Hard Disk Drive Geometry

🔷 What is Disk Geometry?

Disk geometry refers to the physical structure and layout of a hard disk, including how data is organized and accessed.

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1. Basic Components of HDD

🔷 1. Platter

• A circular disk surface where data is stored

• Made of rigid material (e.g., aluminum)

• Coated with magnetic material to store bits

👉 A disk may have multiple platters

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🔷 2. Surface

• Each platter has two sides (top and bottom)

• Each side is called a surface

• Data is stored on both surfaces

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🔷 3. Spindle

• Central rod that holds platters together

• Connected to a motor

• Rotates the platters at constant speed

📌 Speed measured in RPM (Rotations Per Minute)

• Typical: 7200 – 15000 RPM

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🔷 4. Disk Head

• Used to read/write data

• One head per surface

• Detects or changes magnetic patterns

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🔷 5. Disk Arm

• Holds the disk head

• Moves across the disk to access different locations

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2. Data Organization on Disk

🔷 1. Tracks

• Concentric circular paths on disk surface

• Each track stores data

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🔷 2. Sectors

• Each track is divided into small units called sectors

• Typical size: 512 bytes

👉 Disk is viewed as:

Array of sectors (0 to n-1)

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🔷 3. Cylinders

• Tracks at the same radius across all platters

• Form a cylinder

👉 Accessing a cylinder avoids extra head movement

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3. Disk Access Time Components

Disk I/O time consists of three main parts:

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🔴 1. Seek Time

• Time to move disk arm to correct track

• Includes:

  • Acceleration

  • Movement

  • Deceleration

  • Settling

👉 Most expensive operation

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🔵 2. Rotational Latency

• Time waiting for sector to rotate under head

• Depends on disk speed

👉 Example:

• 10,000 RPM → ~6 ms per rotation

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🟢 3. Transfer Time

• Time to actually read/write data

• Happens once sector is under head

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🎯 Total Disk Access Time

Access Time = Seek Time + Rotational Latency + Transfer Time

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4. Disk Operation Example

To read a block:

1. Move head to correct track (seek)

2. Wait for correct sector (rotation delay - latency)

3. Read/write data (transfer)

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5. Important Optimization Details

🔷 1. Track Skew

• Sectors are slightly offset between tracks

👉 Why?

• When moving to next track:

• Disk rotates during head movement

• Prevents missing next sector

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🔷 2. Zoned Bit Recording (Multi-Zone)

• Outer tracks have:

  • More space

  • More sectors

👉 Disk divided into zones

• Each zone has same sectors per track

• Outer zones have more sectors

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🔷 3. Disk Cache (Track Buffer)

• Small memory (8–16 MB) inside disk

• Stores recently read/written data

📌 Benefits:

• Faster repeated access

• Improves performance

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🔴 Write Policies

Type	Description
Write-through	Data written to disk before completion
Write-back	Data stored in cache first

👉 Write-back is faster but risky (power failure)

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6. Key Insights

• Disk performance depends heavily on:

  • Head movement

  • Rotation delay

• Sequential access is:
  👉 Much faster than random access

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7. Summary

• HDD consists of:

  • Platters, surfaces, tracks, sectors

• Access time = seek + rotation + transfer

• Optimizations:

• Track skew

• Zoning

• Disk cache

Problem 1: Basic Disk Access Time

Given:

• Seek time = 8 ms

• Rotational speed = 7200 RPM

• Transfer size = 4 KB

• Transfer rate = 100 MB/s

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🔷 Find:

👉 Total disk access time

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✏️ Solution:

Step 1: Convert RPM → Rotation time

$$\text{Rotation time} = \frac{60}{7200} = 0.0083 \text{ sec} = 8.33 \text{ ms}$$

👉 Average rotational latency = half rotation:

$$= \frac{8.33}{2} = 4.17 \text{ ms}$$

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Step 2: Transfer Time

$$\text{Transfer time} = \frac{4 \text{ KB}}{100 \text{ MB/s}}$$

Convert:

• 4 KB = 4096 bytes

• 100 MB/s = $100\times 1024\times 1024$

$$= \frac{4096}{104857600} \approx 0.000039 \text{ sec} = 0.039 \text{ ms}$$

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Step 3: Total Access Time

$$\text{Total} = \text{Seek} + \text{Rotation} + \text{Transfer}$$
$$= 8 + 4.17 + 0.039 \approx \boxed{12.21 \text{ ms}}$$

Problem 2: Worst-Case vs Best-Case Access Time

🔷 Given:

• Disk speed = 6000 RPM

• Average seek time = 6 ms

• Maximum seek time = 12 ms

• Transfer time = 0.05 ms

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🔷 Find:

1. Best-case access time

2. Worst-case access time

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✏️ Solution:

Step 1: Rotation Time

$$\text{Rotation time} = \frac{60}{6000} = 0.01 \text{ sec} = 10 \text{ ms}$$

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🔹 Best Case

• Seek time = 0 ms (already on track)

• Rotational latency = 0 ms (sector just arrived)

$$\text{Best case} = 0 + 0 + 0.05 = \boxed{0.05 \text{ ms}}$$

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🔹 Worst Case

• Seek time = 12 ms

• Rotational latency = full rotation = 10 ms

$$\text{Worst case} = 12 + 10 + 0.05 = \boxed{22.05 \text{ ms}}$$

Problem 3: Sequential vs Random Access 

🔷 Given:

• Disk speed = 7200 RPM

• Average seek time = 9 ms

• Transfer rate = 120 MB/s

• Block size = 4 KB

• Number of blocks = 100

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🔷 Find:

👉 Total time for:

1. Sequential read of 100 blocks

2. Random read of 100 blocks

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✏️ Solution:

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🔹 Step 1: Rotation Time

$$\text{Rotation time} = \frac{60}{7200} = 8.33 \text{ ms}$$

👉 Average rotational latency:

$$= 4.17 \text{ ms}$$

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🔹 Step 2: Transfer Time per Block

$$4 \text{ KB} = 4096 \text{ bytes}$$
$$\text{Transfer time} = \frac{4096}{120 \times 1024 \times 1024} \approx 0.000032 \text{ sec} = 0.032 \text{ ms}$$

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Case 1: Sequential Access

👉 Key idea:

• Only one seek + one rotation

• Then continuous transfer

Total Transfer:

$$100 \times 0.032 = 3.2 \text{ ms}$$

Total Time:

$$= 9 + 4.17 + 3.2 = \boxed{16.37 \text{ ms}}$$

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Case 2: Random Access

👉 Key idea:

• Each block requires:

  • Seek + rotation + transfer

Time per block:

$$= 9 + 4.17 + 0.032 = 13.202 \text{ ms}$$

Total:

$$100 \times 13.202 = \boxed{1320.2 \text{ ms}}$$

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Final Insight 

Access Type	Time
Sequential	16.37 ms
Random	1320.2 ms (~1.3 sec)

👉 Sequential access is orders of magnitude faster than random access