Deceleration Distances

Sometimes, deciding when to pull the power back and slow down feels like a game of chicken. I wanted to do some math to find out the hard numbers.

I often turn base and final at 160kts, but maintaining 250kts as long as possible is advantageous. How close to the turn can we wait before slowing down?

The aircraft's speed tape provides a trend indicator. The trend indicator is a magenta arrow that points up or down from the speed pointer and indicates what the speed will be in 10 seconds. Reducing speed at a rate of at 5kts per 10 seconds is comfortable and typical. 5kts per 10 seconds can be achieved while descending about 1100 to 500fpm at idle power.

Our equation for speed as a function of time is as follows: $$v_{\text{fin}}=r(t)+v_{\text{init}}$$ Where:
\(v_{\text{init}}\) is the initial speed,
\(v_{\text{fin}}\) is the final speed,
\(r\) is the rate of deceleration in \(\frac{\text{kts}}{\text{sec}}\),
and \(t\) is time in seconds.

Solving for \(t\) we can determine how long slowing from 250kts to 160kts at 5kts per 10secs will take. $$t=\frac{v_{\text{fin}}-v_{\text{init}}}{r}=\frac{160\text{kts}-250\text{kts}}{\frac{-5\text{kts}}{10\text{sec}}}=\frac{-90\text{kts}}{-0.5\frac{\text{kts}}{\text{sec}}}=180\text{sec}$$ To determine how much distance this will cover requires an integral. $$\int_{t=0\text{ sec}}^{180} r(t)+v_{\text{init}}\,dt$$ $$\Big[r\frac{t^2}{2}+v_{\text{init}}(t)\Big]_{180}-\Big[r\frac{t^2}{2}+v_{\text{init}}(t)\Big]_{0}$$ $$\Big[r\frac{(180\text{sec})^2}{2}+v_{\text{init}}(180\text{sec})\Big]-\Big[r\frac{(0\text{sec})^2}{2}+v_{\text{init}}(0\text{sec})\Big]$$ $$r\frac{(180\text{sec})^2}{2}+v_{\text{init}}(180\text{sec})$$ $$=\frac{-5\text{kts}}{10\text{sec}}\cdot\frac{(180\text{sec})^2}{2}+250\text{kts}(180\text{sec})$$ $$=\frac{-5\text{kts}}{10\text{sec}}\cdot\frac{(180\text{sec})^2}{2}+250\text{kts}(t)$$ $$=\frac{-5\text{nm}}{\text{hr}}\cdot\frac{1}{10\text{sec}}\cdot\frac{(180\text{sec})^2}{2}+\frac{250\text{nm}}{\text{hr}}\cdot180\text{sec}$$ $$=\frac{-5\text{nm}}{3600\text{sec}}\cdot\frac{1}{10\text{sec}}\cdot\frac{(180\text{sec})^2}{2}+\frac{250\text{nm}}{3600\text{sec}}\cdot180\text{sec}$$ $$=-2.25\frac{\text{nm }\text{sec}^2}{\text{sec}^2}+12.5\frac{\text{nm }\text{sec}}{\text{sec}}$$ $$=10.25\text{nm}$$

It takes 10nm to slow from 250kts to 160kts if the speed trend vector points down 5kts.

Here is a table of different speed reductions and rates that might be interesting: