6.2 Meshed Gears (Çarklar), Concentric Pulleys, Belt Directions & Turn Ratios

Meshed Gears (Çarklar), Concentric Pulleys, Belt Directions & Turn Ratios

Quick Method: Solve every gear and pulley problem (Çarklar ve Kasnaklar) in two separate passes: Pass 1 (Direction): Trace + (clockwise) and - (counterclockwise) across contacts—flipping sign at external gear teeth and crossed (X) belts, and keeping the same sign across open (=) belts and concentric axles. Pass 2 (Turn Count): Apply $n_1 \cdot r_1 = n_2 \cdot r_2$ across teeth/belts and $n_{\text{inner}} = n_{\text{outer}}$ on shared concentric axles.

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1. The Four Mechanical Coupling Laws

Every connection between two rotating wheels in a TR-YÖS figure belongs to one of four fundamental mechanical couplings:

Law 1: Externally Meshed Gears (Dıştan Dişli Çarklar) & Tangent Friction Wheels

When two gears mesh tooth-to-tooth on their outer rims, their contact edges must move in the same linear direction, forcing their centers to rotate in opposite angular directions (zıt yönlü):

• If Gear 1 turns clockwise (saat yönünde, CW), Gear 2 turns counterclockwise (saat yönünün tersine, CCW).
• In a single linear chain of $k$ externally meshed gears ($G_1 - G_2 - G_3 - \dots - G_k$):
  • All odd-indexed gears ($G_1, G_3, G_5, \dots$) rotate in the same direction.
  • All even-indexed gears ($G_2, G_4, G_6, \dots$) rotate in the opposite direction.

Law 2: Open (Direct) Belt Pulleys (Düz Bağlı Kasnaklar)

When an uncrossed belt wraps around the outer rims of two pulleys, the top strand pulls the top of both pulleys the same way. Both pulleys rotate in the exact same direction (aynı yönlü).

Law 3: Crossed (Figure-8) Belt Pulleys (Çapraz Bağlı Kasnaklar)

When the belt crosses over itself in an X shape between the two pulleys, the top edge of the driving pulley feeds into the bottom edge of the driven pulley. Therefore, crossed-belt pulleys rotate in opposite directions (zıt yönlü), behaving directionally just like externally meshed gears.

Law 4: Concentric / Co-Axial Wheels (Eş Merkezli / Çakışık Çarklar)

When a smaller gear (or pulley) is bolted concentrically onto the face of a larger gear on the same central axle, they form a single rigid body:

• They rotate in the exact same direction ($\text{direction}_A = \text{direction}_B$).
• They complete the exact same number of revolutions ($n_A = n_B$), regardless of their radii or tooth counts!

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2. The Fundamental Turn-Ratio Conservation Equation

Whenever Wheel 1 drives Wheel 2 on a separate axle via meshed teeth or a non-slipping belt, the linear arc length $s = 2\pi r \cdot n$ (or total number of teeth passing the contact point) is identical for both wheels. Consequently, the number of revolutions $n$ is inversely proportional to the radius $r$ or tooth count $T$:

$$n_1 \cdot r_1 = n_2 \cdot r_2 \quad \Longleftrightarrow \quad n_1 \cdot T_1 = n_2 \cdot T_2$$

Intermediate Gear Cancellation in Pure Series Chains

In a pure chain of externally meshed gears $G_1 - G_2 - G_3 - \dots - G_k$ (with no concentric step-up gears in between):

$$n_1 \cdot T_1 = n_2 \cdot T_2 = n_3 \cdot T_3 = \dots = n_k \cdot T_k$$

Notice that all intermediate gears ($G_2, G_3, \dots, G_{k-1}$) act purely as idler gears: their radii and tooth counts cancel out completely! The turn ratio between the first and last gear depends only on $T_1$ and $T_k$: $n_k = n_1 \cdot (T_1 / T_k)$.

