SOUTH ESSEX ENGINEERING
Much of what ypou have posted is not accurate so gere is some additional information which gives details of a hypothetical test application. In this hypothetical the flywheel speed is exceedingly modest yet the energy stored is potentially substantial. Keep in mind the section which notes that the KE is a function of the square of the speed. I trust you will find this of interest.
How it works: This system utilizes a flywheel, an induction mechanism and an overdrive transmission, among other things. The induction device is based on well developed existing eddy current technology and is envisioned as two independent co-axial, shaft mounted, bearing supported, shaded male and female rotating mechanical sections separated by an air gap, encompassed within a coil, operating on the principles of induction resulting from a controllable electric current across the coil. As noted later herein, a friction device could be substituted but with less efficiency. The electric current to the coil is varied in proportion to demand which instantly governs the strength of the flux density across the air gap between the rotating sections. The higher the current voltage to the coil the higher the flux density becomes and the greater the attraction across the air gap between the shaded parts. And, thusly, the throughput of torque is instantly set to any level from zero to full unit rating with no friction losses or the need for other parts. The great advantage to this is that this feature meters the throughput of the torque by enabling the two halves of the induction device to rotate at their own speed while transmitting torque between them as a direct function of controlled magnetic flux density (read voltage).
While in an operating range in the power-from-the-vehicle flow direction, the input torque to the induction device is always greater than or equal to the output torque. The induction device in concert with the gear system provides controllable torque variation without waste between this energy source, which is at a higher level, to that of the load, the flywheel, which is at a lower energy level. By varying the flux density (voltage) the level of transmitted torque can be instantly managed and controlled in real time.
When the energy flow is reversed, that is, from the flywheel to the vehicle, the issue is that of speed inasmuch as the over-drive gearing now acts as a speed reducer and a torque multiplier. As noted, the charged flywheel operating range is fast as compared to the vehicle operating range which is slow. Therefore, in order to accelerate the car, high torque is needed not high speed. The coupling of the induction device to the gear system performs perfectly to transform the high speed of the flywheel into the torque needed to accelerate the "heavy" car and therefore must be operated on a speed priority basis. Again, the induction device, through control of the flux density via voltage supervision, provides for this requirement. It should be recalled what the basic problem is the fact that the source energy in either power flow direction has its speed retarded while the receiving mass has its speed advanced. The key is that by increasing the voltage across the coil of the induction device we can change the overall effective gear ratio. Therefore, as the source slows the induction device via increasing flux density seamlessly changes the gear ratio the system sees.
Functioned off the vehicle's brake and throttle systems, the device conducts fully managed torque levels in each direction. As triggered and controlled, the induction traction feature of the device provides for variable and controlled power to and from the energy storage flywheel; and provides for, in effect, an infinitely variable transmission of power, first in the direction of the flywheel system (as a load) to store energy and subsequently, reversing the process, back from the flywheel system to the drive wheels of the vehicle (as a load).
Let's look further into the overall "gear ratio" of this system. The largest ratio available from the ground to the flywheel would be in a lock-up condition between the car wheels and the flywheel. Assuming the induction device is locked up; this would be axle rpm times differential ratio (1 to 3.5) times the overdrive gear ratio (1 to 6) which is 21 to 1. Say the induction device voltage is adjusted in stepless increments so that a series of output speeds, as a percent of input, are realized through the induction device. We first have: (given) 21 to 1 at 100%, and at 90% we have 18.9 to 1, at 80% we have 16.8 to 1, at 70% we have 14.7 to 1, at 60% we have 12.6 to 1, at 50% we have 10.5 to one, at 40% we have 8.4 to 1, at 30% we have 6.3 to 1, at 20% we have 4.2 to 1, at 10% we have 2.1 to 1, and so on.
Now, if we take a similar look at only the overdrive gear ratios combined with the induction device we have: 6 to 1, at 100%, 5.4 at 90%, 4.8 to 1 at 80%, 4.2 at 70%, 3.6 at 60%, 3 to1 at 50%, 2.4 to 1 at 40%, 1.8 to 1 at 30%, 1.2 to 1 at 20%, and 0.6 to 1 at 10%, and so on. In practice, being that the induction device acts as a traction machine, the actual ratios realized will be a function of many factors, but the effective result will be the same, seamless uninterrupted energy flow.
To visualize the system in action let us look at what takes place:
Car is traveling at some speed.
