I believe the Breguet aircraft cruise-range equation is another important example of this, and why it is crucial that in-flight refueling, or re-charging, or co-thrusting by supply module craft, be introduced. We waste colossal (exponential) amounts of fuel by trying to carry it all with the aircraft. The technology for safe reliable automatic in-flight refueling or recharging his well proven for both manned operation (NASA Dryden, USAF) and UAVs to manned( eg DARPA).

It would be a telling calculation to estimate how much Geoff fuel and pollution could be saved if aircraft just carried or reasonable amount of fuel and refueled in-flight, saving both of fuel they were carrying in the airframe weight and complexity that that required which then meant smaller jet engines which means less weight which means less fuel which means less airframe…

I wonder if sufficient microwave energy can be transmitted from a ground based power station to supplement a fuel-powered jet? You’d have to carry batteries though.

It seems unlikely that classical and medieval generals (or anyone at that time) had the mathematical knowledge to optimize this with the harmonic, though it's possible that it could have been worked out by combining many different generals' trial and error.

I agree - very unlikely. To be fair to them though, this is only a good strategy in this very restricted toy model. The real optimal strategy is “steal food from peasants.” That doesn’t sound as good as “Harmonic Donkey Transfer” though.

I suppose the analogy equivalent would be gravity assists and ramjets. Though for the harm done to fit, it would have to be something more like the drives in that Doctor Who episode in which Ten got possessed by a sentient star.

It's obfuscated by the author. The rule is "once you have room in the rest of the carts to split the load of one of them between them then do it, and turn back the empty cart".

It's very simple rule and I'm sure 5 year old could discover it with ease. But probably wouldn't name it "harmonic".

You’re right, it DOES turn out to be quite simple, which I think would be really interesting if it was optimal. Haha it certainly wasn’t obvious to me though, until I wrote it out explicitly.

I wonder if a modern analog in villages on the warpath in Ukraine would be something like: "borscht with extra paprika and poison." Depends on whether the Russian generals are good at the Harmonic series and studied their math?

The original blog I linked to is great and talks about this a little bit -- I think the advent of railroads just broke this scaling completely, because the train can carry so much more than a donkey. Although, the train still eats fuel, so similar laws apply?

Since I’m looking at food logistics for the Rev War & 1812 for my American Food history podcast (mostly from the food production side & how it altered non-soldier consumption)

This titanic effort into donkey-economics is most welcome.

I got a similar improvement over pairwise transfer using a different protocol.

Transferring food between donkeys takes a fair amount of work, so probably has to be done every evening in the camp rather than whenever they've traveled L0/n. So let's suppose that transfers happen at regular intervals.

I need a variable for the transfer rate. For simplicity, it will be N, defined as follows. If transfers happen every distance l, let N=L0/l. If transfers instead happen every time t, let N=f0/δt. The two-donkey protocol is then N=2.

So at each transfer, the donkeys have eaten 1/N of the food they are carrying, so 1/N of the donkeys must be released so that the remaining (N-1)/N donkeys, now carrying (N-1)/N of the food they had at the last transfer, all now have full loads.

(You are making the simplifying assumption that transfers can be done at any time. I am making the different simplifying assumption that we always have so many donkeys that, when we release 1/N of the remaining donkeys, rounding off to a whole number introduces negligible error.)

So to travel a distance L0, the donkeys must make N transfers, and thus have ((N-1)/N)^N as many donkeys left at the end as when they started. Put another way, if you want to have D (full) donkeys after traveling a distance L, then you must start out with D(N/(N-1))^(NL/L0).

So the key is that ratio B(N)=(N/(N-1))^N. It turns out that B(2)=4 is pretty bad. Observe:

B(2)=4

B(3)=3.375

B(4)=3.16049382716049

B(6)=2.985984

B(8)=2.91028536804653

B(12)=2.84094437661549

B(16)=2.80840396557645

B(30)=2.76501635584221

B(60)=2.7412858509404

B(100)=2.73199902642904. This is probably the practical limit in that it represents doing a transfer every night on wagons that can carry 100 days' worth of food, and is pretty close to the theoretical limit of e=2.71828.

