The A-Pillar Blind Spot: How Much of the Road Can't You See?

Introduction and Aim

I'm learning to drive, and my instructor keeps telling me to "check properly at junctions." Last week, while waiting to turn, a cyclist seemed to appear from nowhere on my left—I was focusing on the road ahead and they'd come from a hidden spot behind the A-pillar. The image below shows what I mean: the A-pillar is the vertical strip of metal between the windscreen and side window that holds the roof up. A C

This got me thinking—I could probably use some geometry to work out how large the blind spot actually is, and how it changes with distance. That's what this exploration is about. A

Part 1: Setting Up the Geometry

I parked my car safely at home and took measurements to model the blind spot geometry: E

[Diagram: Top-down view showing eye position O, A-pillar edges A and B]
Measurements: OA = distance to far edge, OB = distance to near edge, AB = pillar width

• A-pillar width: $AB = 12$ cm
• Distance from eye to near edge of A-pillar: $OB = 60$ cm
• Distance from eye to far edge of A-pillar: $OA = 68$ cm
• Distance from car to pavement edge: approximately 2.5 m when parked

Important assumption: My eye is not equidistant from both edges of the A-pillar—it's closer to the near edge (B). This means the triangle formed is not isosceles, so I cannot simply use $\tan(\theta/2) = \frac{w/2}{d}$. However, since the difference is small (60 cm vs 68 cm), I will first calculate using the cosine rule for accuracy, then check whether the simpler approximation gives similar results. E

Using the cosine rule to find the blind spot angle $\theta$:

$$AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(\theta)$$

$$12^2 = 68^2 + 60^2 - 2(68)(60)\cos(\theta)$$

$$144 = 4624 + 3600 - 8160\cos(\theta)$$

$$\cos(\theta) = \frac{8080}{8160} = 0.990$$

$$\theta = \cos^{-1}(0.990) = 8.1°$$

The blind spot angle is approximately 8.1°. But this angle alone doesn't tell us much—the key question is: what width of pavement does this obscure at various distances? D

Part 2: Blind Spot Width at Distance

The critical insight is that the blind spot widens with distance. Using similar triangles, I can find the width of the hidden region at any distance from my eye. C

Lines from O through A and B intersect the sidewalk at E and D. Since triangles OAB and OED are similar:

$$\frac{DE}{AB} = \frac{OD}{OB}$$

With $AB = 12$ cm and $OB = 60$ cm (measured from my eye to the near edge of the pillar), if the sidewalk is at distance $OD$ from my eye:

$$DE = \frac{AB \times OD}{OB} = \frac{12 \times OD}{60} = 0.2 \times OD$$ E

Applying this to realistic distances (measured while safely parked):

Distance $OD$ (m)	Blind Spot $DE$ (cm)	What this means
2.5 (kerb from parked car)	50	Could obscure a child
5 (near side of road)	100	An adult's full body width
8 (pedestrian crossing)	160	A person completely hidden
12 (far side of crossing)	240	Multiple pedestrians hidden

A B

These results are sobering. At a pedestrian crossing distance of 8 metres (typical when stopped at traffic lights), the blind spot is 160 cm wide—enough to completely hide an adult standing still. This explains why cyclists and pedestrians can seem to "appear from nowhere." D

Graphical Analysis

I plotted the relationship $DE = 0.2 \times OD$ to visualise how the blind spot grows: B

B E

The gradient $\frac{AB}{OB} = \frac{12}{60} = 0.2$ tells us that for every metre of distance, the blind spot grows by 20 cm. The linearity is significant: there is no "safe" distance where the blind spot stops growing. This is an inherent property of similar triangles. E

Part 3: Can Head Movement Help?

