# Crossing the desert (in a steam powered car)

1. Jun 22, 2004

### ceptimus

By one of those inexplicable quirks of puzzle land, it is necessary that you cross a desert in a steam-engine powered car.

Your steam car is very reliable, but it can't carry enough coal to cross the whole desert without refuelling. Fully loaded with coal, it can travel just half way across. The car has a very efficient condenser, so water for the boiler is not a problem. You also have adequate supplies of food and drinking water.

So you hatch a cunning plan... You will drive out some distance into the desert, leave some coal there, and then drive back to the starting point. You aim is to establish fuel depots out in the desert, enabling you to refuel on subsequent journeys. By this means, you hope to drive out further into the desert than would otherwise be possible.

There is another constraint. You have just 8 full loads of coal at your starting point, and there is no prospect of more being delivered till next year.

You decide to get as far as possible in your car, and (if necessary) abandon it and walk whatever distance remains. The desert is 1000 miles wide.

2. Jun 22, 2004

### Gokul43201

Staff Emeritus
The "immediate" guess would be 1/8 of 1000 or a 125 mile walk. I'm sure there's a way to make this smaller; most likely 0.

Also, this is wasteful, and you end up not using all the coal.

Last edited: Jun 22, 2004
3. Jun 22, 2004

### jcsd

1) you drive an eight of the way out into the desert and drop a half-load of fuel there and drive back.

2)Then you drive to the eighth of the way point again, pick up half the fuel you left there, drive out to the quarter way point, drop a half load of fuel there and drive back, picking up the fuel you left at the 1/8 of the way way point to get you back.

3),4) repeat steps 1) and 2)

5) drive out to the quarter way point pick up a half load of fuel, dump at the 3/8 point, return (which means you'll have to pick up the rest of the fuel at the quarte way point

6),7) repeat steps 1 and 2)

Unfortuantely my stategy means that you only get 7/8 of the way through the desert and have to walk 125 miles through the desert where you will invariably die.

4. Jun 22, 2004

### Gokul43201

Staff Emeritus
If you go out 2/15 ths each time you do better - no wastage. That way you make 90% by car and have to walk the last 100.

EDIT : Mistake here...will correct.

First trip : go out 2/15ths, leave 7/15ths of the load and return. Repeat. At the end of 7 such round trips and 1 one-way trip, you have 4 loads. Now from this point, head out 1/7th, drop off 3/7ths of the load and return. Make 3 round trips and 1 one-way trip. Now you have 2 loads at the point x=2/15 + 1/7. Finally head out 1/6ths from here, drop off 1/3 load and return. Take the rest and get back here and you have 1 full load. Use this to go half way.

Grand total = 94.2857%

A paltry 57 miles to walk. Nice evening stroll, wot ?

I think this can be improved...

Last edited: Jun 22, 2004
5. Jun 22, 2004

### jcsd

I suspected that dividing the desrt into eight points was not the most effeicent way.

6. Jun 22, 2004

### Hurkyl

Staff Emeritus
Bah, I can only cross 101.09% of the desert. Y'all waste too much coal.

BTW, when I heard this one, the setup was that you could leave drums of gasoline in the desert to fuel a normal vehicle.

7. Jun 22, 2004

### Gokul43201

Staff Emeritus
Hurkyl, you poor creature. Now you have to walk back 11 miles.

Say no more.

8. Jun 22, 2004

### Gokul43201

Staff Emeritus
1/30 + 1/26 + 1/22 + 1/18 + 1/14 + 1/10 + 1/6 + 1/2 = 1.01

Now, I can sleep.

When I wake I vow to fnd another solution to the paint mixing problem...I'm sure I had one.

Last edited: Jun 22, 2004
9. Jun 23, 2004

### jcsd

One thing can I ask, Ceptimus: are you THE Fred Dibnah, or just a big fan of the man?

10. Jun 23, 2004

### Gokul43201

Staff Emeritus
$$(n_i - 1)f_{i+1} + \frac {1+f_{i+1}}{2} = n_{i+1} ~~i=0..7$$

$$n_i : total~loads~of~coal~deposited~at~i'th~station~;~~n_0=8$$

$$f_i : fraction~of~load~deposited~per~trip~at~i'th~station$$

$$set~~n_{i+1} = n_{i} - 1$$

$$then~~f_{i+1} = \frac {2n_i - 3}{2n_i - 1}$$

$$and~~d_{i,i+1} = \frac {1-f_{i+1}}{4}$$

Last edited: Jun 23, 2004
11. Jun 23, 2004

### Gokul43201

Staff Emeritus
Hmmm... a steam-powered car, eh ?

12. Jun 23, 2004

### jcsd

If Ceptimus is Fred Dibnah I imagine his computer is also steam-powered.

13. Jun 23, 2004

### NateTG

So, let l be the distance that the steam car travels on one load, then the total distance traveled is:
$$l \times \sum_{i=1}{t}\frac{1}{(t-1) \times 2+1}$$
so a total of about 1010 miles. Enough to cross the desert.

Last edited: Jun 23, 2004