Electron Density Map Parameter Calculator

Size an electron density map before you calculate it: grid spacing, FFT dimensions, reflection counts and file size.

🗺️ Electron Density Map Parameter Calculator
Å
Å
Å
Å
Recommended grid spacing
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FFT grid dimensions
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Total grid points
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Map file size
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Unit cell volume
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Reflections in sphere
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Unique reflections
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Nyquist spacing (d/2)
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Step-by-step working

🗺️ What are Electron Density Map Parameters?

An electron density map is the three-dimensional picture of scattering density that comes out of a crystallographic Fourier transform. Before you can compute one, several practical numbers have to be fixed: how finely to sample the density onto a grid, what grid dimensions the FFT will actually accept, how many independent reflections the data contain, and how much disk space the resulting map will occupy. Those quantities all follow from three inputs, the unit cell dimensions, the resolution limit of the data, and the crystal's Laue symmetry.

Getting them right matters in day-to-day work. A grid that is too coarse produces a lumpy map where peak heights shift depending on where grid points happen to land, which makes automatic peak picking and real-space refinement unreliable. A grid dimension that is not built from small prime factors will be rejected outright by an FFT routine, or silently padded to a bigger size that wastes memory. And the unique reflection count is what you check a dataset's completeness against, and what sets the data-to-parameter ratio that decides whether you can refine anisotropic displacement parameters at all.

A common misconception is that a finer map grid improves resolution. It does not. Resolution is fixed by how far out in reciprocal space you actually measured intensities. Sampling more finely only interpolates the same band-limited function more smoothly. The sampling theorem says a spacing of d/2 already retains every bit of information in the map, and the reason crystallographers use d/3 instead is purely practical, it makes contoured features and peak heights insensitive to grid placement. A second misconception is that the theoretical reflection count is what you should measure. Real datasets always fall short of it, typically by 5 to 20 percent, because of the beamstop shadow, overloads, and the geometric blind region.

This calculator ties all of that together. Enter your cell, resolution and symmetry, and it returns the recommended grid spacing, the FFT-friendly grid dimensions rounded up to the next 2, 3, 5, 7-smooth even number on each axis, the total number of grid points, the map file size for a standard 4-byte-per-point CCP4 or MRC map, and both the total and symmetry-unique reflection counts. Two charts show how the reflection count explodes as you push to higher resolution, and how the map file size responds to your choice of sampling factor.

📐 Formula

Δ = dmin ÷ f    and    Nunique = 4πV ÷ (3 dmin³ nLaue)
Δ = recommended map grid spacing (Å)
dmin = resolution limit of the data (Å)
f = grid sampling factor, 2 is the Nyquist minimum and 3 is the standard choice
V = unit cell volume, taken here as a × b × c for 90° cell angles (ų)
nLaue = order of the Laue group, 2 triclinic, 4 monoclinic, 8 orthorhombic, 16 tetragonal 4/mmm, 24 hexagonal 6/mmm, 48 cubic m-3m
Grid axes: nx = next even 2, 3, 5, 7-smooth integer ≥ a ÷ Δ, and likewise for ny and nz
File size: nx × ny × nz × 4 bytes for a 32-bit float map
Example: a = b = 79.1 Å, c = 37.9 Å, d = 2.0 Å, f = 3 → Δ = 0.6667 Å and a 120 × 120 × 60 grid.

📖 How to Use This Calculator

Steps

1
Enter the unit cell edges. Type the three unit cell edge lengths a, b and c in angstroms, assuming 90 degree cell angles.
2
Set the resolution and sampling. Enter the resolution limit d in angstroms and choose the grid sampling factor, normally 3.
3
Choose the Laue class and read the results. Select the Laue symmetry class, then read the grid spacing, FFT grid dimensions, reflection counts and map file size.

💡 Example Calculations

Example 1 — Tetragonal Protein Crystal at 2.0 Å

a = b = 79.1 Å, c = 37.9 Å, dmin = 2.0 Å, sampling factor 3, Laue 4/mmm

1
V = 79.1 × 79.1 × 37.9 = 237,133.1 ų, and Δ = 2.0 ÷ 3 = 0.6667 Å
2
a ÷ Δ = 118.65 rounds up to 120, and c ÷ Δ = 56.85 rounds up to 60, giving a 120 × 120 × 60 grid
3
864,000 grid points at 4 bytes each is 3.30 MB, and (4/3)πV ÷ 8 = 124,163 reflections, so 124,163 ÷ 16 = 7,760 unique
Grid spacing 0.6667 Å, grid 120 × 120 × 60, 7,760 unique reflections, 3.30 MB map
Try this example →