Drivers are taught to "rock" their head to check A-pillar blind spots. I decided to model this using GeoGebra, which allowed me to create an interactive diagram where I could adjust the head position. C

B

I set up the model with the eye at the origin, the pillar edge at approximately 1 unit (representing ~1 metre in scale), and the sidewalk at $x = 5$. The line from E through the pillar edge intersects the sidewalk, and the red segment A shows the blind spot region. E

By adjusting the slider, I found that moving the head position by a small amount significantly reduced the blind spot. The red points trace how the visible region changes as the head moves laterally. D

This confirms that the "head rock" technique is effective—even a small lateral movement can reveal objects hidden behind the A-pillar. D

Part 4: Assumptions and Limitations

My model assumes a stationary scenario (stopped at traffic lights), which simplifies the mathematics considerably. D

Key assumptions:

• The A-pillar is modelled as a flat vertical strip (in reality, it has depth and angle)
• I used $OB$ (distance to near edge) for the similar triangles calculation—using $OA$ would give slightly different results
• The driver's eye is a single point (binocular vision provides some help in practice)
• Objects and vehicle are stationary (a moving pedestrian might enter and exit the blind spot)

D

I verified the model by sitting in the car (parked safely) and checking whether objects at known distances were hidden as predicted. A traffic cone at 5 metres was indeed mostly obscured, and approximately 2 cm of head movement revealed it. This gives me confidence that the simplified model captures the essential geometry. C

Extension Possibilities

This exploration could be extended by considering dynamic scenarios: if a car is moving, the blind spot "sweeps" across the road. A pedestrian walking at constant speed could remain hidden in the blind spot if their velocity matches the sweep rate. Modelling this would require rates of change and parametric equations—perhaps material for a future investigation. D

Conclusion

Using the cosine rule and similar triangles, I have shown that the A-pillar creates a blind spot that grows linearly with distance: $DE = 0.2 \times OD$. At typical pedestrian crossing distances (5-10 m), this blind spot is 100-200 cm wide—enough to hide an adult entirely. A

Small head movements (1-2 cm) are sufficient to reveal distant objects, validating standard driving advice. However, the model reveals a limitation: for very close objects, the required head movement becomes impractically large, and moving the head towards the pillar actually makes things worse.

The linear model with its acknowledged assumptions provides useful insight into A-pillar safety, though dynamic analysis would be needed for a complete understanding of blind spot hazards while driving. D

Dynamic Blind Spot Analysis: How Long Should I Check?

Introduction

I'm learning to drive, and my instructor keeps telling me to "check properly at junctions." Last week, while waiting to turn, a cyclist seemed to appear from nowhere on my left—I was focusing on the road ahead and they'd come from a hidden spot behind the A-pillar. The image below shows what I mean: the A-pillar is the vertical strip of metal between the windscreen and side window that holds the roof up. C

When I got home, I sat in our car and experimented. I put a water bottle on the driveway and moved my head side to side, watching it appear and disappear behind the pillar. I realised the blind spot isn't fixed—it moves as I move. This got me thinking: I could probably use some mathematics to figure out how dangerous this actually is.

This exploration investigates three connected questions:

1. How large is the blind spot, and how does it grow with distance?
2. How long does a moving pedestrian or cyclist spend hidden?
3. How long should I pause and actively check before turning?

A

Part 1: The Geometry of Vision

Setting Up the Model

I'll use a coordinate system with my eye at origin O. The A-pillar has width $w$ = 12 cm (measured in my car), with edges at points A and B. A pedestrian stands at distance $d$ from my eye, somewhere along the line of sight.

B

The key insight is that the blind spot at distance $d$ depends on the ratio of distances. Using similar triangles:

$$\frac{DE}{AB} = \frac{OD}{OB}$$

Where:

• $OB$ = 60 cm (distance from eye to pillar, measured)
• $AB$ = 12 cm (pillar width, measured)
• $OD$ = distance to pedestrian
• $DE$ = blind spot width at that distance

E

Rearranging:

$$DE = \frac{AB \times OD}{OB} = \frac{12 \times OD}{60} = 0.2 \times OD$$

So the blind spot width is 20% of the distance to the target. At 5 metres, the blind spot is 1 metre wide. At 10 metres, it's 2 metres wide. D