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3. Patterned Sector Wheels & Load Lifting

Patterned Wheel Appearance After Fractional Turns (Daire Dilimi Görünümü)

Many TR-YÖS items divide a gear into $S$ equal pie sectors (eş daire dilimi—typically $S = 4, 6,$ or $8$) with one or two shaded wedges, and ask for its visual appearance after a driving gear completes $n_1$ turns:

1. Compute the driven wheel's exact turn count $n_2$ as a mixed fraction: $n_2 = k + \frac{a}{b}$, where $k \in \mathbb{Z}$ is the whole number of full turns.
2. Discard $k$ completely: Every integer revolution ($360^\circ$) restores the wheel to its initial visual state.
3. Convert the fractional turn $\frac{a}{b}$ into sector steps: $\Delta\text{sectors} = \frac{a}{b} \times S$.
4. Shift every shaded wedge by $\Delta\text{sectors}$ positions in the driven wheel's rotation direction.

Lifting Hanging Loads & Rolling Wheels (Yüklerin Yükselmesi ve Yuvarlanma)

• Cord Wound Around a Drum of Radius $r$: When a pulley of radius $r$ completes $n$ revolutions, it winds or unwinds a vertical rope length equal to $n$ circumferences:

$$h = n \cdot (2\pi r)$$

• Differential Co-Axial Winch: If Load $X$ hangs from an outer drum of radius $R$ and Load $Y$ hangs on the opposite side from a concentric inner drum of radius $r$, one revolution moves $X$ by $2\pi R$ and $Y$ by $2\pi r$ in opposite vertical directions, changing their vertical separation by $\Delta h = 2\pi(R + r) \cdot n$.

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Summary Reference Table of Mechanical Couplings

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Worked Examples

Worked Example 6.3: Compound Gear & Concentric Pulley Train

Problem: Pulley $A$ ($r_A = 6\text{ cm}$) is connected by a crossed belt to Pulley $B$ ($r_B = 4\text{ cm}$). Gear $C$ ($r_C = 8\text{ cm}$) is mounted concentrically on the same axle as Pulley $B$, and meshes externally with Gear $D$ ($r_D = 3\text{ cm}$). If Pulley $A$ rotates 2 turns clockwise (CW), find the number of turns and rotation direction of Gear $D$.

Solution:

1. Pass 1 — Trace Directions:
   • Pulley $A$: CW.
   • $A \to B$ (crossed belt reverses direction): Pulley $B$ turns CCW.
   • $B \to C$ (concentric axle preserves direction): Gear $C$ turns CCW.
   • $C \to D$ (external gear mesh reverses direction): Gear $D$ turns CW.
2. Pass 2 — Compute Revolution Counts:
   • Across the belt from $A$ to $B$:

$$n_A \cdot r_A = n_B \cdot r_B \implies 2 \cdot 6 = n_B \cdot 4 \implies n_B = 3\text{ turns}$$

• On the shared concentric axle of $B$ and $C$:

$$n_C = n_B = 3\text{ turns}$$

• Across the meshed teeth from $C$ to $D$:

$$n_C \cdot r_C = n_D \cdot r_D \implies 3 \cdot 8 = n_D \cdot 3 \implies n_D = 8\text{ turns}$$

1. Conclusion: Gear $D$ completes 8 turns clockwise (CW).

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Worked Example 6.4: Sector Pattern After Fractional Revolutions

Problem: Gear $X$ has $45$ teeth and meshes externally with Gear $Y$, which has $12$ teeth. Gear $Y$ is divided into $8$ equal numbered sectors ($1$ through $8$ arranged clockwise), with Sector $1$ currently at the top-right position. If Gear $X$ rotates $\frac{2}{3}$ of a turn counterclockwise (CCW), how many sector positions and in what direction do the sectors of Gear $Y$ shift?

