WO2007046722A1

WO2007046722A1 - Continuously variable transmission (cvt) through gears

Info

Publication number: WO2007046722A1
Application number: PCT/PT2006/000024
Authority: WO
Inventors: João Armando Soledade CHAMPLON
Original assignee: Champlon Joao Armando Soledade
Priority date: 2005-10-18
Filing date: 2006-10-17
Publication date: 2007-04-26
Also published as: PT103369A

Abstract

Continuously variable transmission, with power transmission through gears. Basically consists of two connected differentials: the input differential and the output differential. The input differential splits the movement by two different sets of gears; the output differential joins the movement again, and sends it to the following devices on the vehicle. Each set of gears defines a different gear ratio: the 'force route', which can be compared to the first gear in a manual gearbox, and the 'speed route', that can be compared to the last gear on a common manual gearbox. Continuous variation is obtained by adjusting the amount of movement that is sent to each route. This adjustment can be made by mechanical or magnetic braking of a disc on each differential, one producing an increase of the final gear ratio, and the other producing a decrease of the final gear ratio.

Description

D E S C R I P T I O N

^Continuously Variable Transmission (CVT) Through Gears"

Existing problem

The transmission to the wheels of the torque produced by a car's engine is, most of the times, done through a gearbox.

The gearbox is used to optimise the power produced by the engine for the road speed.

In this way, when it's necessary to climb a hill, the gear ratio is changed in such a way that the engine rotates more times for each rotation of the road wheel, and the torque on the road wheel becomes larger.

On the other hand, when there is too much torque applied to the road wheel, the gear ratio is changed in such a way that the rotation of the engine can be reduced keeping the road speed.

Selecting the gear ratio, when using manual gearboxes, is a decision of the driver. With automatic gearboxes, that decision is made by systems with which the driver has frequently none interaction.

Among the disadvantages of those systems, one can name the loss of power during the gear shifts, and the unnecessary consumption of gas, by inadequate selection of gear.

The search for solving that situation, and for minimizing the consumption, has lead to different approaches.

One of the efforts have been on tuning the control and switching systems of automatic gearboxes (with a discreet number of gear ratios) ; other efforts have been done searching ways of transmitting torque without using a finite number of gear ratios, that is, looking for systems that may be capable of always keep the "perfect" adaptation to optimise the power produced by the engine for the road speed.

These last systems, usually called continuously variable transmissions, are now being used on several cars.

The solutions used are based in mechanisms that use friction, and those approaches are limited in the power transmitted, generate heating problems, and have low performance if compared with geared transmissions.

The actual solution reports a continuously variable transmission through gears which allows the transmission of large amounts of power, and doesn't have the problems of friction solutions.

System Components 1 - Transmission

The final gear ratio, that is, the ratio between the torque measured on the output of the system, and the torque in the input, varies continuously from zero to a maximum that depends on the way the system is build.

Zero torque on the output means that, despite the fact that movement exist on the input, there is no movement on the output. This is what happens when no action is applied by the control (there no braking action) , and the road wheels are stopped.

Maximum ratio is obtained when braking one shaft is forcing the movement to completely (or almost completely) , pass through the "speed route" .

This braking action mentioned before, can result in a complete stop of the shaft, or limiting his rotation to a maximum allowed value, depending on the braking system used.

To completely understand how this system works, it's necessary to have present and understand how a differential works .

A differential receives the torque from the engine in the ring gear, through the pinion shaft and input pinion. The movement is delivered to the axle shafts and to the wheels. The movement is transmitted from the ring gear to the pinion gears, which in turn transmit it to the side gears and axle shafts.

The schematic of a differential is represented in figure 1, where (2) points to the ring gear, and (1) and (3) point to the axle shafts.

The equation that establishes the rotation relation between ring gear (2) , represented by letter C (in turns by time unit) , and the axle shafts rotation (1) and (3) represented by Si and Sj, is:

2xC = Si + Sj .

It's known and easy to deduce from the equation, if Si and Sj have the same rotation speed (in turns by time unit) , then that speed is equal to the C speed. If one of Si or Sj is stopped, then the other will have double speed of C.

The system presented in this document, is based on the connection of two differentials, as represented on figure 2.

One is called the input differential, and the other is called the output differential.

The engine engages with input differential ring gear (4) , and the output torque is delivered through the output differential ring gear (13) .

Shafts (5) and (6) transmit movement to shafts (12) and

(11) , through gear (7) with R2 teeth and gear (9) with R4 teeth, and gear (8) with Rl teeth and gear (10) with R3 teeth, respectively.

Two particular situations can immediately be observed:

A) Figure 3. When shaft (6) or shaft (11) is braked, all the movement goes through shafts (5) and (12) , and the output rotation S can be expressed in relation to input rotation E, by.

S = - (R2/R4)xE (Minus sign means sense rotation changed) .