The flywheel is idle.
The speed differential between the car interface element and the flywheel are at a maximum.
The driver intends to significantly slow or stop the car and applies the brakes.
(The system works to slow the car as though the brakes were actually applied. Panic braking overrides the system.)
The system energized the induction device in proportion to the magnitude of the brake signal or call.
The induction device or modulator throughputs a torque in proportion to the call which rises from zero on a sine wave curve form.
Eventually, a peak ratio begins to descend across the modulator, say from hypothetically 50 to 1, which engenders a high torque because of this high slip angle, a condition well known in the field.
The car begins to slow and the flywheel begins to accelerate.
As a result, the speed differential is caused to lessen across the modulator, closing the ratio and decreasing the transmitted torque
If desired, the driver compensate by increasing the brake signal (call).
The voltage at the coil is thus raised by the system controls.
The flux density is raised across the modulator.
An increased torque is reestablished which again spreads the speed ratio across the modulator which raises output speed to keep it progressively ahead of the flywheel speed. Within limits, these last few steps are automatically repeated over and over by the system so long as transferable energy is available and a call is maintained across the modulator. The more aggressive the call the more aggressive the energy transfer, speeding the flywheel and slowing the car. A similar process is applied to recover the energy from the flywheel. However be reminded that the gear ratios become somewhat reversed while the objective is the same. Now from this it may become visible just how the system changes the effective gear ratio. It can also be seen that the combination of the induction device and the overdrive gearing form an infinitely variable transmission able to smoothly manage output power under all conditions.
Some analytical calculations follow, and this analysis will include only certain aspects of the filed patent application in order to reduce the need to explain each element of a fully functional installed system. These omitted aspects while conducive to a workable system do not add to a clear understanding of the energy-transformation mechanism.
For this analysis we present a hypothetical automobile, supported flywheel, gear system and induction device as a framework for discussion. The car has the following characteristics: weight, 3200 pounds, mass 99.38 slugs (see calculations); tires, 24 inches in outside diameter and circumference of 6.28 feet (see calculations); rear wheel drive w/driveshaft; rear differential gear ratio, 3.5 to 1; approximate drive shaft RPM at 15 and 30 MPH respectively: 736 and 1471 (see calculations); trial flywheel weight, 100 pounds (rimmed) neglect spokes, radius of gyration 0.5 feet, mass, 3.1 slugs (see calculations); trial gear box ratio 1 to 6; induction device, stationary coil eddy current clutch with DC excitation.
Assumptions and stipulations: such controls as required are provided; the system is appointed, installed, connected and arranged with all aspects that may be required in an actual working system; on demand, power flow is conducted from the car drive shaft (as a hypothetical point of system connection for the purposes of this trial analysis) into the induction modulator and from there through the overdrive gearing to the flywheel which is thereby accelerated as needed; the flywheel stored energy is similarly returned in the reverse path to reaccelerate the car. The induction device, as controlled, throughputs from zero to full torque rating; as noted, to facilitate calculation of energy transfer, car speed set points are arbitrarily established at 15 and 30 MPH with flywheel charging beginning at (44ft/sec) 30 MPH (car) and ending at (22 ft/sec) 15 MPH (car);
Formulae and calculations:
For the car traveling in a strait line, Kinetic Energy, (KE) = ½ M V^2,
Where, (M) = Mass = W/g = weight in pounds / gravitational constant of 32.2 feet/second^2;
where, (V) = strait-line speed in feet/second
Car @30, KE = ½ (99.38) (44) ^2 = 96200 Ft-Lbs.
Car @15, KE = ½ (99.38) (22) ^2 = 24050 Ft-Lbs.
For the Flywheel, traveling in its circular motion, KE = ½ (Moment of Inertia) x (omega) ^2
where, Moment of Inertia = Mr^2 = Mass times radius of gyration (r) squared, and
where, in this case, r = mean radius for the rim of the flywheel = 0.5 feet
M = mass = W/g, (same as above), and
omega (w) = angular speed in radians/second = 2pi x revolutions/second, and
where, pi = 3.14159 ...
substituting: Flywheel KE = ½(m) (r^2) (w^2)
= ½(3.1) (.5) ^2(w) ^2, where omega (w), is yet to be determined.
