Distance / time / speed calculations ?

[Teroc]

Anyone got any really neat ways of working out these problems mentally ?
Im using a method wherby I memorise each 5 minute increment as a fraction up to 60 minutes and go from there. Its a bit long winded and prone to Errors though.
Gotta be an easier way !!

[GT]

Not entirely sure what you mean! Try, though, 6 minute increments. Sounds a bit odd a first but, as 6 minutes is 1/10th of an hour, in that time, the distance travelled will be 1/10th of your ground speed.

For example, with a GS of 130kts, you will travel 13nm every 6 minutes. Now, if you have a calibrated distance marker at hand, mark it, hold it, or whatever at 13nm. This length represents 6 minutes flying time.

Advance this along your planned route on your chart and you have the time to your destination, in multiples of six minutes, without ever needing to know the distance to it. I find this method particularly useful for planning diversions in the air.

Hope this helps, hasn't confused you, and others agree. I'm sure people will come up with other methods.

Regards, GT.

[Jumbo Jockey]

Teroc: the following explanation looks complicated, but follow it through slowly and practice it a few times, and you'll be doing d/s/t calculations in your head quickly before you know it:

Don't forget the big difference between written maths and mental maths: With pencil&paper/calculator handy, you're trying to reduce to a minimum the number of steps towards the final answer. In mental maths, whilst there is admittedly an element of time pressure, the trick is to break down the calculation into lots of small, easy, and easily memorable stages.

For example, at 190 kts, how much time to go 30 nm? In written maths, the calculation is 30 divided by 190 and then multiply by 60 to turn the answer into minutes, which is great if you have a whizz wheel or a calculator to hand, and indeed an autopilot to fly the aeroplane whilst you fiddle with them. Mentally, try and break the calculation down into manageable chunks. One useful trick to achieve that is to move decimal places around: 190 kts is 190 nm in 60 minutes, or 19 nm in 6 minutes or even 1.9 nm in 0.6 minutes if you really want to be pedantic. Since most people can also halve or double with reasonable accuracy, it's reasonably easy to say that the same 190 kts (or 19 nm in 6 mins) is also (doubling each time) 38 nm in 12 minutes, and 76 nm in 24 minutes. It's also (halving) 9.5 nm in 3 minutes. So your thought process to do the sum goes like this (and bear in mind we're not looking for tenth-of-a-second accuracy here, get within half a minute and you're doing well!

19 nm in 6 minutes is the first memory item. Halving the above gives 9.5 nm in 3 mins, which you have to hold in the brainbox long enough to add it to 19 in 6, making 28.5 nm in 9 mins, which is now the figure to remember. Only a mile and a half to find! Again, shift some decimals around: 28.5 in 9 mins is pretty close to 30 nm in 9 mins, or 3 nm in 0.9 mins, so why not halve that and call it 1.5 nm in roughly half a minute. Final score = 30 nm in about 9.5 minutes. (The answer by calculator is 9 minutes 28 seconds, which ought to be close enough for government work).

There's a lot more steps in doing it in your head like that, but each step isn't difficult and is also easily remembered for carrying forward to the next step in the calculation.

Like I said before, it sounds complex when you write it down, but read it through slowly and keep practicing and you'll find it works well.

Good luck!

[Pete-O-Heat]

Im with GT. I was taught to use 10nm markers, but found 6min markers to be alot more accurate and simple to use. Suggest though to work from a positive fix after TOC to destination (as opposed to working back), because as you all know, the climb NEVER goes as planned.

[foghorn]

Good advice, JJ, here's an alternative that works for me.

What you have to do is look at the numbers in the problem, and find a sum that is easy to do from a times table. Then you adjust the answer from your easy sum to get close to the result. The real knack in this is getting the feel for how much this adjust ment should be.

Take our 190kts, 30nm example as before. An easy sum here is:

30 x 6 = 180

so at 180 knots we do 6 lots of 30nm per hour. So how many minutes for one 'lot':

1 hour = 60 mins = 10 mins
------ -------
6 6

So at 180 knots we will take 10 minutes to do 30nm.

But we're really going at 190 knots so we need to take a bit off the time... it's only 10 knots faster out of 180 knots, and we have ten minutes to play with, so let's estimate a figure to subtract. 0.5 look like it will be near enough:

10 minutes approx - 0.5 minutes guessed adjustment = 9.5 minutes!!!

If you want to be really clever about working out the adjustment you can see that I did a simple ratio in my head to get the adjustment 'n':

10 knots = n minutes adjustment
--------- --------------------
180 knots 10 minutes approximation

n = 10 x 10 = 10/18 ~ 0.5!!!
-------
180

Do a few of these and you get a feel for the numbers, and when you can approximate.

[This message has been edited by foghorn (edited 23 February 2001).]

[Teroc]

Great advice guys. Thanks.
Keep em coming.

Teroc.

[TimC]

Have a look on the OASC Maths thread on the mil pilots forum. If you want a load of questions to go through, then have a look on my website www.timc.clara.net (http://www.timc.clara.net) .

[Jumbo Jockey]

Teroc - You might also have a look at Martyn Smith's little book "Diversion Planning", which is very good for showing you how to do most of the routine navigational problems in your head. There are only 21 pages in it (single-sided pages, at that) and it's very readable. Before you ask, no my name isn't Martyn Smith! I think it's one of the best books I've seen on the subject, £8.50 and I think you'll get it from Transair.