Example 2 — The Same Crystal Pushed to 1.5 Å

a = b = 79.1 Å, c = 37.9 Å, dmin = 1.5 Å, sampling factor 3, Laue 4/mmm

1
Δ = 1.5 ÷ 3 = 0.5000 Å, so a ÷ Δ = 158.2 rounds up to 160 and c ÷ Δ = 75.8 rounds up to 80
2
The grid becomes 160 × 160 × 80 = 2,048,000 points, which is 7.81 MB, 2.4 times the 2.0 Å map
3
Reflections rise as 1/d³, so 294,311 total and 18,394 unique, about 2.4 times the 2.0 Å count
Grid 160 × 160 × 80, 18,394 unique reflections, 7.81 MB map
Try this example →

Example 3 — Small-Molecule Cubic Cell at 0.8 Å

a = b = c = 10 Å, dmin = 0.8 Å, sampling factor 3, Laue m-3m

1
V = 1,000 ų and Δ = 0.8 ÷ 3 = 0.2667 Å
2
10 ÷ 0.2667 = 37.5 rounds up to 40 on every axis, so the grid is 40 × 40 × 40 = 64,000 points, only 0.24 MB
3
(4/3)π × 1,000 ÷ 0.512 = 8,181 reflections, and cubic symmetry divides that by 48 to give 170 unique
Grid 40 × 40 × 40, 170 unique reflections, 0.24 MB map
Try this example →

Example 4 — Large Cubic Cell at Low Resolution

a = b = c = 200 Å, dmin = 3.5 Å, sampling factor 3, Laue m-3m

1
V = 8,000,000 ų and Δ = 3.5 ÷ 3 = 1.1667 Å
2
200 ÷ 1.1667 = 171.43, and the next even 2, 3, 5, 7-smooth integer is 180, so the grid is 180 × 180 × 180
3
5,832,000 grid points is 22.25 MB, while 781,582 reflections divided by 48 gives 16,283 unique
Grid 180 × 180 × 180, 16,283 unique reflections, 22.25 MB map
Try this example →

❓ Frequently Asked Questions

What grid spacing should an electron density map use?+
Use the resolution limit divided by 3 as the default. At 2.0 angstrom resolution that means a grid spacing of about 0.67 angstroms. Dividing by 2 is the Nyquist minimum and loses map detail, while dividing by 4 or more is used for density modification and averaging where smooth interpolation matters.
Why must FFT grid dimensions have small prime factors?+
The fast Fourier transform is only fast when the transform length factorizes into small primes. Crystallographic FFT code typically supports factors of 2, 3, 5 and 7 only, so a dimension such as 179 (a prime) is rejected or padded. This calculator rounds each axis up to the next even number built from those factors.
How many unique reflections does a crystal give at a given resolution?+
The count is approximately (4/3) pi V divided by d cubed, then divided by the order of the Laue group. For a 79.1 by 79.1 by 37.9 angstrom tetragonal cell at 2.0 angstroms with Laue class 4/mmm, that gives about 124,163 total lattice points and 7,760 unique reflections.
What is the Laue class multiplier for?+
Symmetry makes many reflections equivalent, so only a fraction of reciprocal space is independent. The order of the Laue group, which already includes the inversion from Friedel's law, tells you that fraction: 2 for triclinic, 8 for orthorhombic, 16 for tetragonal 4/mmm, and 48 for cubic m-3m.
How big will my electron density map file be?+
Multiply the three grid dimensions together and multiply by 4 bytes per point, which is the standard 32-bit float used in CCP4 and MRC map files. A 120 by 120 by 60 grid holds 864,000 points and occupies about 3.3 megabytes plus a small header.
Does a finer map grid improve resolution?+
No. Resolution is set entirely by how far out in reciprocal space you measured data. A finer grid only samples the same density function more densely, which makes contours smoother and interpolation more accurate. Beyond about d/4 the extra points carry no new information at all.
What is Nyquist sampling in this context?+
The map is a Fourier series band-limited at 1/d, so by the sampling theorem a spacing of d/2 is the coarsest that preserves all the information. In practice d/3 is used instead, because Nyquist sampling makes peak heights and contour shapes sensitive to exactly where the grid points happen to fall.
Why is the real reflection count lower than the calculated one?+
The formula counts every reciprocal lattice point in a perfect sphere. Real datasets exclude very low angle reflections behind the beamstop, lose overloaded strong spots, miss the blind region set by the rotation geometry, and reject outliers, so a complete dataset typically records 80 to 95 percent of the theoretical count.
Does this calculator work for non-orthogonal unit cells?+
It assumes all three cell angles are 90 degrees, so the volume is a times b times c. Monoclinic and triclinic cells have a smaller true volume than that product, and every result that depends on volume, meaning the reflection counts, scales down in proportion. Grid spacing and grid dimensions remain correct.
How does resolution change map and data size?+
Both scale with the inverse cube of the resolution limit. Halving d from 3.0 to 1.5 angstroms multiplies the unique reflection count by eight and multiplies the number of grid points by eight as well, since the required grid spacing also halves along every axis.
What resolution counts as high resolution?+
For macromolecules, better than 1.5 angstroms is high resolution and resolves individual atoms, 1.5 to 2.5 angstroms is the routine working range where side chains are clear, and worse than 3.0 angstroms shows only backbone and secondary structure. Small-molecule crystallography routinely reaches 0.8 angstroms.
Why is 4 bytes per grid point assumed?+
CCP4 and MRC format maps store density as 32-bit floating point numbers by default, which is 4 bytes each. Some programs offer a 1-byte or 2-byte integer mode that shrinks the file four-fold or two-fold, at the cost of precision in the stored density values.
Should the grid cover the whole unit cell or just the asymmetric unit?+
This calculator sizes a full unit cell map, which is what most FFT programs produce and what graphics programs expect. Storing only the asymmetric unit shrinks the file by roughly the order of the space group, but the viewer then has to expand it by symmetry before it can be contoured around a model.

