Buffalo Base Reflectivity May 1999 How do we

Buffalo Base Reflectivity May 1999 How do we interpret this image? Brock Estimates J. E. Black

Interpreting NEXRAD Base Reflectivity Images John Black Brock University • Two kinds of bird densities • Volume density from NEXRAD base reflectivity images • Total bird numbers • Bird current density (flow) • Bird Current Density from NEXRAD Doppler images • A Turbine Scenario • Spring Migration Estimates J. E. Black

Two Kinds of Bird Density • In working with birds on radar there are two sorts of density to consider • Bird volume density, the number of birds in a cubic kilometer of space and • Bird current density, the number of birds crossing a square kilometer per hour Estimates J. E. Black

Some Weather Radar Migration Facts • Nocturnal passerine migrants do not flock. On the Buffalo weather radar at night I have only once seen evidence of a flock of birds, perhaps shorebirds or ducks heading north. • I have heard of diurnal hawk migration being seen on a radar on the Texas Gulf coast. Estimates J. E. Black

Volume Density • The base reflectivity DBZ provides a measure of the bird volume density, that is how many birds are found in a cubic kilometer. • As a rough guide • volume density = 3* 10(DBZ/10) Estimates J. E. Black

Buffalo Base Reflectivity May 1999 Brock Estimates J. E. Black

Effects of Earth Curvature Estimates J. E. Black

Estimate of Bird Numbers in Radar Image Density of 10 cm 2 Birds = 3. 0*Z 20 km 30 -360 m 75 km 160 km 360 -1600 m 1600 -4300 m Add these densities(weighted) to get total of 140 birds in a column 1 km square which extends from 30 to 4300 m (can average density at each range) Estimates J. E. Black

What if all the birds aloft landed 1 km 140 birds 4. 27 km 140 birds/km**2 Estimates J. E. Black ground

A Bird Landing Scenario What if the 140 birds /km 2 were to land peacefully? A little thought shows there would be about one bird on the ground every 84 meters, almost only one bird if you walked the length of a football field in any direction, not a lot of birds. Estimates J. E. Black

Buffalo Base Reflectivity May 1999 Brock Estimates J. E. Black

Total birds in radar range May 15/16, 1999. (Birds Contributing to Echoes in Image) Total Birds/km 2 = 140 Approximate Radius of Echoes = 150 km Area Covered by Echoes = 70, 000 km 2 Total birds in radar range = 9, 900, 000 Estimates J. E. Black

Bird Current Density or Flow • The product of the bird speed (km/hour) and the bird volume density(birds/km 3) gives the bird current density in units of birds crossing a km 2 per hour. • This density is what we need to look at if we are considering the impact of wind turbines on birds in flight Estimates J. E. Black

Buffalo Doppler Velocity Fall Average velocity at each range (height) Estimates J. E. Black

Turbine Scenarios. • The number of birds crossing the area swept out by the wind turbine per hour is given by the current density at the height of the turbine times the area swept out by the turbine in km 2. • The number of birds actually hit by the turbine blades is then given by this number times the probability a bird passing through the sweep area will be hit by the blades Estimates J. E. Black

Estimate What if 100 birds per cubic kilometer approach a turbine at 50 km/hour. Then there is a current density of 5000 birds/km squared/hour. For a turbine of radius 27 meters the area swept out by the turbine is approximately 0. 0023 km squared so about 11 birds cross the turbine area per hour. What we do not know is what percentage of these birds would suffer a fatal collision with the blades. Estimates J. E. Black

Flight density or Station Passage Rate • In studying migration the number of birds passing overhead is of interest. This means adding up the birds in a column, as we did before, to get birds per kilometer squared. Then the product of this with the speed give the total birds crossing a kilometer through the observer per hour. Estimates J. E. Black

Flow May 15/16, 1999. Birds Contributing to Echoes in Image Total Birds/km 2 = 140 Approximate Speed of Birds = 50 km / hr Flight Density =140*50=7000 birds per km/hr Station Passage Rate (10 hr night) = 7000*10 =70, 000 birds/km/night Birds crossing through the entire radar range over the night = 70000*300 = 21 million! Estimates J. E. Black

Flow Scenario What if migrating birds are stopped from crossing from the shore to the lake for 1 hour? This could happen if bad weather conditions were present over the lake such as rain. Then, from the flow arguments, we see about 7000 birds could pile up along a km of Lakeshore in one hour. Now we have about 7 birds every meter we walk along the shore. Under these conditions the suitability of the shore for birds becomes crucial. Often the birds are exhausted and hungry when landing. The same sort of arguments apply to birds forced by weather to land after crossing the lake, or even just after a full nights migration. . Estimates J. E. Black

Worst case turbine scenario If all birds in the vertical column were forced down to heights intercepted by the wind turbines The width of a typical wind turbine is about 0. 045 kilometers. So if 7000 birds cross 1 km in an hour 315 will cross 0. 045 km in an hour. What is not known is what percentage of these would have fatal collisions with the blades? Estimates J. E. Black

Spring 1999 Estimates J. E. Black

Estimates J. E. Black

Vertical Radars • Vertical radars measure the bird current density. If the speed is known the volume density can be calculated from the formula vol. density = current density/speed • With a vertical radar adding the current density(weighted) at the various heights gives the total flow or flight density. Estimates J. E. Black

Volume density from Z • One can show that the • average volume density of birds(birds/km 3)*average radar cross -section of the birds (cm 2) = 28*Z • Bird cross-sections range from 10 to 20 cm 2, they depend on bird size, aspect and radar wavelength in a complicated way. • For 10 cm 2 birds we have bird volume density = 3*Z Estimates J. E. Black

Velocity from Weather Radar Doppler Images • Often the images are corrupted. • Not sure why • If not corrupted then one can do a fit at a given range to get the velocity average at that range (and the velocity for the corresponding height) VAD Estimates J. E. Black