Nameplate megawatts are not useful AI compute

The method

1. For each constraint, work out how many racks of the target platform it permits, from the hall’s own ratings, the platform’s published rack power and the redundancy the hall keeps.
2. Treat a pass-or-fail gate (can one rack position be fed at all? can the cooling architecture take this platform?) as zero racks until it passes.
3. Take the minimum. That constraint binds; the others have headroom.
4. Relieve the binding constraint, price and date the relief, and solve again. Repeating this is the headroom ladder.

Counts are floored to whole racks, never rounded up. Each relation is public on its own page; start from the constraint reference. A figure’s evidence class travels with it, and a result carries the weakest class that went into it (how the classes work).

Worked example: a synthetic reference hall

This is the engine’s published reference hall. It is synthetic: invented to demonstrate the method, not a customer asset, and every figure below is modelled or assumed unless its basis says otherwise. The grid agreement is 12 MW contracted (14 MVA firm), the site peak is 7.4 MW and the existing IT load 4.9 MW. The target platform is gb300_nvl72, in the scenario “Hybrid DLC on retained plant, grid only”.

As found, the hall carries 0 racks, bound by Rack feed / tap-off rating. After the costed ladder it carries 26, bound next by grid_firm_capacity.

Source: the engine’s published model pack (schema gridforge/model-pack/2, generated 2026-09-14). Constraints the pack does not express as a count: Floor loading, Cooling architecture per-rack capability, Achievable facility water temperature. The full document is on the worked-example page.

Read the table from the top: whichever row holds the smallest number sets the hall’s capacity, however large the contracted figure and however many racks the other rows would allow. That is why the same hall can hold 12 MW of contract and run 0 racks.

Why total power cannot answer the question

Dividing usable power by rack power gives how many racks you could feed if power were the only limit. It says nothing about whether any one position can be fed. The figure below applies the public rack-feed relation to a second, invented hall: with 63 A tap-off units no rack above about 41 kW can be fed, so the count is zero there even though total power would allow dozens.

What this does not cover

• It is a screening method, not a design, an approval or a statement that a hall can be loaded. Protection, commissioning, structural survey and the utility’s own conditions are outside it.
• The reference hall is synthetic and the models have not been calibrated against measured data from a real hall: the calibration ledger is empty (see the evidence page).
• Where an input is not supplied, a labelled library default is used and listed in a data request. A missing value is never treated as zero and never produces a date.
• Lead times and costs are ranges from a reference library until a supplier quotation replaces them. They are not commitments.
• Time to Power sells no equipment. Relief options are modelled so a buyer can compare them, not offered for sale.

Run it on your own hall

1. 01
   Free capacity checkFree, no card

   Seven numbers you already know, free. It names the constraint likely to bind first and what it had to assume.

2. 02
   Density Screen€4,500

   5 working days from a complete intake

   One hall, with the first six rungs of the ladder and the inputs nobody has measured.