Attempting to Quantify Alex Honnold's Taipei 101 Climb

So Alex Honnold is apparently going to free solo Taipei 101 in about 10 days. Which is completely insane.

I got curious: what would that look like in Sendsville points? This is totally unscientific and just for fun—more of a thought experiment than anything else. But it's a way to wrap my head around just how absurd this climb is.

Best of luck to Alex 🤞

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Assumptions (rough estimates)

I'm making some big simplifications here:

• I'm treating the movement like bouldering in terms of difficulty
• Honnold compared the climbing to doing the Rostrum (5.11c) twice in a row , which in bouldering terms I'd ballpark around V3–V4—technical, balancey, insecure, but not absolute limit climbing
• Sendsville assumes boulder problems are 4 meters tall
• Taipei 101 is roughly 508 meters

Source: Climbing.com - "Fruit, Custom Shoes, a Cave, and More Ways Alex Honnold Is Gearing Up for His Live Free Solo"

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Bouldering points

In Sendsville:

V3 = 10 points
V4 = 14 points

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Scaling for height

If we use Sendsville's 4-meter assumption:

508 m ÷ 4 m ≈ 127 boulder problems

So you could think of this as:

Doing 127 consecutive V3–V4 boulders... where falling off any of them means you die.

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Total points (ballpark)

At V3:

127 × 10 ≈ 1,270 points

At V4:

127 × 14 ≈ 1,778 points

So somewhere around 1,300–1,800 Sendsville points.

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Some perspective

That's a ridiculous amount of sustained climbing. And this estimate doesn't even touch:

• The exposure
• The mental game
• Physical fatigue over 500 meters
• The actual consequences

This is the stuff Honnold does that the rest of us can't even comprehend.

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Wrap-up

This was just a fun way to put an absurd climb into a framework I could actually wrap my head around. It's a wild thing to attempt and I'll definitely be watching, hoping Alex pulls it off safely.

Good luck! Stay safe out there.