Calculate mass for car and flywheel:
Car mass = W/g = 3200/32.2 = 99.38 slugs
FW mass = 100/32.2 = 3.1 slugs
Calculate circumference (c) of the car tire:
C = (pi) (D) = 3.14159(2 Ft.) = 6.28 Ft
Determine speed of drive shaft @ 30 & 15 MPH:
30MPH = 44 Ft/sec and 15 MPH = 22 Ft/sec over the ground and
44 x 60 = 2640 feet/ minute @ 30 MPH,
and 22 x 60 = 1320 feet/minute @ 15 MPH
Now, find car drive wheel RPM:
If we take the FPM rates above and divide by tire circumference (6.28 Ft) we get axel RPMs:
2640/6.28 = 420.38 @ 30, and 209.92 @ 15
Convert this to driveshaft RPMs:
Recall that the differential gear ratio is 3.5 to 1 therefore:
420.38 x 3.5 = 1471.33, and 734.72 RPMs respectively at the drive shaft.
Converting to revolutions per second we have:
1471.33/60 = 24.52 RPS @ 30 MPH
and 734.72/60 = 12.25 RPS @ 15 MPH
Now, for flywheel KE (above), we need (w), omega, in radians/sec,
Multiplying RPS by 2pi we get:
12.25 (2) (3.14159) = 76.94 Rad/sec @ 15 MPH, and,
24.52 (2) (3.14159) = 154.06 Rad/sec @ 30 MPH
Now recall that the power flows through the overdrive gearing at a 1 to 6 ratio from the car driveshaft to the flywheel, and importantly in a 6 to 1 ratio from the flywheel to the car driveshaft.
Therefore, multiplying the above by 6 we get theoretical (assuming modulator at a1:1 final ratio) flywheel speed in Radians/sec:
76.94 (6) = 461.64 Rad/sec @ 15 MPH
(Note: Flywheel RPM = 4408 (safe))
And,
(154.06) (6) = 924.38 Rad/sec @ 30 MPH
(Note: Flywheel RPM = 8827.18 RPM (safe))
From above: Flywheel KE = ½(m) (r^2) (w^2) = ½(3.1) (.5) ^2(w) ^2
Substituting, we get:
(½) (3.1) (.5) ^2 (461.64) ^2 = 82580.7 Ft-Lbs. @ 15 MPH equivalent shaft speed
And,
(1/2) (3.1) (.5) ^2 (924.38) ^2 = 331110 Ft-Lbs. @ 30 MPH equivalent shaft speed
Analysis:
Recall that the energy of motion in the car at 30 MPH was a maximum of 96,200 Ft-Lbs and 24,050 Ft-Lbs @15 MPH. It is important to note that the system can store no more energy than that which is available from the source. In the subject example, we have presented a trial scenario which could if available store between 82,580.7 and 331,110 Ft-Lbs of energy. In these trial calculations we arbitrarily selected a 100 pound flywheel and a 1 to 6 overdrive gear ratio. Either or both can be altered (lessened) in which case we would have less weight in the flywheel and/or less speed in the flywheel. Each of these present a favorable design outlook. In conclusion, the subject system as assumed has far more capacity than needed under the assumptions offered. Massaging the numbers for a more perfect design is left here to others.
In this trial examination, the mass of the car was about 33 times the mass of the selected flywheel and what this flywheel mechanism does is manage energy transfer over time between these two masses.
A further review of how this system works helps one visualize precisely what is taking place when the system is in operation. Lets say while approaching a red traffic signal, the initial conditions would be the car traveling at some level of speed and the energy storage system in a dormant state. In preparation to stop the car the driver initiates normal braking action and the instrumentation of the storage system energizes the induction modulator coil in a direct relationship to driver input and the modulator outputs a proportional torque which turns the overdrive gearing and begins to accelerate the flywheel transferring energy from the car to the flywheel and thus slowing the car.
With certain exceptions, as long as the input shaft speed (source) to the energized modulator is greater than its output speed the flywheel will continue to be accelerated thereby storing more and more energy, and continuing to slow the car. Integrated system controls hold the friction braking process out of initiation unless or until prescribed parameters dictate their initiation. System components and controls isolate the flywheel from the remainder of the system when no further input torque is available in the cycle. That is to say, when the flywheel speed and the input shaft speed to the flywheel begin to approach the same rate as happens at the end of the charging process. Without completely stating the process the energy transfer back to the car is similar.
