The square root is the whole story
A parachute holds a steady speed when its drag equals the weight hanging under it, and drag goes as the square of speed. Rearrange and the descent rate is the square root of twice the weight over the air density times the drag coefficient times the area. Every counterintuitive thing about parachutes falls out of that square root.
Double the mass under a given canopy and the rate goes up by 1.41, not by 2. At the defaults — 14 oz on a 25.8 in flat canopy at a drag coefficient of 0.9 — the rate is 15 ft/s; put 28 oz under the same canopy and it is 21.2 ft/s. Halve the canopy diameter instead and the rate does double, to 30 ft/s, because area goes as the square of diameter and the two squares cancel. And to halve a descent rate you need four times the area, which means twice the diameter: 51.6 in instead of 25.8.
What a slow descent costs in ground
From 1,100 ft at 15 ft/s the rocket is in the air for 73 seconds. In a 9 mph wind, which is 13.2 ft/s, that is 968 ft of drift — nearly a fifth of a mile, and if you do not know in advance which way the wind will be from, the circle of that radius covers 68 acres.
Split the descent and the picture changes. Come down from 1,100 ft to 500 ft at 45 ft/s, then switch to the main: the fast leg takes 13.3 seconds and drifts 176 ft, the main leg takes 33.3 seconds and drifts 440 ft, for 616 ft total. That is 64 percent of the single-canopy drift and the rocket still lands at 15 ft/s. The reason it works is that the drift is proportional to the time, and putting the fast part at the top removes the majority of the time while removing none of the softness at the bottom.
The small canopy for 45 ft/s is 8.6 in across at the same drag coefficient, which is a handkerchief. That is why it works and also why it is easy to underestimate.
Which area is your drag coefficient referenced to
This is the single most common way to get a parachute calculation wrong by a factor, and it is not a subtle error. A canopy has two areas: the flat spread-out area you get by laying it on the floor and measuring across, and the smaller projected area of the inflated dome in flight. A drag coefficient quoted against the projected area is much larger than one quoted against the flat area, because it has to describe the same force over less area.
This page defines diameter as the flat one and expects a coefficient to match. If you take a coefficient from a source that means the projected area and feed it to a flat diameter, the answer comes out wrong in a direction that looks entirely plausible. Check which one your figure belongs to before you cut anything.
What the model leaves out, and it is not small
Steady-state descent under a fully inflated canopy is the easy part of a recovery and the only part this describes. The opening is not modelled: there is a period of freefall before the canopy fills, there is a shock as it does, and both of them happen at speeds well above the descent rate. The swing is not modelled either, nor is a canopy that fouls or only partly opens.
The drift figure is the softer of the two. It multiplies one wind speed by a descent time, and the wind is neither steady nor the same at 1,000 ft as at 50. Ground-level wind understates what the rocket sees up high, sometimes badly. Treat 616 ft as a way of comparing two arrangements rather than as the radius of a search.
Questions people ask
What descent rate should a model rocket land at?
That is your call and possibly your range rules, not this page. What the page will do is tell you what any rate costs. Slower always means a bigger canopy and more time in the air, and the time is what the wind works on: at 9 mph, every extra second of descent is another 13 ft downwind. A rate that gets the rocket back in one piece and inside the field is a trade between those two, and the numbers for both are printed side by side so you can make it with figures rather than by feel.
Why does a heavier rocket not need a proportionally bigger parachute?
Because the descent rate goes as the square root of the load. Doubling the mass under a fixed canopy multiplies the rate by 1.41, so to hold the original rate you need twice the area, which is only 1.41 times the diameter. A 14 oz rocket wanting 15 ft/s takes a 25.8 in canopy; 28 oz at the same rate takes 36.5 in, not 51.6.
Does a drogue and main really cut the drift that much?
At the defaults it takes 968 ft down to 616, which is a third off. The mechanism is straightforward: drift is wind speed times time in the air, and time is altitude divided by rate. Descending the top 600 ft at 45 ft/s instead of 15 removes 40 seconds of hanging about. The lower you set the main, the more you save and the less margin there is if it is late. That trade is one for your own judgement and your range rules, and the page will run any figures you put in it.
What drag coefficient should I use for a parachute?
Whatever your canopy maker publishes, and check which area it is referenced to before using it. This page works in flat spread-out area, so it wants a coefficient quoted against that. The 0.9 in the field is a placeholder, not a specification of any canopy. If you have flown the canopy and timed a descent from a known altitude, you can solve for your own coefficient by adjusting it here until the page reproduces the rate you measured.
How much bigger does the field need to be than the drift?
That is a range and club question, not a physics one, and this page will not answer it. What it gives you is the drift for a stated wind and the area of the circle that drift describes if the wind direction is unknown — 616 ft and 27 acres at the defaults. Whether that fits inside where you are allowed to fly, and what margin your site asks for on top, is set by whoever runs the site.