For L/L0=10, at the right end of your graphs, the harmonic donkey protocol needs to start out with over 10000 donkeys—most of the transfers look more like the B(12) or B(30) or B(100) case than the B(2) case. 4^10 = 1048576, 3^10 = 59049—even the six-donkey protocol is giving you an almost 18-fold improvement over the two-donkey protocol.

So yes, I believe in your last graph. It's an exponential graph with base (4/e).

Thank you so much for this detailed comment! This is an excellent parametrization of the problem. Is there a meaningful difference between the equal-time transfer and the harmonic transfer protocol? In the continuum limit I feel like there shouldn’t be…

What about grass on the side of the road? I think herd animals tend be good at "in-situ" resource collection. Also water probably weighs as much as food.

As long as the donkeys consume positive food on average, I think grazing is simply a change to the constant delta, in which case the same scaling law applies.

In my experience with long-distance hiking, water has to be sourced locally. That's no problem in wet areas, and in dry areas you're planning your route around hopping between water sources.

The Mongol range was far greater than any other military until the British Empire. Their solution? Soldiers drank the milk of the mares, which could produce milk for roughly 10 days on standard grass before they need a period of grazing. Mongols planned military campaigns around first securing foraging within 1-2 days of the target city, often splitting forces per available foraging.

That would be an interesting equation, but heuristics seemed to work well enough.

There should be straight-forward extension to the polar explorer problem using dogs and/or horses, which are eaten when other supplies are exhausted. It's a little more complicated because the task of hauling sleds is transferred from the animals to the explorers as the animals are consumed. Unfortunately, the margin of my brainpower is too small to contain the derivation of this.

Another way to carry food energy is to store it in adipose tissue (fat). I wonder if this was ever used as an intentional strategy.

It is a net loss to fatten up a soldier who dies though, so you would need to factor in loss rates in addition to the obvious question of fighting effectiveness with increased adiposity.

I believe the Breguet aircraft cruise-range equation is another important example of this, and why it is crucial that in-flight refueling, or re-charging, or co-thrusting by supply module craft, be introduced. We waste colossal (exponential) amounts of fuel by trying to carry it all with the aircraft. The technology for safe reliable automatic in-flight refueling or recharging his well proven for both manned operation (NASA Dryden, USAF) and UAVs to manned( eg DARPA).

It would be a telling calculation to estimate how much Geoff fuel and pollution could be saved if aircraft just carried or reasonable amount of fuel and refueled in-flight, saving both of fuel they were carrying in the airframe weight and complexity that that required which then meant smaller jet engines which means less weight which means less fuel which means less airframe…

I wonder if sufficient microwave energy can be transmitted from a ground based power station to supplement a fuel-powered jet? You’d have to carry batteries though.

It seems unlikely that classical and medieval generals (or anyone at that time) had the mathematical knowledge to optimize this with the harmonic, though it's possible that it could have been worked out by combining many different generals' trial and error.

I agree - very unlikely. To be fair to them though, this is only a good strategy in this very restricted toy model. The real optimal strategy is “steal food from peasants.” That doesn’t sound as good as “Harmonic Donkey Transfer” though.

I suppose the analogy equivalent would be gravity assists and ramjets. Though for the harm done to fit, it would have to be something more like the drives in that Doctor Who episode in which Ten got possessed by a sentient star.

That’s a really good analogy. The “peasant assist” maneuver. It’s why your armies have to pass close to towns.

It's obfuscated by the author. The rule is "once you have room in the rest of the carts to split the load of one of them between them then do it, and turn back the empty cart".

It's very simple rule and I'm sure 5 year old could discover it with ease. But probably wouldn't name it "harmonic".

You’re right, it DOES turn out to be quite simple, which I think would be really interesting if it was optimal. Haha it certainly wasn’t obvious to me though, until I wrote it out explicitly.

I wonder if a modern analog in villages on the warpath in Ukraine would be something like: "borscht with extra paprika and poison." Depends on whether the Russian generals are good at the Harmonic series and studied their math?

The original blog I linked to is great and talks about this a little bit -- I think the advent of railroads just broke this scaling completely, because the train can carry so much more than a donkey. Although, the train still eats fuel, so similar laws apply?

Since I’m looking at food logistics for the Rev War & 1812 for my American Food history podcast (mostly from the food production side & how it altered non-soldier consumption)

This titanic effort into donkey-economics is most welcome.

