Dune Sandworm Intercept Calculator

How long until the worm arrives? Can the riders mount before it passes?

Starting values — Dune Sandworm Intercept Calculator

Worm speed (m/s)	22
Target distance (m)	1800
Fremen speed (m/s)	18
Approach distance (m)	900
Worm length (m)	120
Setup delay (s)	0

Change any control above and the results recompute from these.

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Lesson

The theory — Dune Sandworm Intercept Calculator

This is an interception problem, and its real content is that the target is not a point. A sandworm has length, so it does not arrive at an instant — it occupies the mount point for a window of time, and the question is not whether you beat it but whether you arrive inside that window.

What each symbol means

v_worm: the worm’s speed, 22 m/s, and the distance it must cover, 1800 m.
L: the worm’s length, 120 m. This is what converts an arrival time into an interval.
v_team: the team’s speed, 18 m/s, over their own approach distance of 900 m.
delay: the setup delay before they can move, added directly to their arrival time.

Where the formula comes from

Each arrival is just distance over speed. The worm’s head reaches the point at 1800 / 22 = 81.82 s; the team arrives at 900 / 18 = 50 s, plus any delay.
The worm’s tail is L behind its head, so it clears the point L / v_worm later: 120 / 22 = 5.45 s after the head, at 87.27 s.
So the opportunity is the interval from 81.82 s to 87.27 s. Arriving at 50 s is early — comfortably inside it, with the whole worm still to pass.

How to read what you see

Three times, each with its division shown: when the head arrives, when the tail finally clears, and when the team gets there. Comparing the third against the first two is the whole answer — and the wording of the verdict is careful about which of the two it beat.

Assumes: Constant speeds in straight lines, both parties starting at the same moment, and a worm that neither accelerates nor changes course. It is a timing model, not a chase: nothing here reacts to anything.
Breaks when: Being early is not automatically safe, and the model cannot tell you so. It compares arrival times only — it has no notion of being seen, of waiting exposed for thirty seconds, or of a worm that alters course because something is standing there. A longer or slower worm widens the window, which reads as good news, but the same change means more time spent alongside a very large animal.

why the worm's own length saves you 🖖

A sandworm is up to hundreds of meters of moving body, not a single point, so missing the head by a few seconds doesn't mean missing the ride — any exposed segment is still hookable. That's why the real deadline isn't when the worm arrives, it's when it finishes passing: arrival time plus body length divided by speed. It's the same forgiveness that lets you catch a moving train through any open door rather than sprinting for the exact front car — the longer or slower the target, the wider your effective catching window.

it's a race between two clocks 🖖

This tool doesn't reward raw speed — it compares two countdowns: how long until the worm's window shuts, and how long until your team is ready and in position. You win the moment the second one finishes first. The single term entirely in your hands is the setup delay, a flat number of seconds added no matter the distances or speeds, so trimming it is often the cheapest way to turn a miss into a catch.

a faster worm hurts more than you'd think 🖖

Your whole deadline is (D + L) / v — target distance plus body length, divided by worm speed. Because speed sits in the denominator, your available time drops as a hyperbola, not a straight line: at a slow crawl a few extra m/s barely register, but near a fast worm the same jump can halve your window. A worm only slightly quicker can flip a comfortable catch into a clean miss.

Problems solved in full

Worm closing at 22 m/s and team crossing sand at 18 m/s 5 steps
A worm 1800 m out, closing at 22 m/s, 120 m from head to tail. Your team waits 900 m from the mount point and crosses sand at 18 m/s, moving the instant the thumper starts. Do you make it, and how fast could that worm be before you don't?
1. Distance over speed puts the worm's head on the mount point at 81.82 s. That is the number most people then race against, and it is the wrong one.
  
  $t_{head} = \frac{D}{v _{w}} = \frac{1800}{22} = 81.82 s$
2. You are not trying to beat the head. A hook lands anywhere along the body, so the last usable instant is when the tail goes past, and that grace is the worm's own length divided by its own speed. Note what that ratio implies: size buys time only in proportion to how slowly the worm is moving.
  
  $\frac{L}{v _{w}} = \frac{120}{22} = 5.45 s ⟹ t_{tail} = 81.82 + 5.45 = 87.27 s$
3. Your side is a delay plus a run. Setup time is dead time before the sprint rather than a drag on it, so it adds to the total instead of scaling it. Here it is 0, and 900 m at 18 m/s is 50 s flat.
  
  $t_{team} = δ + \frac{d}{u} = 0 + \frac{900}{18} = 50 s$
4. Subtract, and the margin is 37.27 s. Put that back on the sand: you are standing on the mount point with the worm's head still 1800 − 22 × 50 = 700 m away, waiting.
  
  $margin = t_{tail} - t_{team} = 87.27 - 50 = 37.27 s$
5. The four inputs never mattered separately. Everything on the worm's side collapses into one time, and everything on yours into another; the verdict is only which of the two is larger. That is why a 200 m worm and a 120 m worm can be the same problem.
  