What grid spacing should an electron density map use?

Use the resolution limit divided by 3 as the default. At 2.0 angstrom resolution that means a grid spacing of about 0.67 angstroms. Dividing by 2 is the Nyquist minimum and loses map detail, while dividing by 4 or more is used for density modification and averaging where smooth interpolation matters.

Why must FFT grid dimensions have small prime factors?

The fast Fourier transform is only fast when the transform length factorizes into small primes. Crystallographic FFT code typically supports factors of 2, 3, 5 and 7 only, so a dimension such as 179 (a prime) is rejected or padded. This calculator rounds each axis up to the next even number built from those factors.

How many unique reflections does a crystal give at a given resolution?

The count is approximately (4/3) pi V divided by d cubed, then divided by the order of the Laue group. For a 79.1 by 79.1 by 37.9 angstrom tetragonal cell at 2.0 angstroms with Laue class 4/mmm, that gives about 124,163 total lattice points and 7,760 unique reflections.

What is the Laue class multiplier for?

Symmetry makes many reflections equivalent, so only a fraction of reciprocal space is independent. The order of the Laue group, which already includes the inversion from Friedel's law, tells you that fraction: 2 for triclinic, 8 for orthorhombic, 16 for tetragonal 4/mmm, and 48 for cubic m-3m.

How big will my electron density map file be?

Multiply the three grid dimensions together and multiply by 4 bytes per point, which is the standard 32-bit float used in CCP4 and MRC map files. A 120 by 120 by 60 grid holds 864,000 points and occupies about 3.3 megabytes plus a small header.

Does a finer map grid improve resolution?

No. Resolution is set entirely by how far out in reciprocal space you measured data. A finer grid only samples the same density function more densely, which makes contours smoother and interpolation more accurate. Beyond about d/4 the extra points carry no new information at all.

What is Nyquist sampling in this context?

The map is a Fourier series band-limited at 1/d, so by the sampling theorem a spacing of d/2 is the coarsest that preserves all the information. In practice d/3 is used instead, because Nyquist sampling makes peak heights and contour shapes sensitive to exactly where the grid points happen to fall.

Why is the real reflection count lower than the calculated one?

The formula counts every reciprocal lattice point in a perfect sphere. Real datasets exclude very low angle reflections behind the beamstop, lose overloaded strong spots, miss the blind region set by the rotation geometry, and reject outliers, so a complete dataset typically records 80 to 95 percent of the theoretical count.

Does this calculator work for non-orthogonal unit cells?

It assumes all three cell angles are 90 degrees, so the volume is a times b times c. Monoclinic and triclinic cells have a smaller true volume than that product, and every result that depends on volume, meaning the reflection counts, scales down in proportion. Grid spacing and grid dimensions remain correct.

How does resolution change map and data size?

Both scale with the inverse cube of the resolution limit. Halving d from 3.0 to 1.5 angstroms multiplies the unique reflection count by eight and multiplies the number of grid points by eight as well, since the required grid spacing also halves along every axis.

What resolution counts as high resolution?

For macromolecules, better than 1.5 angstroms is high resolution and resolves individual atoms, 1.5 to 2.5 angstroms is the routine working range where side chains are clear, and worse than 3.0 angstroms shows only backbone and secondary structure. Small-molecule crystallography routinely reaches 0.8 angstroms.

Why is 4 bytes per grid point assumed?

CCP4 and MRC format maps store density as 32-bit floating point numbers by default, which is 4 bytes each. Some programs offer a 1-byte or 2-byte integer mode that shrinks the file four-fold or two-fold, at the cost of precision in the stored density values.