Buoy wind accuracy

[StormSkeptic]

Curious question. How accurate are the wind measurements of data buoys in a hurricane, especially a major? Do the high seas block the wind at times and cause the wind to be under reported?

[Tampa Bay Hurricane]

yes High seas may cause significant underestimations of wind IMO...

[vbhoutex]

yes High seas may cause significant underestimations of wind IMO...

At the same time I would expect that the wind could also be over reported when there are huge waves lifting the bouy higher into the wind???? My guess is they are relatively accurate or at least it all averages out.

[MiamiensisWx]

There is also the matter of marine winds versus winds over land, which depends on the amount of friction.

[Margie]

NHC TPC references buoy obs all the time in the discussions and they are also referenced in the final reports.

In 2004 and 2005 it appeared that several key buoys were knocked out or set adrift by some of the more powerful storms and weren't available for the remainder of the season (I think Emily got one in the Carribean).

[terstorm1012]

NHC TPC references buoy obs all the time in the discussions and they are also referenced in the final reports.

In 2004 and 2005 it appeared that several key buoys were knocked out or set adrift by some of the more powerful storms and weren't available for the remainder of the season (I think Emily got one in the Carribean).

Katrina and Rita killed most in the Gulf

[f5]

NHC TPC references buoy obs all the time in the discussions and they are also referenced in the final reports.

In 2004 and 2005 it appeared that several key buoys were knocked out or set adrift by some of the more powerful storms and weren't available for the remainder of the season (I think Emily got one in the Carribean).

Katrina and Rita killed most in the Gulf

Bouys can't withstand a category 5

[WxGuy1]

One thing to remember as well is that as the buoy rides up and down a big wave, it is not vertical. I'd imagine that you'd have something like Vbuoy = Vreal * cos(theta) where theta is the angle of the sea off of horizontal. For example -- suppose there's a massive wave, which causes the buoy to tilt 30 degrees. This means that the anemometer is no longer horizontal, and it's actually measuring the real wind times cos(30 deg), or less than the real wind (since cos(30deg) < 1). Of course, this is a simplification since it assumes that the wind is entirely horizontal near the surface, which is wouldn't be since the waves pretty much satisfy an impermeability condition, meaning that there could be an appreciable vertical component to the wind that follows the waves (strongest right above the water, becoming increasingly less important with height in the boundary layer).

And actually, I'd think that the winds would be higher on top of a wave than on the bottom of the wave not so much because the effects of friction would be less at 10m above *mean* sea level (top of wave) than at 10m below mean sea level (bottom of wave), but because of the flow would have to speed up as it rounds the crests of the waves (and weakens as it rounds the valleys). This is similar to how the flow increases on top of an airplane wing -- because the surface is curved, the air immediately on top of the wing moves faster than the air below the wing. Credit goes to Bernoulli... Of course, this assumes an invicid boundary condition, which is not satisfied as the air-sea interface. Point? The effects of wave action on the buoy-reported wind speeds is complicated. I think the only thing we can do is assume that the various effects average themselves out...

[Tampa Bay Hurricane]

One thing to remember as well is that as the buoy rides up and down a big wave, it is not vertical. I'd imagine that you'd have something like Vbuoy = Vreal * cos(theta) where theta is the angle of the sea off of horizontal. For example -- suppose there's a massive wave, which causes the buoy to tilt 30 degrees. This means that the anemometer is no longer horizontal, and it's actually measuring the real wind times cos(30 deg), or less than the real wind (since cos(30deg) < 1). Of course, this is a simplification since it assumes that the wind is entirely horizontal near the surface, which is wouldn't be since the waves pretty much satisfy an impermeability condition, meaning that there could be an appreciable vertical component to the wind that follows the waves (strongest right above the water, becoming increasingly less important with height in the boundary layer).

And actually, I'd think that the winds would be higher on top of a wave than on the bottom of the wave not so much because the effects of friction would be less at 10m above *mean* sea level (top of wave) than at 10m below mean sea level (bottom of wave), but because of the flow would have to speed up as it rounds the crests of the waves (and weakens as it rounds the valleys). This is similar to how the flow increases on top of an airplane wing -- because the surface is curved, the air immediately on top of the wing moves faster than the air below the wing. Credit goes to Bernoulli... Of course, this assumes an invicid boundary condition, which is not satisfied as the air-sea interface. Point? The effects of wave action on the buoy-reported wind speeds is complicated. I think the only thing we can do is assume that the various effects average themselves out...

Thank you. Very interesting discussion.