  $margin = \frac{D + L}{v _{w}} - (δ + \frac{d}{u})$
Answer

37.27 s of margin: you are early, not lucky. Turn the inequality round and it names a ceiling. At this geometry the intercept survives any worm up to 1920 ÷ 50 = 38.4 m/s and fails above it, and the tool never prints that number because it only ever scores the worm you gave it. What happens to your grace on the way up to that ceiling is the part worth keeping: the body window is L ÷ v, so the same 120 m worm that hands you 5.45 s at 22 m/s hands you 3.13 s at 38.4 m/s. The worm's length is worth least exactly when the worm is fastest. Setup delay is harsher still, because it is subtracted before the division rather than after: carry 45 s of it and the ceiling falls from 38.4 m/s to 1920 ÷ 95 = 20.2 m/s, under the 22 m/s worm you started with. Same worm, same sand, and the run is now unwinnable. Nothing moved but your watch.

$v_{w}^{m a x} = \frac{D + L}{δ + d / u} = \frac{1920}{50} = 38.4 m/s$
The Fast worm state of 45 m/s and a team 900 m out 6 steps
The Fast worm state: 45 m/s, 200 m long, 1500 m from the mount point, your team still 900 m out at 18 m/s with no setup delay. Bigger worm, shorter chase than the last problem. Show that neither helps, and price the two ways out.
1. 45 m/s eats 1500 m in 33.33 s, and the 200 m body then buys 4.44 s. That is less grace than the 120 m body bought at 22 m/s, despite being 67% longer. Grace is L ÷ v, and the denominator won.
  
  $t_{head} = \frac{1500}{45} = 33.33 s, \frac{L}{v _{w}} = \frac{200}{45} = 4.44 s$
2. So the boarding window shuts at 37.78 s.
  
  $t_{tail} = 33.333 + 4.444 = 37.78 s$
3. Your side has not changed at all: 900 m at 18 m/s is the same 50 s it was in the first problem. Every difference between the two runs sits on the worm's side of the ledger.
  
  $t_{team} = 0 + \frac{900}{18} = 50 s$
4. You are 12.22 s late, which is not a near miss. In those 12.22 s the worm covers 550 m, so the tail you were reaching for is half a kilometre downrange by the time your boots reach the mount point.
  
  $margin = 37.78 - 50 = - 12.22 s$
5. One way out is speed. The distance is fixed and the window is fixed, so the requirement is 900 m inside 37.78 s. That is 32% above 18 m/s, and speed is the one quantity on your side you cannot simply decide to have.
  
  $u_{m i n} = \frac{d}{( D + L ) / v _{w}} = \frac{900}{37.78} = 23.82 m/s$
6. The other way out is geometry. Keep 18 m/s and your reachable radius is speed times window: 680 m, which is 220 m nearer than where you chose to wait. That choice is made before the thumper drops, not after.
  
  $d_{m a x} = u \cdot \frac{D + L}{v _{w}} = 18 \times 37.78 = 680 m$
Answer

The run misses by 12.22 s, and both rescues are priced at the same 32%. That is not a coincidence. The ceiling for this geometry is (1500 + 200) ÷ 50 = 34 m/s, so this worm sits 45 ÷ 34 = 1.32 times over it, and dividing the margin equation through by that ratio makes it appear everywhere at once: the speed you would need is 18 × 1.32 = 23.82 m/s, and the distance you can afford is 900 ÷ 1.32 = 680 m. A single number, how far past its ceiling the worm is, prices every fix simultaneously. It also settles which fix is real. You choose where to wait; you do not choose how fast you run. So the ambush point is the only lever that ever moves, and moving it the full 220 m brings the margin to exactly 0 — a dead heat, and a dead heat is not a boarding.

$\frac{u _{m i n}}{u} = \frac{d}{u} \cdot \frac{v _{w}}{D + L} = \frac{v _{w}}{v _{w}^{m a x}} = \frac{45}{34} = 1.32$

References (2)

The premise: Frank Herbert, Dune. First published 1965; Ace Books edition, ISBN 978-0-441-17271-9.
The mathematics underneath the fiction — interception and pursuit treated properly: Paul J. Nahin, Chases and Escapes: The Mathematics of Pursuit and Evasion. Princeton University Press, 2007. ISBN 978-0-691-15501-2.

By the 42Math team · Updated 2026-09-03

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

All Tidal Zone lessons →

Example problems

Clean intercept - With no setup delay, the Fremen team reaches the mount point at 50 seconds — well before the worm (arriving at 81.8s) even finishes passing at 87.3s. That's a comfortable 37-second boarding margin, plenty of time to plant the hooks.
Late setup - Same speeds and distances as a clean intercept, but a 45-second setup delay pushes the team's arrival to 95 seconds — 7.7 seconds after the worm's tail has already cleared the mount point. The worm doesn't get faster; the team just runs out of runway.
Fast worm - At 45 m/s this worm covers 1,500m in just 33.3 seconds, and even its extra-long 200m body only buys 4.4 seconds of grace before the tail clears at 37.8s — nowhere near enough for a 50-second Fremen approach. Some worms are simply too fast to catch on foot.