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 velocity as a function of time is as follows: $$v(t)=\left(\frac{1\text{hr}}{3600\text{sec}}\right)(r \cdot t+v_{init})$$ Where:
\(v_{\text{init}}\) is the initial speed in \(\text{kts}=\frac{NM}{hr}\),
\(r\) is the rate of acceleration in \(\frac{\text{kts}}{\text{sec}}\),
and \(t\) is time in seconds.

Because this function is a function of \(t\) which has units of seconds, the factor \(\frac{1\text{hr}}{3600\text{sec}}\) is required to make \(v(t)\) resolve to units of NM per second.

To determine how much distance this will cover requires an integral. $$\int v(t)\,dt=\int \left(\frac{1\text{hr}}{3600\text{sec}}\right)(r \cdot t+v_{init})\,dt=\left(\frac{1\text{hr}}{3600\text{sec}}\right)\left(r\cdot \frac{t^2}{2}+v_{init}\cdot t\right)$$

The last piece is to figure out how much time it takes to change from one speed to another. We can use the following equation: $$t=\frac{v_{\text{final}}-v_{\text{init}}}{r}$$ Where:
\(v_\text{final}\) is the final target speed in \(\text{kts}\).

Let's plug these together: $$\left(\frac{1\text{hr}}{3600\text{sec}}\right)\left(r\cdot \frac{(\frac{v_{\text{final}}-v_{\text{init}}}{r})^2}{2}+v_{init}\cdot \left(\frac{v_{\text{final}}-v_{\text{init}}}{r}\right)\right)$$ $$=\left(\frac{1\text{hr}}{3600\text{sec}}\right)\left(r\cdot \frac{(\frac{v_{\text{final}}-v_{\text{init}}}{r})^2}{2}+v_\text{init}\cdot \left(\frac{v_{\text{final}}-v_{\text{init}}}{r}\right)\right)$$ $$ =\left(\frac{1\text{hr}}{3600\text{sec}}\right)\left(\frac{v_\text{final}^2-2(v_\text{final})(v_\text{init})+v_\text{init}^2}{2r}+v_\text{init}\cdot\left(\frac{v_{\text{final}}-v_{\text{init}}}{r}\right)\right) $$ $$ =\left(\frac{1\text{hr}}{3600\text{sec}}\right) \left( \frac{v_\text{final}^2}{2r} -\frac{v_\text{final}\cdot v_\text{init}}{r} +\frac{v_\text{init}^2}{2r} +\frac{v_\text{init}\cdot v_\text{final}}{r} -\frac{v_\text{init}^2}{r} \right) $$ $$ =\left(\frac{1\text{hr}}{3600\text{sec}}\right) \left( \frac{v_\text{final}^2}{2r} -\frac{v_\text{init}^2}{2r} \right) $$ $$ =\frac{v_\text{final}^2-v_\text{init}^2}{7200\frac{\text{hr}}{\text{sec}}\cdot r} $$ Here is our generalized equation which can be used for any initial speed, target speed, and rate of change.

Let's see how long it will take to slow from 250kts to 160kts at a rate of 5kts per 10 seconds: $$ =\frac{(160\text{kts})^2-(250\text{kts})^2}{7200\frac{\text{hr}}{\text{sec}}\cdot (-0.5\frac{\text{kts}}{\text{sec}})}$$ $$=10.25\text{NM}$$

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:

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