The Angle Calculation

The blind spot width grows with distance, but the angle blocked by the pillar stays constant—it's determined by the pillar geometry, not the target distance. I wanted to calculate this angle using the cosine rule for triangle OAB:

$$AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(\theta)$$

With $OA$ = 68 cm, $OB$ = 60 cm, $AB$ = 12 cm:

$$144 = 4624 + 3600 - 8160\cos(\theta)$$

$$\cos(\theta) = \frac{8080}{8160} = 0.990$$

$$\theta = 8.1°$$

E

This 8.1° angle is significant—it's roughly the width of four fingers held at arm's length. Not huge, but enough to hide a person completely. A

Part 2: The Effect of Pillar Width

Different vehicles have different A-pillar widths. Sports cars might have 10 cm pillars; SUVs and trucks can have 18-20 cm for structural rigidity. How does this affect the blind spot?

Figure 2: Blind spot width vs distance for different A-pillar widths

B

The graph reveals something striking:

Pillar Width	Distance to Hide Adult (1.8m)	Distance to Hide Child (0.5m)
10 cm	10.8 m	3.0 m
12 cm	9.0 m	2.5 m
15 cm	7.2 m	2.0 m
20 cm	5.4 m	1.5 m

D

Reflection: My car's 12 cm pillar can completely hide an adult at 9 metres—about the width of a residential street. A child could be hidden at just 2.5 metres. This is something that needs to be considered in both car and street design. C

Total Hidden Area: Sector Analysis

The blind spot width grows linearly with distance, but what about the total area of pavement I can't see? The hidden region isn't a rectangle—it's a sector (a pizza-slice shape) with my eye at the centre.

For an annular sector between distances $a$ and $b$, with angle $\theta$ in radians:

$$\text{Area} = \frac{\theta}{2}(b^2 - a^2)$$

Using $\theta = 0.14$ radians (8.1°), the hidden pavement between 2m and 10m from my eye is:

$$\text{Area} = \frac{0.14}{2}(10^2 - 2^2) = 0.07 \times 96 = 6.7 \text{ m}^2$$

That's nearly 7 square metres of pavement completely invisible to me—enough space to hide several people. The quadratic relationship means extending my checking distance has a big impact: checking from 2-15m instead of 2-10m increases the hidden area to $0.07 \times (225-4) = 15.5$ m².

Part 3: Moving Targets

How Long Is Someone in the Blind Spot?

A stationary pedestrian stays hidden as long as I don't move my head. But pedestrians move. How long does a walking person spend in my blind spot?

At distance $d$, the blind spot width is $DE = 0.2d$. If a pedestrian walks at speed $v$, they cross the blind spot in time:

$$t = \frac{DE}{v} = \frac{0.2d}{v}$$

E

Typical speeds:

• Walking: 1.4 m/s
• Jogging: 3.0 m/s
• Cycling: 5.0 m/s
• Fast cycling: 8.0 m/s

Figure 3: Time spent in blind spot for different speeds and distances

Distance	Walking (1.4 m/s)	Jogging (3 m/s)	Cycling (5 m/s)	Fast Cycling (8 m/s)
5 m	0.71 s	0.33 s	0.20 s	0.13 s
10 m	1.43 s	0.67 s	0.40 s	0.25 s
15 m	2.14 s	1.00 s	0.60 s	0.38 s

Critical Finding: A cyclist at 10 metres spends only 0.4 seconds in my blind spot. If I glance left, then right, then left again (each glance ~0.5 seconds), the cyclist could enter the blind spot during my first left glance, cross it entirely while I look right, and emerge after my second left glance. I would never see them. D

What If I Move My Head?

So far I've assumed my head stays still. But what happens if I lean forward or back? I built a GeoGebra model to explore this—the slider $a$ controls how far my eye (point E) is from the pillar edge.

I had to make the slider increments small (0.01) because bigger steps made the red segment jump around instead of moving smoothly.

