# How many giants are needed to pull this sleigh?

In my world, there is a race of giants, with an average height of 5 meters. Let's assume that every part of their body is proportional to that of a "normal" human being.

I want these giants to pull a sleigh, and on this sleigh to place a block of iron with the following dimensions: 5 meters (width), 7,5 meters (length) and 10 meters (high). The sleigh would have a design similar to those used by ancient egyptians to move the building blocks of the pyramids. The giants would be placed in front of the sleigh in a fashion similar to that of the cinematic great entrance of Cleopatra in Rome in the "Cleopatra" movie of 1963 (see link). They should be moving the sleigh around on a grassy, earthy, flat terrain.

My question is two-fold:

1. What would the best dimensions be for this sleigh?
2. How many giants would be needed to move this sleigh without getting it stuck?
• There isn't actually a consensus in the scientific community on how the pyramids were built. Quite fascinating research came out a few years ago about wetting the ground to reduce friction, for example. I can't tell from your image link which method you have in mind. Could you add a summary of the method to your question? The particular method you choose could include or disqualify entire answers. Alternatively, you could specify that you are okay with ANY non-disproven theory still being considered by mainstream science (based on your tags). It will be a non-trivial problem, that's for sure. – type_outcast May 27 '17 at 20:55
• Also, since I already checked the math, hopefully this will save someone else a minute or two: a $\rho{_{Fe}} \times 5 \text{m} \times 7.5 \text{m} \times 10 \text{m} = 2,950 \text{t}$. (That's metric tons, of course.) – type_outcast May 27 '17 at 21:01
• This depends too heavily on the definitions of "giant", "normal human being", and the build of the giants. Even amongst "average" humans, strength differences can be quite massive. – Aify May 27 '17 at 22:21
• Whoever voted to close, why would you want to close a question that can actually be calculated concretely as "primarily opinion based"? I mean no offense of course, but I do think this is a good question that at least deserves a chance to be improved if there is a legitimate concern. – type_outcast May 27 '17 at 23:20
• The keyword is “what is the best…” I think, that triggers the Primary Opinion Based voters. – JDługosz May 28 '17 at 7:13

First, a few quick facts and assumptions (yes, I'll try to differentiate between the two as best I can):

Mass of Cargo: (Fact) $\rho{_{Fe}} \times 5 \text{m} \times 7.5 \text{m} \times 10 \text{m} = 2.95 \times 10^6\text{ kg} = 2,950 \text{ t}$

Mass of Sled: The sled needs to essentially support the weight of a small building. Non-negligible but difficult to calculate without some serious engineering. (Can't just "scale up" a utility trailer, as the sled needs to support its own weight that increases non-linearly with the mass of the load.)

For now, I'm going to take an educated guess at about 1/3 the load, based on, well, utility trailers, if you must know. Thus I'll call it 1,050 t to give us an even 4,000 t for total mass (basically GWVR/GVM for those familiar with towing). (Assumption)

Giant's Mass: Quasi-realistic giants (as opposed to just big-ish humans) are always problematic due to the cube-square law, which basically means as giants get bigger, their volume (and hence mass) increases with the cube of height, but the surface area and only increases with the square of height (or width, doesn't matter).

It's a fascinating topic you can read about, but the gist is, your five-meter giants are going to need some magic, I'm afraid. When "juiced up" to proportionate strength, assuming you are OSHA compliant ('cause I mean, who isn't, right?), a human should push no more than 225 N. My average human is 90 kg ≈ approx 90ℓ and stands 180 cm tall. Since the giant's whole weight is involved, we need to do some more math. The cube-square law gives us:

$$V_{giant} = V_{human} ( \frac{h_2}{h_1} )^3 = 90\ell ( \frac{500\text{cm}}{180 \text{cm}} )^3 = 1,929\ell \approx 1,929 \text{kg}$$

Dividing 1,929 ÷ 90 = 21.4 gives us an upper bound on how much more force a giant could exert in a horizontal push than a human. I say "upper bound" because the legs are significantly involved, and the strength of those would not scale as much as the mass. The giants had to be hand-waved anyway, so you can tune this "scaling" to suit your story the best.

Final answer? Each giant can exert a force of 225 N × 21.4 = 4,815 N. Oh, and this is another big assumption, not due to the math so much as the fact that five meter giants wouldn't be able to stand up without some magic, as above.

Height of force: I'm going to spare you the vector sums and assume the "push" (pull) force from the giants is exactly horizontal. In reality, the angle of force will differ depending on the giant and how far they are from the sled, but not by a whole lot.

Type of sled: I'll do a diagram, but my sled will be rolled on logs, due to grassland not being suitable for the sand-wetting technique proposed c.2014.

Friction (Rolling Resistance): This cracking read (PDF) puts mountain bike tyres on level grass at a rolling resistance of 0.005 to 0.010. Rough-but-elastic logs on rough terrain will be worse, but I'll be very generous and go with the low end of the scale at 0.005.

Logs: Are going to have to be tough, and an average diameter of 50 cm will do for discussion's sake. Thus we have ten logs if the load is oriented as described in the question.

# Calculating Friction on Each Log

Friction due to the coefficient of rolling resistance is given as:

$$F = C_{rr}N = 0.005 \times 4\times 10^7 \text{N} = 200,000 \text{N}$$

The second term is the normal force, or weight of the rig on each wheel in this case, but it's the same on all ten wheels anyway. Thus ignoring heat losses in the ropes and elsewhere, all giants combined need to pull with a force of two hundred thousand Newtons to get the cart to start moving.

Drum roll...

$$200,000 \text{N} \div \frac{4,815 \text{N}}{\text{giant}} = 42 \text{ giants}$$

# 42 giants

Yep, bring a towel. Despite the fact that you gave us quite a well-defined question, I still had to make a lot of assumptions. I think my math holds up, but if anyone has any suggestions to make it a bit more elegant, accurate without blowing the answer length right out of the water, (or outright correct any mistakes I might have made—it's late, I'm tired, and I do so love to wake up to "wrong by order-of-magnitude"-type comments :-), please let me know.

• The fact that you had to make so many assumptions means that the question was not well-defined enough. This is probably why it has a PoB CV on it. – Aify May 28 '17 at 0:52
• It's a bit sad to live in a world where random people ask strangers to do their highschool level physics for them – Raditz_35 May 28 '17 at 5:21
• @aify: the only assumption I see here is related to the cube-square law, which would make these giants implausible in real life. Type_outcast handwaved that away (and well). So "so many assumptions" is an exaggeration. But if people still think they needed clarification, they could ask in the comments before voting to close the question. At least that's what it says on the WBSE tutorial. – Pedro Gabriel May 28 '17 at 7:16
• @type_outcast: I selected your answer, because you were really the only one that tried to help me out here. I was going to give you some bounty points for it, but I'm afraid the question will be closed before I'll be able to. So take my deepest thanks. – Pedro Gabriel May 28 '17 at 8:05
• @JDługosz You're quite right. The answer got a bit longer than I was expecting. I'll edit out some of the inline 'Jax. Is that mobile alignment thing a known bug, though? Because that's rather horrible looking indeed. :-) – type_outcast May 28 '17 at 9:19