The function of the induction device is analogous to the friction brakes of an ordinary car or truck. With such friction brakes one aspect rotates while the other is fixed. When in operation a friction system attempts to lock the moving part to the stationary part which causes considerable disagreement between them turning the energy in the rotating part into the heat of friction. As with auto brakes, the subject induction device when energized causes an infinitely variable attraction across an air gap between its two halves but instead of one being free and one being fixed, both are free to turn although this turning is alternately resisted by the inertia of the masses attached to opposite ends of the system, one being the car the other the flywheel. In operation, the rate of transfer of energy in each direction is directly proportional to the level of excitation of the induction coil which is controlled by ordinary driver actions while monitored and supervised by system controls. While the calculations noted herein were based on 15 and 30 MPH earmarks, the actual usable operating range will be more encompassing. Importantly, as noted, flywheel speeds are low within the system as described.
Additional calculations:
Let us consider what happens during a flywheel charging cycle using this system:
Let the cycle time be 3 seconds
First, we must recognize that the induction device is analogous to a variable torque slip clutch but without friction losses. For such equipment, the torque range is from zero to full name plate rating for continuous service but as much as 2 - ½ times name plate rating for intermittent service. The subject system is in the latter class. Regarding this induction modulator, it should be kept in mind that a friction device could be substituted for the induction unit.
So, to further explore this machine, let us select some reasonable level of torque within the range available to the machine. Arbitrarily, we pick 75 Ft-Lbs. This tells us that this is the amount of force coming through the induction device so long as the torque supplied to the device is greater than 75 Ft-Lbs. We don't know how much torque it takes to overcome the power flow disadvantage presented by the hypothetical 1 to 6 overdrive gearing so we need to make an estimate. Let us say 10 Ft-Lbs. (recall that no such losses are incurred when the power flow is reversed as the gearing becomes a 6 to 1 reduction drive,)
That leaves a net force to accelerate the flywheel of 65 Ft-Lbs. The cycle time is 3 seconds, the period to charge the flywheel and to slow the car before the friction brakes come into play.
Find the speed to which the flywheel is accelerated by this torque (L = 65 Ft=Lbs.)
First determine moment of inertia (I) for the flywheel:
Where m = mass, and r = radius of gyration
I = 1/2 (m) (r)^2
m = 3.1, and r = .5 Ft
I = ½ (3.1) (.5)^2 = 0.3875 slug-ft^2
Now find acceleration for the flywheel:
Let (L) represent torque, (a) represent angular acceleration, and (I) represent moment of inertia
Where L = (I) (a), and a = L/I = 65/.3875 = 167.7 rad/sec^2
Now find the angular distance traveled (s) in the time (t) of 3 seconds:
(s) = wt + ½ (a) (t)^2, when w = 0 rad/sec, s = ½ (a) (t)^2 = (.5) (167.7) (3)^2 =
754.65 radian
Now find the final angular speed (V) of the flywheel:
The change in speed V, (omega2 – omega1) = angular acceleration (a) time (t)
with a in rad/sec^2, t in seconds, and V in rad/sec
V = (167.7) x (3) = 503.1 rad/sec
Then, flywheel KE = ½(I) (V)^2
Where: I = 0.3875 slug-ft^2
Then, KE = ½ (0.3875) (503.1)^2 = 49040 Ft-Lbs
Find flywheel RPS and RPM:
Where Rev/second = radians/sec / 2pi, and rev/min = rev/sec x 60
503.1/2 pi = 80.1 rev/Sec, or 4804.2 RPM)
Now let us attempt to visualize what result we get when we return this energy to the car. We will assume that the system components are arranged in the most advantageous configuration.
The energy drained from the flywheel is fully controlled by the induction device; this modulation of the available flywheel energy gives full control of the available energy as to rate and duration and thereby quantity. This exceedingly high degree of control provides seamless recovered energy application to re-accelerate the car up until the car speed equals the equivalent proportional residual flywheel speed exhausting the transferable energy and completing the return cycle. During this return cycle, the passed energy goes through the 1 to 6 gearing which lowers its speed and raises its torque. This is of great importance as the ratio of the involved trial masses is about 33 to 1, as noted above. It is obvious that for the system to work this disparity in mass has to be accounted for and it is; when the receiving mass is low the system provides the speed needed; conversely when the receiving mass is high the system provides low speed and high torque required to motivate the greater mass of the car.
Very truly yours,
/S/
JOHN JONES, PE
President of South Essex Engineering
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