Solution:

1. Find the rotation direction of Gear $Y$: Since $X$ turns CCW and meshes externally with $Y$, Gear $Y$ turns clockwise (CW).
2. Calculate the revolution count $n_Y$:

$$n_X \cdot T_X = n_Y \cdot T_Y \implies \frac{2}{3} \cdot 45 = n_Y \cdot 12$$ $$30 = 12 n_Y \implies n_Y = \frac{30}{12} = \frac{5}{2} = 2 + \frac{1}{2}\text{ turns}$$

1. Discard whole turns and convert the fractional part to sectors: Drop the $2$ full turns. The remaining $\frac{1}{2}$ turn CW shifts every sector by:

$$\Delta\text{sectors} = \frac{1}{2} \times 8 = 4\text{ sectors CW}$$

(Equivalently, every shaded sector rotates $180^\circ$ to the diametrically opposite position.)

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Worked Example 6.4B: Rolling Spool and Rack-and-Pinion Displacement (Kremayer Dişli ve Yuvarlanan Makara)

Problem: A double spool consists of an outer wheel of radius $R = 20\text{ cm}$ and a concentric inner drum of radius $r = 8\text{ cm}$. The outer wheel rolls without slipping toward the right (sağa doğru) along a horizontal floor for $n = 2.5$ revolutions, while a rope wound over the top of the inner drum is attached to a hanging block $M$ over a fixed corner pulley to the left of the spool; the taut horizontal rope leaves the top of the inner drum toward that pulley. How far does the center of the spool move horizontally, and by how much does block $M$ move? (Take $\pi = 3$.)

Solution:

1. Horizontal translation of the rolling center ($x_{\text{center}}$): As the outer wheel of radius $R = 20\text{ cm}$ rolls $n = 2.5$ turns clockwise to the right, its axle advances horizontally by:

$$x_{\text{center}} = n \cdot (2\pi R) = 2.5 \cdot (2 \cdot 3 \cdot 20) = 2.5 \cdot 120 = 300\text{ cm}$$

1. Winding action of the concentric inner drum ($x_{\text{wind}}$): Because the inner drum ($r = 8\text{ cm}$) is concentric with the outer wheel, it also rotates $n = 2.5$ turns clockwise, winding in:

$$x_{\text{wind}} = n \cdot (2\pi r) = 2.5 \cdot (2 \cdot 3 \cdot 8) = 2.5 \cdot 48 = 120\text{ cm}$$

1. Total rope displacement for block $M$: Since the center moves $300\text{ cm}$ to the right AND the top of the inner drum winds $120\text{ cm}$ in the same rightward direction, block $M$ is pulled upward by:

$$h_M = x_{\text{center}} + x_{\text{wind}} = 300 + 120 = 420\text{ cm}$$

(Note: If the rope had exited from the bottom of the inner drum, the winding would oppose the forward roll, giving $h_M = 300 - 120 = 180\text{ cm}$.)

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Common TR-YÖS Gear & Pulley Traps to Avoid

• Trap 1 — Using $n_1 r_1 = n_2 r_2$ on Concentric Wheels: Concentric wheels are welded together ($n_1 = n_2$). Never multiply their turns by their radii against each other; only apply $n_1 r_1 = n_2 r_2$ where a belt or gear tooth bridges two different axles.
• Trap 2 — Wasting Time on Intermediate Idler Gears: In a simple chain $G_1 - G_2 - G_3 - G_4$ with no concentric shafts, ignore the tooth counts of $G_2$ and $G_3$ completely; jump straight to $n_1 T_1 = n_4 T_4$ and count odd/even positions for direction.
• Trap 3 — Shifting Sectors in the Driving Wheel's Direction: Always determine the driven wheel's own rotation direction first before shifting its colored wedges.

Review the Method

To find the visual sector pattern of a divided wheel after n revolutions, discard the integer part of n and advance each colored sector in the wheel's rotation direction by the fractional part multiplied by the total number of equal sectors.