<p>The winds are at a maximum velocity Vmax in they eye wall, a thin annulus immediately around the eye. With increasing distance the winds fall off. If a hurricane was a Rankine vortex, the wind velocity at radius R from the center of the eye would be V(R) = Vmax X (Reye/R), where Reye is the radius of the eye. In reality friction usually prevents a hurricane from being a perfect Rankine vortex. Instead it typically a modified Rankine vortex, with V(R) = Vmax X (Reye/R)^n, where n is typically in the range 1/2 &lt; n &lt; 3/4 (perhaps n = 2/3 is the best overall average value). Of course this is a radially-smoothed-out wind velocity distribution. In the approach of an actual hurricane, wind does not increase monotonically as R decreases. But V(R) = Vmax X (Reye/R)^2/3 typically approximates the secular trend of its increase.</p>

CC BY-SA 3.0
2017-09-15T12:19:04.780Z
Perhaps it would be useful to add a "lay summary" to this answer, since the OP specified that they were not a meterologist. If I follow the formula correctly, you are saying that the answer to their first question is that the larger the eye, the stronger the winds just outside it, and that the falloff further from the eye is a bit less than inversely proportional to the distance?
Semidiurnal Simon
CC BY-SA 3.0
2017-09-16T19:12:03.433Z
The wind speed is a function of the ratio Reye/R, not of Reye itself. Typically in a modified Rankine vortex, n is assumed constant (not varying as a function of Reye/R). Accounting for possible variations of n would make the formula more complex, and assuming constant n usually is sufficiently accurate for at least approximating the secular trend. If n = 1 angular momentum is conserved within a hurricane itself. If n < 1 the hurricane must impart some angular momentum to the Earth to conserve total angular momentum, thus altering Earth's rotation very slightly.
Jack Denur
CC BY-SA 3.0
2017-09-16T19:14:03.870Z
So the wind speed at the eye wall does not vary with the size of the eye?

Either way, since the OP specifically said they were "not a meterologist", I think a lay explanation - based in English rather than maths - would be helpful :-)
Semidiurnal Simon
CC BY-SA 3.0
2017-09-16T19:23:25.640Z
There is no definite relation between a hurricane's maximum wind speed and the size of the eye or the total size of a hurricane. Large hurricanes have at least approximately the same probability distribution for maximum wind speeds as small ones. Wind speed as a function of distance R from the center depends on the ratio Reye/R, not Reye by itself or R by itself. Conservation of angular momentum within a hurricane itself implies wind speed V inversely proportional to R. Due to friction there is some loss of angular momentum to the Earth as winds spiral inwards.
Jack Denur
CC BY-SA 3.0
2017-09-12T22:56:38.120Z
2017-09-12T22:56:38.120Z
Jack Denur
<p>I am not a meteorologist but watching Hurricane Irma (2017), I have a couple questions about how hurricanes keep their formation based on geographical landmarks.</p>

<ol>
<li><p>How does the radius of the hurricane's eye affect the winds felt by
those at different distances from the eye?</p></li>
<li><p>When a hurricane reaches land, how does the land affect the eye's
formation?</p></li>
</ol>

CC BY-SA 3.0
2017-10-16T12:54:24.517Z
That's two different questions. Please split them. The answer that's already here answers your first question, so make the *land* issue a new question.
Jan Doggen
<tropical-cyclone>
Hurricane eye on landfall and relation to wind speeds
ButtahNBred
# Hurricane eye on landfall and relation to wind speeds

(Asked by: username_0 on 2017-09-09)

I am not a meteorologist but watching Hurricane Irma (2017), I have a couple questions about how hurricanes keep their formation based on geographical landmarks.

1. How does the radius of the hurricane's eye affect the winds felt by
   those at different distances from the eye?
2. When a hurricane reaches land, how does the land affect the eye's
   formation?

**Tags:** tropical-cyclone

> **Comments:**
> **username_1**: That's two different questions. Please split them. The answer that's already here answers your first question, so make the *land* issue a new question.  

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## 1 answers

### Answer by username_2 on 2017-09-12. Score: 0 | Accepted answer

The winds are at a maximum velocity Vmax in they eye wall, a thin annulus immediately around the eye. With increasing distance the winds fall off. If a hurricane was a Rankine vortex, the wind velocity at radius R from the center of the eye would be V(R) = Vmax X (Reye/R), where Reye is the radius of the eye. In reality friction usually prevents a hurricane from being a perfect Rankine vortex. Instead it typically a modified Rankine vortex, with V(R) = Vmax X (Reye/R)^n, where n is typically in the range 1/2 < n < 3/4 (perhaps n = 2/3 is the best overall average value). Of course this is a radially-smoothed-out wind velocity distribution. In the approach of an actual hurricane, wind does not increase monotonically as R decreases. But V(R) = Vmax X (Reye/R)^2/3 typically approximates the secular trend of its increase.

> **Comments:**
> **username_3**: Perhaps it would be useful to add a "lay summary" to this answer, since the OP specified that they were not a meterologist. If I follow the formula correctly, you are saying that the answer to their first question is that the larger the eye, the stronger the winds just outside it, and that the falloff further from the eye is a bit less than inversely proportional to the distance?  
> **username_2**: The wind speed is a function of the ratio Reye/R, not of Reye itself. Typically in a modified Rankine vortex, n is assumed constant (not varying as a function of Reye/R). Accounting for possible variations of n would make the formula more complex, and assuming constant n usually is sufficiently accurate for at least approximating the secular trend. If n = 1 angular momentum is conserved within a hurricane itself. If n < 1 the hurricane must impart some angular momentum to the Earth to conserve total angular momentum, thus altering Earth's rotation very slightly.  
> **username_3**: So the wind speed at the eye wall does not vary with the size of the eye?  Either way, since the OP specifically said they were "not a meterologist", I think a lay explanation - based in English rather than maths - would be helpful :-)  
> **username_2**: There is no definite relation between a hurricane's maximum wind speed and the size of the eye or the total size of a hurricane. Large hurricanes have at least approximately the same probability distribution for maximum wind speeds as small ones. Wind speed as a function of distance R from the center depends on the ratio Reye/R, not Reye by itself or R by itself. Conservation of angular momentum within a hurricane itself implies wind speed V inversely proportional to R. Due to friction there is some loss of angular momentum to the Earth as winds spiral inwards.  

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