Thank you, we here at Maximum Effort, Minimum Reward are thrilled to bring you the latest research into optimized donkey logistics.

I got a similar improvement over pairwise transfer using a different protocol.

Transferring food between donkeys takes a fair amount of work, so probably has to be done every evening in the camp rather than whenever they've traveled L0/n. So let's suppose that transfers happen at regular intervals.

I need a variable for the transfer rate. For simplicity, it will be N, defined as follows. If transfers happen every distance l, let N=L0/l. If transfers instead happen every time t, let N=f0/δt. The two-donkey protocol is then N=2.

So at each transfer, the donkeys have eaten 1/N of the food they are carrying, so 1/N of the donkeys must be released so that the remaining (N-1)/N donkeys, now carrying (N-1)/N of the food they had at the last transfer, all now have full loads.

(You are making the simplifying assumption that transfers can be done at any time. I am making the different simplifying assumption that we always have so many donkeys that, when we release 1/N of the remaining donkeys, rounding off to a whole number introduces negligible error.)

So to travel a distance L0, the donkeys must make N transfers, and thus have ((N-1)/N)^N as many donkeys left at the end as when they started. Put another way, if you want to have D (full) donkeys after traveling a distance L, then you must start out with D(N/(N-1))^(NL/L0).

So the key is that ratio B(N)=(N/(N-1))^N. It turns out that B(2)=4 is pretty bad. Observe:

B(2)=4

B(3)=3.375

B(4)=3.16049382716049

B(6)=2.985984

B(8)=2.91028536804653

B(12)=2.84094437661549

B(16)=2.80840396557645

B(30)=2.76501635584221

B(60)=2.7412858509404

B(100)=2.73199902642904. This is probably the practical limit in that it represents doing a transfer every night on wagons that can carry 100 days' worth of food, and is pretty close to the theoretical limit of e=2.71828.

For L/L0=10, at the right end of your graphs, the harmonic donkey protocol needs to start out with over 10000 donkeys—most of the transfers look more like the B(12) or B(30) or B(100) case than the B(2) case. 4^10 = 1048576, 3^10 = 59049—even the six-donkey protocol is giving you an almost 18-fold improvement over the two-donkey protocol.

So yes, I believe in your last graph. It's an exponential graph with base (4/e).

Thank you so much for this detailed comment! This is an excellent parametrization of the problem. Is there a meaningful difference between the equal-time transfer and the harmonic transfer protocol? In the continuum limit I feel like there shouldn’t be…

What about grass on the side of the road? I think herd animals tend be good at "in-situ" resource collection. Also water probably weighs as much as food.

As long as the donkeys consume positive food on average, I think grazing is simply a change to the constant delta, in which case the same scaling law applies.

In my experience with long-distance hiking, water has to be sourced locally. That's no problem in wet areas, and in dry areas you're planning your route around hopping between water sources.

The Mongol range was far greater than any other military until the British Empire. Their solution? Soldiers drank the milk of the mares, which could produce milk for roughly 10 days on standard grass before they need a period of grazing. Mongols planned military campaigns around first securing foraging within 1-2 days of the target city, often splitting forces per available foraging.

That would be an interesting equation, but heuristics seemed to work well enough.

There should be straight-forward extension to the polar explorer problem using dogs and/or horses, which are eaten when other supplies are exhausted. It's a little more complicated because the task of hauling sleds is transferred from the animals to the explorers as the animals are consumed. Unfortunately, the margin of my brainpower is too small to contain the derivation of this.

Another way to carry food energy is to store it in adipose tissue (fat). I wonder if this was ever used as an intentional strategy.

It is a net loss to fatten up a soldier who dies though, so you would need to factor in loss rates in addition to the obvious question of fighting effectiveness with increased adiposity.

Do donkeys get fat?

Haha probably! I googled "fat donkey" and got a bunch of chubby donkey pictures.

IINM, this is known in mathematics as the "Jeep problem": https://en.wikipedia.org/wiki/Jeep_problem

Oh it does exist already! I looked (not very hard) and didn’t find this previously. Thank you!

I bet I can still publish in a purely donkey-focused journal ;)

Can't wait for the relativistic version!