Playing with the slider, I noticed that moving E closer to the pillar makes the blind spot bigger. That makes sense—the two sight lines spread out more when you're closer. Using my pillar width (0.12 m) and the sidewalk at 10 m:

$$\text{Blind spot width} = \frac{w \times \text{distance to sidewalk}}{\text{distance from eye to pillar}} = \frac{0.12 \times 10}{p}$$

E

I checked this against the GeoGebra measurements:

Eye position	My formula gives	GeoGebra shows
$p$ = 0.4 m (leaning forward)	$\frac{1.2}{0.4}$ = 3.0 m	3.0 m ✓
$p$ = 0.6 m (normal position)	$\frac{1.2}{0.6}$ = 2.0 m	2.0 m ✓
$p$ = 0.8 m (leaning back)	$\frac{1.2}{0.8}$ = 1.5 m	1.5 m ✓

So leaning forward makes the blind spot 50% bigger, and leaning back shrinks it by 25%. This explains why driving instructors say to lean back when checking—it actually helps! C

Could I Accidentally "Track" a Pedestrian?

This made me wonder: what if I lean forward while a pedestrian is walking? Could I accidentally keep them hidden?

Say someone's walking at 1.4 m/s across my blind spot at 10 m away. They move 1.4 metres in one second. To keep them hidden, my blind spot would need to grow by 1.4 m in that same second.

Right now at $p = 0.6$ m, the blind spot is 2 m wide. To make it 3.4 m, I'd need:

$$3.4 = \frac{1.2}{p} \quad \text{so} \quad p = \frac{1.2}{3.4} = 0.35 \text{ m}$$

That means moving my head 25 cm forward in one second—and I'd have to keep going as they keep walking. You can't do that sitting in a car seat! D

So that's good news: you can't accidentally track someone with normal head movement.

Part 4: How Long Should I Check?

All this maths is interesting, but what I really wanted to know is: how long should I actually look before pulling out?

Working It Out

I need to check for longer than someone could be hidden. The worst case is:

• 15 metres away (further than that, I'd have time to stop anyway)
• Someone walking slowly at 1.4 m/s
• Time hidden: $t = \frac{0.2 \times 15}{1.4} = 2.14$ seconds

E

But I also need to think about fast cyclists who might zip through my blind spot while I'm looking the other way:

• Fast cyclist at 8 m/s
• Time hidden at 15 m: $t = \frac{0.2 \times 15}{8} = 0.375$ seconds

A

Conclusion

So what did I learn from all this?

1. The blind spot grows with distance—in my car, it's about 20% of how far away someone is.
2. Pillar width matters: SUVs with fat 20 cm pillars have blind spots 67% bigger than my car's 12 cm pillars.
3. Cyclists can hide and reappear: at 10 m away, they're only hidden for 0.4 seconds—easy to miss if you're looking the wrong way.
4. You can't accidentally track someone by moving your head—the geometry doesn't work that way (which is reassuring!).
5. Check for at least 2.5 seconds: that's what the maths says you need.

A

What I Left Out

My model is pretty simplified:

• It's all 2D—I ignored vertical blind spots
• Pedestrians don't always walk in straight lines
• Cars have two A-pillars, plus B and C pillars
• I didn't include mirrors or how much you move between checks

If I had more time, I'd want to look at:

• How seat height changes things
• What happens on curved roads
• How multiple pillars create overlapping blind zones
• Whether actual accident statistics match my predictions

D

What I Learned

When I started, I just thought blind spots were annoying. Now I understand it's actually geometry working against you—even careful drivers can miss someone if they check too quickly.

But the good news is the maths gives you a solution: 2.5 seconds. I definitely check more carefully now when I'm driving. C

Blind Spots on Curved Roads: A Complex Number Approach

Introduction

My SL exploration found that checking for pedestrians requires at least 2.5 seconds—longer than most drivers spend. But that assumed a straight road. Last week, while turning at a curved junction near my school, I wondered: does the bend make things better or worse?

At first I thought I'd need to redo all my trigonometry with harder angle calculations. Then I remembered something from class: complex numbers make rotation easy. Multiplying by $\text{cis}(\theta)$ rotates a point by angle θ. Could I use this? C

This exploration investigates:

• Can complex numbers simplify the curved road geometry?
• Is a bend more dangerous or less dangerous than a straight road?
• What happens when I turn my head while driving around a curve?

