Predict the total vision-token count for a tiled high-resolution image

The answer is **2,560 vision tokens**. Work it in two steps. First count the encoder passes. A 3x3 grid means 3 rows times 3 columns, which is 9 tiles. The downscaled global view is one more pass on top of those. So 9 + 1 = 10 passes total. Second, multiply by the per-pass token count. Each tile emits 256 tokens, and the global view also emits 256. Since every pass costs the same here, the total is just 10 x 256 = 2,560 tokens. The reason this matters: a single high-resolution image just added 2,560 tokens to the prompt. That is larger than many full text questions a user would type. The image, not the text, dominates the context for this request. The two slips to avoid are reading 3x3 as 3 tiles or 6 tiles instead of 9, and forgetting the global view. Both make you undercount. Count the grid as rows times columns, add the global view, then multiply by the per-pass token count.

**2,560 vision tokens.** The calculation is a tile count times a per-tile token block. The tile count comes from the grid plus the global view. $$\text{passes} = (3 \times 3) + 1 = 10$$ A 3x3 grid is 9 tiles. The downscaled global view is one additional encoder pass, giving 10 passes. Each pass emits a fixed 256-token block. $$\text{tokens} = 10 \times 256 = 2{,}560$$ The structure to remember: AnyRes tiling makes token count scale with tile count. And tile count scales with resolution. Here the global view adds a constant one block on top of the grid, which is the part people forget. The takeaway is about magnitude. 2,560 tokens for one image is more than a long paragraph of text. This is why VLM teams budget image resolution per use case rather than always tiling at maximum. Each step up in the tiling grid adds a full block of tokens per new tile, and those tokens flow through prefill and the KV cache exactly like text. A 4x4 grid plus global view would be 17 x 256 = 4,352 tokens, so the cost climbs fast as you add resolution.