I believe you can go down to as few as

22

sprinklers.

Approach:

Instead of using circles, I was looking at the puzzle in terms of hexagons. The reasoning behind this is that you will always have overlap between circles, and you want to minimize the overlap. The tightest packing is achieved for circles where the center points form regular hexagons.

Necessary formulas:

Area of a hexagon with a side length s: $A=3/2 \; \sqrt{3} s^2$
Area of a hexagon with apothem a: $A = 3/2 \; \sqrt{3} 4/3 a^2 = 2 \sqrt{3} a^2 $

The area will be tiled by small hexagons which are circumscribed by a circle of radius 1m. The area of those hexagons is $A_s = 1/2 \; \sqrt{3}\; (1m)^2 = \sqrt{3}/2 \; m^2$.

Then:

The tiling formed by the hexagons will again be a hexagon. To fill the circle, we need this to be the hexagon of which the inscribed circle equals our circular lawn. This is the hexagon with the area $A_l = 2 \sqrt{3} (4m)^2 = 32 \sqrt{3} m^2$

Therefore, the fewest number of sprinklers is:

$\frac{32 \sqrt{3}}{3/2 \sqrt{3}} = \frac{64}{3} \approx 21.33$ The smallest whole number larger than this is 22.

I am not in possesion of a graphical proof yet. At least this should be a lower bound.

I believe you cannot go further down then

22

sprinklers.

I can't show a solution that it's possible to do it with this number yet. But here is my reasoning why I believe it's not possible to go any lower.

Approach:

Instead of using circles, I was looking at the puzzle in terms of hexagons. The reasoning behind this is that you will always have overlap between circles, and you want to minimize the overlap. The tightest packing is achieved for circles where the center points form regular hexagons.

Necessary formulas:

Area of a hexagon with a side length s: $A=3/2 \; \sqrt{3} s^2$
Area of a hexagon with apothem a: $A = 3/2 \; \sqrt{3} 4/3 a^2 = 2 \sqrt{3} a^2 $

The area will be tiled by small hexagons which are circumscribed by a circle of radius 1m. The area of those hexagons is $A_s = 1/2 \; \sqrt{3}\; (1m)^2 = \sqrt{3}/2 \; m^2$.

Then:

The tiling formed by the hexagons will again be a hexagon. To fill the circle, we need this to be the hexagon of which the inscribed circle equals our circular lawn. This is the hexagon with the area $A_l = 2 \sqrt{3} (4m)^2 = 32 \sqrt{3} m^2$

Therefore, the fewest number of sprinklers is:

$\frac{32 \sqrt{3}}{3/2 \sqrt{3}} = \frac{64}{3} \approx 21.33$ The smallest whole number larger than this is 22.