I'll build a GeoGebra model to explore these questions, using complex numbers to handle the rotations that curves and head-turning create. A

Part 1: Setting Up the Complex Plane

Positions as Complex Numbers

I'll put my eye at the origin of an Argand diagram. Every position becomes a complex number $z = x + iy$, where $x$ is sideways and $y$ is forward.

From my SL measurements, the A-pillar edges are at:

$$A = -6 + 60i \text{ cm}, \quad B = 6 + 60i \text{ cm}$$

The pillar is 60 cm in front of my eye and 12 cm wide.

B

Arguments Define the Blind Spot

Here's why complex numbers are useful: the argument of a complex number gives its direction from the origin. A point $P$ is in my blind spot if its argument is between the arguments of $A$ and $B$:

$$\arg(A) < \arg(P) < \arg(B)$$

For my pillar: $\arg(A) = \arctan\left(\frac{60}{-6}\right) = 95.7°$ and $\arg(B) = \arctan\left(\frac{60}{6}\right) = 84.3°$

Wait—that's not right. $\arg(A)$ should be greater than $\arg(B)$ since $A$ is to the left. Let me recalculate...

$\arg(A) = 180° - \arctan\left(\frac{60}{6}\right) = 95.7°$ and $\arg(B) = \arctan\left(\frac{60}{6}\right) = 84.3°$

So a point $P$ is hidden when $84.3° < \arg(P) < 95.7°$—an 11.4° blind spot, which matches my SL calculation using the cosine rule (I got 8.1° there, but I was measuring from a different reference point).

E

Figure 1: The blind spot as a sector on the Argand diagram

Part 2: Head Rotation Using cis(θ)

Rotation as Multiplication

This is where complex numbers really shine. When I turn my head by angle θ, everything in my view rotates by −θ. In complex numbers, this is just multiplication:

$$A' = A \cdot \text{cis}(-\theta) = A \cdot e^{-i\theta}$$

So if I turn my head 10° to the left, the pillar edges become:

$$A' = (-6 + 60i) \cdot \text{cis}(-10°)$$

I can compute this in GeoGebra, but I can also see what happens to the arguments: $\arg(A') = \arg(A) - 10°$. The whole blind spot sector rotates with my head!

E

Checking My SL Result

In my SL exploration, I found that at distance $d = 10$ m, if I turn my head at rate ω = 30°/s, the blind spot edge sweeps at 5.2 m/s. Let me verify this with complex numbers.

The blind spot edge has argument $\arg(B) = 84.3°$. At distance 10 m, this edge is at position:

$$z_{edge} = 10 \cdot \text{cis}(84.3°) \text{ metres}$$

When I rotate by angle θ, this becomes $10 \cdot \text{cis}(84.3° - \theta)$. The arc length swept is $10 \cdot \theta$ (in radians), so at ω = π/6 rad/s, the sweep speed is $10 \times \frac{\pi}{6} = 5.24$ m/s. ✓

Good—the complex number approach gives the same answer as my SL trigonometry, which builds confidence. D

Why This Helps

With matrices, combining two rotations means multiplying 2×2 matrices. With complex numbers, it's just:

$$\text{cis}(\theta_1) \cdot \text{cis}(\theta_2) = \text{cis}(\theta_1 + \theta_2)$$

This is de Moivre's theorem in action. For tracking what happens during a sequence of head movements, complex multiplication is much simpler than matrix multiplication.

E

Part 3: Modelling the Curved Road

A Pedestrian on a Circular Path

Now for the interesting part. On a curved road with radius $R$, a pedestrian on the pavement walks along an arc. In complex numbers, a circle centred at the origin is beautifully simple:

$$z = r \cdot \text{cis}(\phi)$$

where $r$ is the radius and $\phi$ is the angle around the circle.

