Efficiency of transport on Military Roads

A 1930s ideal of motorisation: Mercedes L3750 towing trailer with a combined load of 8 tonnes.

A 1930s ideal of motorisation: Mercedes L3750 towing trailer with a combined load of 8 tonnes.

In the second edition of "Supplying War" by van Creveld, in the chapter "Logistics in perspective" (p234 in my edition) he discusses the how armies attempt to move forward while supplying themselves and how far they are able to move forward. He makes the point that the distance that vehicles can theoretical travel is never achieved in reality.

For instance:
A horse drawn wagon travelling 20 miles a day, with a load of 1 ton, pulled by 4 horses each consuming 20lb of fodder a day, gives a maximum distance before using up its entire payload as 20 X 2,240 /80 = 560 miles of which only 120 miles has ever been used in practice ie. 22%

A 5 ton WWII lorry loaded with nothing but fuel could travel at least 5,000 miles before using all its fuel. However at most only 10% of this has been used as armies seemed unable to support themselves more than 500 miles from their bases using lorries so there are greater losses with using lorries than with horses.

What is it that stops an army supplying itself over a distance of 1,000 miles as Rommel tried to do in the Western Desert? An individual lorry can drive that far although it may take some time, so you would think that adding further lorries would allow them to deliver a fixed load at any distance. Of course, there is some element of the law of diminishing returns at great distances, as fuel starts to take up an increasing proportion of the load but at 1,000 miles you should still have 4/5 of your payload available so not too much of a drop.

My starting point in this investigation was the 1941 edition of the Field Service Pocket Book to see how armies really moved by lorry. This stated that:

  • average speed = 30 mph (this is the speed needed to be used by the driver on most stretches of road)
  • miles in the hour = 10-15 mih (this is the actual distance travelled in an hour when allowance is made for other traffic on the road, rest stops, minor breakdowns and traffic crossing the road.)
  • vehicles per mile = 20 (which relates to a 90 yard spacing between vehicles to allow for air attack)
  • vehicles passing a point in one hour = 200 (vehicles per mile x their rate of progress 10 mih)
  • travel per day = 150 miles
  • length of working day = 12 hours
  • number of lorries used = 2,400

So this gives us "at the end of the road in one day" when using 1 ton standard sized vehicles, an effort in ton miles of:

tonnage (number of vehicles passing a point x number of hours of operation x tonnage) x distance (150 miles) or (200 x 12 x 1) x 150 = 360,000 ton mile

so if we want to deliver a fixed amount at a set distance it works out like this:

360,000 / distance = tonnage   or    360,000 / tonnage = distance

  • 50 miles = 7,200 tons
  • 100 miles = 3,600
  • 150 miles = 2,400
  • 300 miles = 1,200
  • 600 miles = 600
  • 2,400 tonnes = 150 miles
  • 4,800 tonnes = 75 miles
  • 9,600 tonnes = 37.5 miles

The first point to note is that this scheme ends up with all the lorries at one end of the journey at the end of the day. So for a continuous supply operation, where they need to do the same the following day, you need to get the lorries back to the starting point by the end of the day and so you have to divide these figures by half so 150 miles translates into 75 to destination and then 75 miles back in one day.

Secondly it is obvious that for every days travel you halve the amount carried, so a better way to express the problem would be in "day marches" (which would be both the journey to the destination and the return journey), so using our 150 miles a day example and a days march of 75 miles:

  • 1 days march = 2,400 tons
  • 2 days march = 1,200 tons (1/2 the original load)
  • 4 days march = 600 tons (1/4 the original load)
  • 6 days march = 400 tons (1/6 the original load)
  • 10 days march = 240 tons (1/10 the original load)
American Civil War ambulances

American Civil War ambulances

From this it is easy to see how distance is eroding the carrying capacity of the fleet. For the Second World War a typical distance at which an army could be sustained by motor vehicles was around 500 miles (see Van Crevelds account of the DAK in North Africa as an example,) which in the scheme above equates to 6 marches (at 75 miles a day = 450 miles.) The interesting point is that horse-drawn armies, like those used in the American Civil War, were held to have a sustainable distance from the railhead of 100 miles (Sherman) and with wagons in convoy managing 30 miles a day, this also works out at 6 marches. The determining factor is purely related to the diminishing effect of the number of cycles or marches. The distance that can be achieved is determined by what can be achieved in each cycle and that is a direct result of the road capacity.  Delivering a greater tonnage can be achieved by using the same number of larger vehicles but to really make a difference, the road has to be upgraded to increase miles in the hour and to reduce the effects of choke points in slowing the traffic.

The German view can be found in Versorgung der Feldheeres" H.Dv.g 90 (the secret annex to the main document) [a captured and translated version can be found at German Documents in RussiaFindbuch 12480 - Beutedokumente der sowjetischen Militäraufklärung  Akte 329 - starts from image 82.]

