The Trapped Deficit \(\Delta t\). The Equation for Transforming Time into Newton-Meters

Many dogmas have grown around advanced asymmetric systems. Academic orthodoxy rejects any innovations that violate traditional, simplified engineering models. Skeptics raise the objection that in classical rotational mechanics, input power (\(P_{in}\)) and output power (\(P_{out}\)) at ideal efficiency (\(\eta = 1\)) must be equal, which at equal angular velocities (\(n_1 = n_2 \implies \omega_1 = \omega_2\)) absolutely forces the equality of torques (\(M_{in} = M_{out}\)):

$$P = M \cdot \omega$$

It is high time to step out of this textbook cage and demonstrate mathematically why this linear model is completely inadequate for asymmetric systems. Let's go to the board.

Step 1: Proof of Mechanical Asymmetry (Vector Geometry)

Classical mechanics considers the circumferential force at the contact point of gears as a vector perfectly perpendicular to the radius. This is a cardinal oversimplification. In a three-axis transmission with an external variable-pitch chain, the contact points and angles of attack (\(\alpha\)) change dynamically as a function of the system's rotation angle (\(\theta\)).

The resultant force (\(F_R\)) transmitted by the chain link to the tooth of the output gear with a larger radius (\(R\)) is not identical to the input force on the smaller radius (\(r\)). Writing this out into vector components, we obtain the actual torque equation:

$$M_{out}(\theta) = F_R(\theta) \cdot R \cdot \cos(\alpha(\theta))$$

Since the system forces rigid kinematic synchronization (equality of angular velocities \(\omega_{in} = \omega_{out}\)), and the gear diameters are extremely different (\(R > r\)), the backward reaction force vector is unable to cancel out the input force in a 1:1 ratio. A mathematical reversal of the lever's path occurs – the integral of the torque over a full rotation cycle (\(2\pi\)) shows a constant, measurable surplus:

$$\Delta M = \frac{1}{2\pi} \int_{0}^{2\pi} \left[ M_{out}(\theta) - M_{in}(\theta) \right] d\theta > 0$$

Step 2: Mathematical Bridge to the Energy from Time Theory

Since \(\Delta M > 0\) while \(\omega = const\), the classical energy balance in three-dimensional space screams of an error in academic assumptions. To close the equation without breaking the laws of physics, we must introduce time (\(t\)) as an active variable of the system (the fourth dimension), not just a passive counter.

The linear velocity of a point on the circumference of the small gear is \(v_1 = \omega \cdot r\), and on the circumference of the large gear is \(v_2 = \omega \cdot R\). Since \(R > r\), the obvious relationship \(v_2 > v_1\) occurs.

The marked control points on both gears return to their starting positions in the exact same time \(T\). This means that in a unit of time, matter on the circumference of the large gear covers the distance:

$$S_2 = v_2 \cdot T = \omega \cdot R \cdot T$$

The difference in linear distances covered by the points in the same time interval \(T\) is:

$$\Delta S = S_2 - S_1 = \omega \cdot T \cdot (R - r)$$

In a classical approach, this would require additional time. Here, time \(T\) is a rigid frame for both velocities. The system generates a local time deficit (\(\Delta t\)) needed to cover this extra distance in space:

$$\Delta t = \frac{\Delta S}{v_2} = \frac{\omega \cdot T \cdot (R - r)}{\omega \cdot R} = T \left(1 - \frac{r}{R}\right)$$

Step 3: The Energy from Time Equivalent Equation

According to Jerzy Żbikowski's Energy from Time Theory, this local deficit (contraction/burning of time \(\Delta t\)) transforms directly into mechanical energy, actualizing a relativistic loss in the temporal structure. We introduce the time-energy transformation constant (\(k_t\)), which determines how many joules are generated by a second of time compressed by geometry:

$$\Delta E = k_t \cdot \Delta t$$

Substituting the calculated time deficit into the power balance of the open system, we get the final, closed equation of the entire process:

$$P_{out} \cdot T = P_{in} \cdot T + k_t \cdot T \left(1 - \frac{r}{R}\right)$$

Dividing both sides by time \(T\), we get the pure formula for the system's output power:

$$P_{out} = P_{in} + k_t \left(1 - \frac{r}{R}\right)$$

Taking into account the fundamental relationship that \(P = M \cdot \omega\) and \(\omega = const\), we can derive the final form for the output torque of the transmission:

$$M_{out} \cdot \omega = M_{in} \cdot \omega + k_t \left(1 - \frac{r}{R}\right) \implies M_{out} = M_{in} + \frac{k_t}{\omega} \left(1 - \frac{r}{R}\right)$$

Summary on the Board

Here is the ready mathematical proof in black and white. The increase in torque (\(\Delta M\)) does not come from "nowhere." It is directly proportional to the transformation constant \(k_t\) and the asymmetry of the radii \(\left(1 - \frac{r}{R}\right)\), and inversely proportional to the angular velocity \(\omega\).

Mathematics does not lie – if a unique geometric system forced material points to cover a longer distance in the same time, the missing time (\(\Delta t\)) had to manifest itself in our three-dimensional world as additional newton-meters.

I am putting down the chalk. What do the skeptics have to say?

Safety Clause: The content, theoretical descriptions, and mathematical proofs regarding the Energy from Time Theory presented on this blog are solely for journalistic, scientific-research, and literary purposes. No analyses, formulas, or conclusions drawn based on the operation of asymmetric mechanical systems constitute investment recommendations, financial advice, or commercial advice within the meaning of applicable law. All decisions related to financial markets, trading, and capital allocation are made entirely at your own risk.