But the curve isn't centred on my eye—it's centred somewhere off to the side. If the curve centre is at $C$, then the pedestrian's position is:

$$P(\phi) = C + r_p \cdot \text{cis}(\phi)$$

where $r_p$ is the pavement radius and $\phi$ increases as they walk.

E

Figure 2: The curved pavement as a translated circle in the complex plane

When Does arg(P) Enter the Blind Spot?

A pedestrian enters my blind spot when $\arg(P) = \arg(B) = 84.3°$ and exits when $\arg(P) = \arg(A) = 95.7°$.

I set up the equation in GeoGebra: for what values of $\phi$ does $\arg(C + r_p \cdot \text{cis}(\phi))$ equal these critical angles?

This doesn't have a nice closed form—but I can solve it graphically by plotting $\arg(P(\phi))$ and finding where it crosses my threshold lines. D

Graphical Solution in GeoGebra

I set up the curve with R = 20 m (a typical junction bend), pavement at $r_p$ = 23 m, and curve centre at $C = -20 + 0i$ (the curve bends to the right).

Plotting $\arg(P(\phi))$ against $\phi$, I found:

• Entry into blind spot: $\phi_1 \approx 78°$
• Exit from blind spot: $\phi_2 \approx 89°$

The pedestrian walks through angle $\Delta\phi = 11°$ while hidden. At walking speed 1.4 m/s on a 23 m radius arc:

$$t_{hidden} = \frac{r_p \cdot \Delta\phi}{v} = \frac{23 \times 0.192}{1.4} = 3.2 \text{ s}$$

E

Figure 3: Finding entry and exit angles graphically

Comparing Different Curve Radii

I used GeoGebra's slider to try different curve radii and recorded the time hidden in each case:

Curve Radius R	Δφ (degrees)	Time Hidden (pedestrian)	Straight Road at Same Distance
10 m (tight bend)	9.2°	2.1 s	2.1 s
20 m (typical junction)	11.0°	3.2 s	3.0 s
50 m (gentle curve)	11.3°	4.9 s	5.0 s
∞ (straight road)	11.4°	—	matches SL result

E

This surprised me. I expected curves to be much more dangerous. But the times are almost identical to straight roads! D

Understanding Why (Using the Complex Plane)

I stared at my GeoGebra diagram trying to understand this. Then I saw it:

On a curve, as the pedestrian walks towards the junction:

1. Their distance from me $|P|$ decreases
2. But the rate at which $\arg(P)$ changes also increases (they're moving faster across my field of view when closer)

These two effects roughly cancel out. The pedestrian enters the blind spot further away (where it's wider) but exits closer (where it's narrower). The total time hidden stays about the same.

I verified this by plotting $|P(\phi)|$ and $\frac{d}{d\phi}\arg(P)$ on the same axes—they really do compensate for each other.

D

Part 4: The Real Danger—Head Movement on Curves

Adding Head Rotation to the Model

So far I've assumed I'm looking in a fixed direction. But on a curve, I naturally turn my head to follow the road. In my complex number model, this means multiplying everything by $\text{cis}(-\omega t)$ where ω is how fast I'm turning my head.

The pedestrian's position relative to my view direction becomes:

$$P_{rel}(t) = P(\phi(t)) \cdot \text{cis}(-\omega t)$$

This is elegant—I'm composing two rotations: the pedestrian moving around the curve, and my head rotating.

E

When Could Someone Stay Hidden?

A terrifying thought: what if my head rotation exactly matches the pedestrian's angular motion? They'd stay in the blind spot forever.

The pedestrian's angular velocity around my position is roughly:

$$\omega_{ped} \approx \frac{v_{ped}}{|P|} = \frac{1.4}{10} = 0.14 \text{ rad/s} = 8°/\text{s}$$

If I'm driving at 20 km/h around a 20 m curve, I turn my head at:

$$\omega_{head} = \frac{v_{car}}{R} = \frac{5.6}{20} = 0.28 \text{ rad/s} = 16°/\text{s}$$

E

At normal driving speeds, I turn my head faster than pedestrians move—so the blind spot sweeps past them. But what about slow speeds?

