Math Bridge: Why can four short hops beat one long hop?

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Math BridgeWSNStruggle-friendly runway

Why can four short hops beat one long hop?

Connect path-loss exponent, hop count, hop length, and total transmit-power ratio for a WSN route.

Packet Pete, the guidePacket Pete guides
The one targetDerive the geometry-only power ratio for equal WSN hops.
The chapter caseA 200 m route split into four 50 m hops at n=2 and n=3.
What it buys youA reason to test multi-hop without hiding relay energy.

A field team has a real problem to settle: Why can four short hops beat one long hop? They must decide what happens before they change equal hops on the device. Predict the direction first.

See the relationship first

The figure reads from left to right. The blue card is equal hops. The middle card uses this page's rule. The green card is hop length. Follow the arrows: set the input, use the rule, then read the result and its unit.

The audit later on checks more than one number. Here, the added model uses the baseline named below and holds every other chapter value fixed. That sentence bridges the fixtures, so the numbers do not change without a reason.

Equal hops changes hop length An input card leads through the page rule to the hop length result. SET INPUT ONE CONTROL USE RULE predict calculate check units READ RESULT
Follow the arrows. A larger path-loss exponent punishes the long hop more strongly, so equal subdivision buys more transmit-power saving before relay costs are added.

Derive the baseline in four moves

  1. 1

    Name the input. The chapter baseline for equal hops is 4.

  2. 2

    Name the rule. R=k(D/k)^n/D^n=k^(1-n) k=4: R(n=2)=0.250 → 6.02 dB; R(n=3)=0.0625 → 12.04 dB

  3. 3

    Put in the chapter value. Set equal hops to 4. The page rule gives hop length as 50.0 m.

  4. 4

    Read the result. Keep m next to the value. Use it only within the limits on this page.

Predict, then change equal hops

Try Predict what happens to hop length. Move one control, calculate, then check your idea.

4
Chapter baseline
Hop length

Observe A larger path-loss exponent punishes the long hop more strongly, so equal subdivision buys more transmit-power saving before relay costs are added. Reset to 4 and compare hop length.

Explain Only equal hops moves here. The other chapter values stay fixed.

Check yourself

What should you do before you trust the result?
Answer: Predict its direction, use the shown rule, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only equal hops moves. Field effects named in the page limits stay fixed.

Honesty note. This teaching model changes only equal hops. It replays the chapter baseline; it is not a field prediction.

1. Start with the physical question

Derive the geometry-only power ratio for equal WSN hops. A reason to test multi-hop without hiding relay energy.

Packet Pete: Keep the units and the model boundary visible from the first line.

2. Name every algebra move

1

Split the distanceEach hop has length D/k.

2

Apply the path lawOne hop needs power proportional to (D/k)^n.

3

Count all transmittersMultiply the one-hop term by k.

4

Cancel the common distancek(D/k)^n/D^n becomes k^(1−n).

5

Convert the ratio to decibelsSaving=10log10(1/R).

3. Reproduce the chapter case

R=k(D/k)^n/D^n=k^(1−n)
k=4: R(n=2)=0.250 → 6.02 dB; R(n=3)=0.0625 → 12.04 dB

The arithmetic reproduces the chapter case while keeping its assumptions explicit.

4. Try the controlling input

TryMove the control and watch every displayed result come from the shown formula.

Equal hops
Hop length
Free-space ratio
Free-space saving
Obstructed ratio
Obstructed saving

ObserveAt four hops, each hop is 50.0 m. The geometry-only ratio is 0.250 at n=2 and 0.063 at n=3.

ExplainA larger path-loss exponent punishes the long hop more strongly, so equal subdivision buys more transmit-power saving before relay costs are added.

Technical boundaries.

This compact engine isolates one relationship; it is not a deployment certificate.

Radios
Receive, forwarding, startup, acknowledgement, and retry energy are omitted
Geometry
Equal hop lengths and one shared path exponent are assumed
Evidence
The result does not prove delivery, latency, or battery life

Measure the real system and reopen the decision when its inputs change.

5. Put relay tax back into the decision

Measure receive and forwarding energy per relay, retry rate, traffic pressure, and idle cost. Multi-hop wins only when the geometric saving exceeds those added costs.

6. Record the route assumption

Keep total distance, hop count, per-hop distance, path exponent, radio state energy, traffic load, fade margin, result, owner, and retest trigger.

7. Check yourself

Why is each hop 50 m?
Answer: The 200 m total is divided into four equal parts: 200/4=50 m.
Why is the n=3 ratio 0.0625?
Answer: Four equal hops give 4^(1−3)=4^−2=1/16=0.0625.
Does 12.04 dB prove four hops use less battery?
Answer: No. It is geometry-only transmit power; relay receive, forwarding, retries, and idle energy can reverse the result.
Honesty boundary.

The worked values are traceable chapter examples or explicitly labelled teaching assumptions.

200 m
Explicit illustrative corridor distance
n=2 and n=3
Free-space and bounded obstructed examples
12.04 dB
Geometry-only transmit-power saving

Correct, not complete: field evidence still decides acceptance.

Return to WSN Routing Fundamentals with the maths unpacked →