

m=(-g/2cos(bita)+g1*cos(alpha))*r
where:
g is weight of plate (=100 kg)
g1 is given weight (=50 kg)
alpha and bita are angle of rotation of each pulley
2. In case 2) weigt of plate g=0.
Figure (energy efficiency1) is in cross section. As shown in figure two pulleys and 100 k.g. plate is arranged. Pulleys are designed so that from points a and a` to points b and b`, radii of pulleys continuously decrease. Now the weight or force of plate apply on point a and a` of two pulleys in vertical downward direction. Tangential component of this force helps pulley to rotate. when point b and b` come in contact with plate, plate traveled .5 cm in vertical downward direction. The gravitational energy plate loose is mgh=.005*50 n-mtr=.25 n-mtr. But the same time work done on pulley or energy got by pulley is mgcos(thita)h=.04*50*cos(thita), where theta is 45. so energy got is .04*50*cos45=.04*50*.7=1.4 n-mtr. .04mtr is periphery of pulley from point a to b. calculation is given only for one pulley. we can lift mass using less energy.
Jigar Y. Patel
energyefficiency@zoomshare.com

Plz visit http://energyefficiency.zoomshare.com

#### Gregory

• Full Member
• Posts: 135
##### Re: Increase efficiency of rotating shaft.
« Reply #1 on: October 09, 2006, 07:20:02 PM »
Hello Jigar,

At first sight I thought You're wrong, but after I realized why and how can it useful.
Thanks!  This is a good idea

But only works good with very low rate of CoF. And the stationary plate must be fixed only in horizontal directions to allow the rotating pulleys to take the force of the plate.
Very good, we can lift the weights for less force. I will try it for myself in a simulation to see what happens.

#### jigar

• Newbie
• Posts: 10
##### Re: Increase efficiency of rotating shaft.
« Reply #2 on: October 10, 2006, 04:41:39 AM »
Thank you. Plz make it as soon as posible.My theory is purely depanding up on free body diagram.

#### lancaIV

• elite_member
• Hero Member
• Posts: 5174
##### Re: Increase efficiency of rotating shaft.
« Reply #3 on: October 10, 2006, 03:15:38 PM »
US4464095,Hydraulic energy converter,Kango Iida(JPN):
only as idea,to compare !

S
dL

#### Gregory

• Full Member
• Posts: 135
##### Re: Increase efficiency of rotating shaft.
« Reply #4 on: October 10, 2006, 03:21:18 PM »
By the way I don't know for 100% sure, that it will work or not...
I will test it, but please be patient because I'm very busy in this field now.

#### Gregory

• Full Member
• Posts: 135
##### Re: Increase efficiency of rotating shaft.
« Reply #5 on: October 16, 2006, 03:57:07 PM »
Hi Jigar,

As I presume after, my first thought was right and the second is very wrong, your too.
The stationary weight can't help to rotate the pulleys, and the reason for this is very simple:
Try to push a wheel / pulley with your hand, while you never take back your hand... Impossible.
In this case your hand hinder the rotation of the pulley. The pulley can't rotate because your hand is there.

The weight excert its force on the pulley, but this same force prevent the rotation.
When the friction is set to a high rate the thing is locked. When the friction is set to a low rate, the wheel will begin to swing like a pendulum, but stops sooner.

There is only one exception:
When the weight is not stationary, but swing up and down, and gives short pushes to the wheel.
However this version has limited usability.

In the picture only the blue plate is allowed to collide with the wheels. The brown rods limit the movement of the plate which only allowed to move up. Purple weights set to 12 Kg, blue plate set to 25 Kg.

Greg

#### jigar

• Newbie
• Posts: 10
##### Re: Increase efficiency of rotating shaft.
« Reply #6 on: October 17, 2006, 05:18:03 AM »
Hi Jigar,

plz view second figure, where I decrease redii of pulleys continueously up to one point. consider minor decrease of diameter In this case I am 100% sure we can create  energy.we can lift mass using less energy. The energy lost by the plate is very less compere to the energy got by pulleys because we give minor decrease in diameter. here pulleys become like a cam. and we can apply free body diagram correctly on this figure. and in this figure also we can get continuous rotation by doing minor change.check this figure in your simulation
As I presume after, my first thought was right and the second is very wrong, your too.
The stationary weight can't help to rotate the pulleys, and the reason for this is very simple:
Try to push a wheel / pulley with your hand, while you never take back your hand... Impossible.
In this case your hand hinder the rotation of the pulley. The pulley can't rotate because your hand is there.

The weight excert its force on the pulley, but this same force prevent the rotation.
When the friction is set to a high rate the thing is locked. When the friction is set to a low rate, the wheel will begin to swing like a pendulum, but stops sooner.

There is only one exception:
When the weight is not stationary, but swing up and down, and gives short pushes to the wheel.
However this version has limited usability.

In the picture only the blue plate is allowed to collide with the wheels. The brown rods limit the movement of the plate which only allowed to move up. Purple weights set to 12 Kg, blue plate set to 25 Kg.

Greg

#### Gregory

• Full Member
• Posts: 135
##### Re: Increase efficiency of rotating shaft.
« Reply #7 on: October 19, 2006, 11:24:00 AM »
Of course, with changing radii the pulleys can rotate some degrees, less than 360 degrees.

But in this case the stationary weight doesn't remain stationary at all. It descends to a lower position while the radii of the pulleys are decreasing. So, at the end of the cycle you must lift the stationary weight back to the initial position, and replace it on the longest radius of the spiralic body. This takes energy. Finally you didn't gain any energy. The result is Zero, minus the friction.

It is analogous to the swing of a pendulum. It can't swing higher by itself.
Or to a roller coaster...

#### jigar

• Newbie
• Posts: 10
##### Re: Increase efficiency of rotating shaft.
« Reply #8 on: October 19, 2006, 11:52:12 AM »
It is not like pendulum.
Consider two off set cams and between these two cams there is  one follower.From the contact point of cam and follower, we continuously decrease redii of cam up to one point.We give minor decrease in redii. we decrease redii at 10:1 ratio. It means if the length of periphery from one point to other point is 10 cm than decrease of redii from one point to other point is 1 cm. For left side cam we decrease redii from right to left. I mean we decrease redii of left side cam in antclockwise direction. For right side cam we decrease redii from left to right. we decrease redii of right side cam in clockwise direction.Now we apply some vertical load or weight from follower to two cams.Tengential componant of this load helps the cams to rotate. Left side cam rotate in clockwise direction and right side cam rotate in anti clockwise direction. Because of minor decrease in redii of cams ,follower come little bit down in vertical direction. But the same time  the load or weight helps cams to rotate on more length of periphery at some angle. We can increase efficiency of rotating shaft. We can lift mass using less energy. And for 360 degree rotation we give rise from last point to first point of cam in less length of periphery.or we can use some external mean to lift weight.But the energy we got on the cams is more than energy lost to lift weight to its initial position.
Of course, with changing radii the pulleys can rotate some degrees, less than 360 degrees.

But in this case the stationary weight doesn't remain stationary at all. It descends to a lower position while the radii of the pulleys are decreasing. So, at the end of the cycle you must lift the stationary weight back to the initial position, and replace it on the longest radius of the spiralic body. This takes energy. Finally you didn't gain any energy. The result is Zero, minus the friction.

It is analogous to the swing of a pendulum. It can't swing higher by itself.
Or to a roller coaster...