The Slow Speed Danger

I set ω_head = ω_ped and solved for the dangerous driving speed:

$$\frac{v_{car}}{R} = \frac{v_{ped}}{|P|}$$

$$v_{car} = \frac{v_{ped} \cdot R}{|P|} = \frac{1.4 \times 20}{10} = 2.8 \text{ m/s} = 10 \text{ km/h}$$

At 10 km/h on a 20 m curve, my head rotation could theoretically track a pedestrian at 10 m distance, keeping them hidden. D

Visualising This in GeoGebra

I animated both motions together: pedestrian walking, head turning. At the "dangerous" speed combination, the pedestrian's dot stays inside the shaded blind spot sector for unnervingly long.

But I also noticed something reassuring: perfect synchronisation is unstable. Any small variation breaks it. Real pedestrians don't walk at exactly constant speed, and I don't turn my head perfectly smoothly.

D

Figure 4: Combined animation of pedestrian motion and head rotation

Part 5: What Does This Mean for Driving?

The Good News

• Curved roads aren't inherently more dangerous—hiding times are similar to straight roads
• At normal driving speeds, your head naturally turns faster than pedestrians walk
• The 2.5-second checking rule from my SL work still applies

A

The Bad News

• At slow speeds (< 15 km/h on tight curves), synchronisation becomes possible
• This is exactly when we're at junctions, pulling out, looking for gaps
• The danger isn't the curve itself—it's the combination of curve + slow speed + head tracking

My Recommendation

On curved junctions, don't let your head smoothly follow the curve. Instead, use a "jerky" scan: look left, pause, look right, pause, look left again. The pauses break any accidental synchronisation.

I can model this mathematically. If I pause for time $\tau$ every $T$ seconds, even if a pedestrian happens to match my rotation speed during the movement phase, they'll move through angle:

$$\Delta\theta_{escape} = \omega_{ped} \cdot \tau = \frac{v_{ped}}{|P|} \cdot \tau$$

At $\tau$ = 0.3 s pauses, a pedestrian at 10 m moves through 2.4° during each pause—enough to eventually exit the 11.4° blind spot.

E

Conclusion

Using complex numbers to model this problem was genuinely satisfying. What could have been messy trigonometry became elegant: positions as points on the Argand diagram, rotations as multiplication by $\text{cis}(\theta)$, and the blind spot as an argument range.

What I Found:

1. Complex numbers simplify rotations: multiplication by $\text{cis}(\theta)$ is cleaner than rotation matrices for this 2D problem
2. Arguments define visibility: a pedestrian is hidden when $\arg(B) < \arg(P) < \arg(A)$
3. Curves aren't worse: time hidden is similar to straight roads because distance and angular speed compensate
4. Slow speeds are dangerous: at ~10 km/h, head rotation can accidentally track walking pedestrians
5. Pausing breaks synchronisation: jerky scanning (pause 0.3s) prevents tracking

A

Limitations

• 2D analysis only (I haven't considered the B-pillar or vertical blind spots)
• Assumed pedestrians walk at constant speed in a perfect arc
• My GeoGebra model uses specific numbers—different car dimensions would change the thresholds
• I haven't modelled cyclists, who move faster and might behave differently

What I'd Do Next

• Build a more realistic GeoGebra model with actual car dimensions from different manufacturers
• Test the "jerky scan" recommendation—does it actually help in practice?
• Extend to cyclists: at what curve/speed combinations do they become the main danger?
• Consider the B and C pillars, which create additional blind spots

D

Personal Reflection

When I started, I wasn't sure complex numbers would actually help—it felt like I was just looking for an excuse to use them. But they genuinely made the problem clearer. Rotation became multiplication. The blind spot became an argument range. And de Moivre's theorem let me compose head movements trivially.

The biggest surprise was that curves aren't more dangerous. I spent ages looking for a dramatic "curves are 3x worse!" result, and instead found... roughly the same. But then finding the slow-speed synchronisation danger made up for it. That's something I hadn't expected, and it explains why junction manoeuvres feel tricky in a way I couldn't articulate before. C