B) Figure 4. When shaft (5) or shaft (12) is braked, all the movement goes through shafts (6) and (11) , and the output rotation S can be expressed in relation to input rotation E, by: S = - (Rl/R3 ) xE .

The situation described in B) is the transmission by the "force route" , whenever the choice of R1/R3 forces the output rotation to be in relation with the input as it would be the situation on a manual gearbox with the first speed engaged.

The situation described in A) is the transmission by the "speed route", whenever the choice of R2/R4 forces the output rotation to be in relation with the input as it would be the situation on a manual gearbox with the last speed engaged.

1.1 - Movement Equations

The equations of the differentials are:

Input differential: Output differential:

2E = Sl + S2 2S = S3 + S4

Where :

E represents the rotation of the ring gear in the input differential ,

S represents the rotation of the ring gear in the output differential ,

51 represents shaft (6) rotation,

52 represents shaft (5) rotation,

53 represents shaft (11) rotation,

54 represents shaft (12)

The rotation relations between gears are quite simple too:

S3 = - (R1/R3) .Sl S4 = - (R2/R4) .S2

We have 4 equations and 5 free variables S, Sl, S2 , S3, S4, so we must express output in relation to input E and one of the free variables. That is:

S=(Sl/2) . ( (R2/R4) - (R1/R3)) - (R2/R4).E (Eq. 1) S=(S2/2) . ( (R2/R4) - (R1/R3)) - (Rl/R3).E S=(S3/2).(1 - (R2/R4) . (R3/R1) ) - (R2/R4).E S=(S4/2).(1 - (R4/R2) . (R1/R3) ) - (Rl/R3).E (Only valid as long as R1/R3 and R2/R4 are different)

Obviously, if we don't force shafts Sl, S2 , S3 and S4 to rotate at predetermined values, the equations will be useful to determine the speed each shaft has, as long as we know the input and output rotations.

In short we can say that, if the forces acting on the system are only coming from the engine and the road wheels, then each one will rotate independently, and they will determine the rotation of shafts Sl, S2 , S3 and S4.

We can compare this situation to a manual gearbox with no gear engaged.

Lets analyse the case where the road wheels are stopped (S=O) and there is E rotation on the input.

Forcing S=O on previous equations, it's possible to express each shaft rotation in relation to the input rotation:

51 = (R2/R4) . 2E/ ( (R2/R4) - (R1/R3) ) (Eq. 2)

52 = -(R1/R3) . 2E/ ( (R2/R4) - (R1/R3) )

53 = - (R1/R3) . (R2/R4) . 2E/ ( (R2/R4) - (R1/R3) )

54 = (R1/R3) . (R2/R4) . 2E/ ( (R2/R4) - (R1/R3) )

Assuming R2/R4 > 1 (S4 speed larger than S2 speed) , and 0 < R1/R3 < 1 (S3 speed less than Sl speed) , we will note that: Sl has a larger speed than 2E, and every shaft is turning .

The value of 2E on shaft Sl happens, as we know from the differential equation, when the opposite shaft (S2) is stopped. What is verified when all the movement is passing through "force route" , and "speed route" is stopped (S2=S4=0) .

On the other side, when all the movement goes through "speed route", the "force route" and Sl will be stopped (Sl=S3=0) .

Eq. 1 describes this behaviour, and a graphical representation can be seen on figure 5. As it was told, we verify the existence of a value for Sl that stops the output S; and if we force down that rotation to the value of 2E, then we stop shafts S2 and S4 , and all the movement will go through the "force route" (E -> Sl -> S3 -> S) .

If we continue braking Sl and force the reduction of speed till zero, the transmission will vary continuously until reaching maximum output speed.

So, when Sl speed goes from 2E to zero, the transmission of engine's torque will change from totally through "force route" to totally through "speed route" . This is a continuously variation of final gear ratio.

Once again, if we compare the situation with a manual gearbox, we can say that by braking Sl shaft we are "changing gears from first toward last" ; and that if no braking is applied, the vehicle will loose road speed, Sl will increase rotation speed, and "gears will go from last toward first" .

However, this is true only when there are forces opposing to the movement . In case where the external forces are adding to the movement (going down a hill, for instance) , we'll want to increase Sl rotation speed, and "bring down the gears" . What can be done by braking one of the shafts in the "speed route", that is S2 or S4.

Let's observe that, at any moment it's established a relation between output S and input E that can be changed by braking any of the shafts Sl, S2, S3 and S4.

So, when the external forces are opposing to the movement, the vehicle will tend to stop and increase Sl rotation speed. If we want to keep final gear ratio, it's necessary to brake Sl in a way that vehicle keeps road speed, and Sl keeps rotation speed.

If the external forces are adding to the movement, the vehicle will tend to increase road speed and increase S4 rotation speed. If we want to keep final gear ratio, it's necessary to brake S4 in a way that vehicle keeps road speed, and S4 keeps rotation speed.