They use a formula to deal with the loading and unloading factors:

L = (14 - 2X) * 225 * Y * Z

  • L = Daily amount carried in ton km
  • ton km = distance (km travelled) and cargo transported for a certain time (1 hour or 1 day)
  • X = number of successful journeys with load
  • Y = number of Small Motor Columns (30 ton) (double this for Large Motor Columns 60 ton)
  • Z = State of roads: 
  • State Highways = 2.5 Main Roads = 1.6 Variable road conditions, (good roads, field roads, steep gradients) = 1 Forest and Field Roads, Steep climbs = 0.7
  • (These figures depend on the time of day, night journeys multiply by 0.7 journey completely in daylight by 1.3)
  • 14 = number of working hours (8 hours allocated to rest period and vehicle maintenance and 2 hours for drivers hot meals)
  • 225 is a constant (average speed 15 km/h of a Small Motor Column and useful journeys 0.5 to account for the return journey. 15 x 30 x 0.5 = 225)
German military lorry in Great War

German military lorry in Great War

 Example 1.
Daily journey using 2 Small Motor Columns:

a) 1 journey L = (14-2 hours) * 225 * 2 (columns) * 1 (average roads) = 5400 ton km. So to deliver 60 tonne in 1 journey = 90 km

b) 2 journey L = (14-2*2 hours) * 225 * 2 (columns) * 1 (average roads) = 4500 ton km. So to deliver 120 tonnes in 2 journeys = 37.5 km

c) 3 journey L = (14 - 2*3 hours) * 225 * 2 (columns) * 1 (average roads) = 3600 ton km. So to deliver 180 tonnes in 3 journeys = 20 km

Example 2.
Every day 6 Small Motor Columns bring up 360 tonnes of ammunition.

How far can they deliver this under normal driving conditions? Carry 180 tonnes so 360 tonnes is 2 moves.  Distance (km) = (14 - 4) * 225 * 6 * 1 divided by 360 tonnes = 37.5 km

Example 3. 
During the battle the average consumption is 3,120 tonnes a day. Army has 8 Large Motor Columns, and 10 divisions each with 6 Small Motor Columns. How far can the Army operate away from the railhead?

a) Army: 8 x 60 tons = 480 tons lift, so to move 3,120 tons from railhead to depot will take 3120/480 = 6.5 journey a day so Distance = (14 -8) * 225 * (8 x 2 for Large Motor Columns) * 1 divided by 3120 = 7 km

b) Each Division receives 312 tonnes a day. 6 x 30 ton Small Motor Columns = 180 tonnes so 2 daily journeys from depot to Divisional Supply Point
so Distance = (14 - 2*2) * 225 * 6 * 1 divided by 312 = 43 km

c) Distance travelled by combat units of artillery batteries from firing position to Divisional Supply Point = 20 km
So Distance = 7 + 43 + 20 = 70 km from railhead to frontline

So you can see from these examples in H.Dv.g.90 that the Germans took into account loading times as a major factor as each Motor Column took an hour to load and then another hour to unload with the help of a Supply Platoon. Also they account for differing road conditions and daylight conditions.

A column of motor lorries on the Contay-Amiens road in September 1916.

A column of motor lorries on the Contay-Amiens road in September 1916.

Their assumed rate of advance is 15 km/h which relates very closely to the British 10 miles in the hour (14 km) and again drivers are travelling at 50 km/h speed to achieve this with a 10 minute stop every hour for a break, fix minor issues on vehicles, close up stragglers, etc. Traffic slows down the average rate, as do defiles such as built up areas, steep terrain, bridges or other terrain features. If heavy traffic is crossing a road of course this more than halves the available capacity of both roads.

Another factor to take into account is that our British series of lorry columns comprising 2,400 lorries, at 20 vehicles to the mile, is 120 miles long. The first lorry is just arriving at the destination after 12 hours and a drive of 150 miles and the last lorry has only been on the road for 3 hours and travelled 30 miles. So the whole transport will actually take 12 + 9 hours = 21 hours.

The capacity of the road itself is close to 2,400 vehicles a day, allowing a few hours a day to make repairs, recover broken down vehicles and deal with the inevitable traffic jams. The only way that this can be increased is to:

  • a) shorten the distance between vehicles - this creates other problems such as concertinaing as well as vulnerability to air attack.
  • b) increase the load of the individual vehicle. A 10 tonne lorry and trailer combination will still fill up one slot in the convoy compared to a 1 tonne truck. But there does come a point when the road is full to capacity and you need another road.
  • c) improving the road to remove things that slow down the traffic or to increase traffic control.

The maximum that can be delivered if it is assumed that a 10 tonne semi trailer is the largest vehicle and that 4,800 vehicles is the maximum number of vehicles on this road (at 45 yards per vehicle with no chance of air attack,) then 7,200,000 ton mile is delivered. This is 48,000 ton at 75 miles. Using normal lorries (5 tonne) this is 24,000 tonnes a day. When large vehicle spacing is required to counter possible air attacks, then the number is 2,400 vehicles per day and normal lorries will deliver 12,000 tonnes at 75 miles. 

US Army semi-trailers in France 1945

US Army semi-trailers in France 1945

This was the American pattern in Normandy 1944, they started with 2.5 tonne 6x6 and 6x4 GMC Lorries loaded for road use to 5 tonnes. This proved too little and so they added trailers which increased the load to 10 tonnes but came at a cost as towing trailers uses more fuel and slows down. The British Pocket Book gives a figure of 20 mph cruising speed and only 8-12 miles in the hour. So in 1945 the US Army started using semi-trailers which could both carry the load and maintain the speed to gain the maximum road capacity for delivering supplies.