To set a correct final gear ratio at any moment, it will be necessary to study and develop careful reading and analysis of data important for making decisions. Obviously it will have to consider the engine rotation, the situation of the vehicle and the intentions of the driver. 2 - Control

Increasing/decreasing final gear ratio can be done through mechanical means. The logic to decide when and how to adjust the ratio will have to be done by a special designed software, that wil have to take in account several things, including the the system's status, the movement of the vehicle and the driver's will. I shall now describe a method for braking the shafts, that is just one among all. This method is a different solution for the simple task of shaft braking, that we easily image like a disc been stopped.

As we saw before, with the road wheels stopped, for a general input E, shaft Sl will rotate according to Eq. 2.

What we are looking for, is to act by braking but, not to do it directly on shaft Sl, neither S2 , S3 or S4. The solution presented now, acts directly on pinion gears, making them stop or reducing their rotation when necessary.

2.1 - Braking pinion gears

Let's consider a differential where pinion gears and side gears are bevel gears with identical number of teeth. For each turn of the differential cage, each axle shaft will make one turn if, and only if, pinion gears do not rotate on their one axes. In this case, differential (and axle shafts) will turn as a fixed block. But, if we stop any of the axle shafts, then for each turn of the differential cage, each pinion gear will make a complete turn on his one axe.

More specifically, if side gears have P teeth and pinion gears have Q teeth, when stopping one axle shaft, each pinion gear will complete P/Q turns on his one axe, for each complete turn of the differential cage.

It's easy to observe also, that stopping one axle shaft will force pinion gears to rotate in opposite sense of what we get by stopping the other axle shaft. So, we may say that pinion gears rotate from -P/Q to +P/Q when stopping one road wheel to stopping the other road wheel.

Let's consider now, that attached to the pinion gears and rotating on the same axe, there are other bevel gears (14) . These two bevel gears engage simultaneously on a third bevel gear (15) , this one turn on an axe coincident with the axle shaft axe. Naturally, this last gear must have a hole for the axle shaft to pass through.

The rotation speed of gear (15) , if measured relatively to the pinion gears axes, is T/R, where T represents gear (15) teeth number, and R represents gear (14) teeth number.

As (14) has the same rotation speed of pinion gears, we can conclude that rotation of gear (15) is given by

-P/QxT/R,

When one of the axle shafts is stopped, and by

÷P/QxT/R

When the other axle shaft is stopped.

Both measured relatively to the pinion gears axes.

Measured relatively to the vehicle axes, we must add (or subtract, it's arbitrary) one, since all the previous values were calculated in relation to one differential cage turn.

In this way, the rotation D relatively to the vehicle axes is given by:

D = I + P/Q x T/R with one axle shaft stopped.

And:

D = I - P/Q x T/R with the other axle shaft stopped.

If pinion gears and side gears have the same number of teeth, and R and T have also the same number of teeth, then D will vary between 0 and 2. Of course other combinations of teeth numbers will produce the same result. As long as P/QXT/R = 1, the result will be the same.

Anyway, what we wanted to make clear, was the fact that we can brake a axle shaft by braking disc (16) , even if not completely. Finally, if we mount on each differential a disc as described above, symmetrically, then we have a complete solution for controlling the transmission:

- in case where external forces oppose to the movement, and we want to increase gear ratio, we brake disc (16) ;

- in case where external forces oppose to the movement, and we want to decrease gear ratio, no action;

- in case where external forces add to the movement, and we want to decrease gear ratio, we brake disc (17) ;

- in case where external forces add to the movement, and we want to increase gear ratio, no action.

It still is necessary to develop logic to control transmission and take the correct decisions to brake discs whenever adequate .

The best braking solution must be determined experimentally, but magnetic braking seems to be a better solution by allowing electric control more easily, no direct friction, thus producing less heat, requiring less maintenance, and a longer life.

Claims

C L A I M SI claim:

1. A continuously variable transmission comprising two differentials in parallel, one receiving the movement through the ring gear (4) , the other sending the movement to the following device through his ring gear (13) , and by engaging each pair of shafts (5) and (12) , and (6) and (11) with gears (7) and (9) , and (8) and (10), with different gear ratios.

2. A continuously variable transmission as in claim 1 where the adjustment of the final gear ratio is obtained by braking discs (16) and (17), one on each differential; each disc is attached to a gear (15) , which in turn is engaged with two other gears (14) solid with the pinion gears shafts, on each differential; braking one disc reduces the final gear ratio, braking the other disc increases the final gear ratio.

3. A continuously variable transmission as in claim 1 where the adjustment of the final gear ratio can also be obtained by mechanical or magnetic braking of two shafts (6) or (11) , and (5) or (12) , one increasing the final gear ratio, the other decreasing the final gear ratio.