If an average human is 5'10" (1.8m), when scaled to be roughly 50 ft (15 m) how much food would it need to remain healthy?

Let's also assume there are three types of this giant; carnivorous, herbivorous and omnivorous. What would be the dietary requirements of each?

  • 1
    $\begingroup$ On an earth-like planet, right? $\endgroup$ – Theraot Apr 26 '17 at 14:48
  • 1
    $\begingroup$ Ignoring what, exactly? $\endgroup$ – Mołot Apr 26 '17 at 14:48
  • 5
    $\begingroup$ For those answering: Note that the giant is x8.6 as tall, making him x630 as massive as our baseline human (since he would scale depth and width as well). $\endgroup$ – Nex Terren Apr 26 '17 at 14:49
  • 7
    $\begingroup$ Also: Reminder to close-voters: The problem cannot be fixed if the OP is not made aware of it. $\endgroup$ – Frostfyre Apr 26 '17 at 14:52
  • 1
    $\begingroup$ Giants like this would only evolve on low gravity Worlds which have suitable flora and fauna to sustain them... is there magic involved? $\endgroup$ – Kilisi Apr 26 '17 at 15:31

The giant is 8.6 times taller than the human. This means that your giant is roughly 630 times heavier than a human. The square cube law is your biggest problem here. Your giant is simply too big to hold his own weight up. Good news for you, your giant is magical and doesn't care about that particular piece of physics.

Fortunately, 630 times greater mass does not mean 630 times greater food. If this relationship were true, large animals would never be able to find enough food, and mice would eat a kernel of corn per week. The real relationship is less than linear.

Kleiber's Law states that the metabolism of a creature is proportional to its body weight to the 0.75 power. GM = C*GW0.75 In this GM and GW are giant metabolism and weight, and C is a constant number depending on the type of animal.

If we say that the average man weighs 80 kilograms and, requires 2000 calories per day we can write this as: $ 2000 \frac{calories}{day} = C*80 Kg$

We can also say that the giant weighs about 50400 Kg. We can solve both equations for C and equate them, and substitute in GW=50400.


Solving for giant metabolism gives us 280000 calories per day, 138 times as much as is needed for an average human.

A pound of beef has roughly 1100 calories. It would take 250 pounds of beef per day to feed this giant. If the giant doesn't eat entrails and bones, that comes out to a whole cow every other day. A pound of grain comes out to roughly 1400 calories. It would eat 200 pounds of grain per day.

Note: I made edits to calories in grain. My original figures were incorrect.

Note that these are rough estimates and do not take into account SOOO many other factors. Wikipedia has a link to Kleiber's original chart. Kleiber didn't see any animals who have 10 times higher metabolism than his law predicted, but he did examine a few who were half or double his predictions.

In conclusion: An omnivorous giant would eat a quarter cow and 100 pounds of grain per day or twice that, or half that.

  • 10
    $\begingroup$ "What will you have for dinner?" "Half a cow and only 30 pounds of grain - I am on a diet, you know" $\endgroup$ – Secespitus Apr 26 '17 at 15:46
  • 9
    $\begingroup$ Not necessarily true that the giant could not hold themselves up. There were two-legged dinosaurs almost that tall. Clearly some structural "architectural" changes would have to take place to a human skeleton and proportions would be different, but a 40+ foot tall biped is indeed physically possible. $\endgroup$ – JBiggs Apr 26 '17 at 17:29
  • 4
    $\begingroup$ Argentinasaurus was 100 tons (albeit 4 legged, but that does mean that at least one leg was carrying at least 25 tons). Some studies mention that Spinosaurus (2-legged carnivorous) could be up to 20 tons. $\endgroup$ – njzk2 Apr 26 '17 at 17:53
  • 1
    $\begingroup$ @BobTheAverage yeah, that's a pretty big giant (and also spinosaurus and friends have tails to help balance. Keeping a body vertical on such a height and just 2 feet is hard) $\endgroup$ – njzk2 Apr 26 '17 at 19:06
  • 1
    $\begingroup$ @JackAidley Can you provide a source? $\endgroup$ – BobTheAverage Apr 27 '17 at 14:33

So, basing these calculations on the dietary guidelines released by the US department of health for 2015-2020, and the x630 mass factor given by @Nex Terren in their comment, these are the recommended daily allowances for a giant:

Calories: 1260000 - 1890000
Protein: 28980 - 32760 g
Carb: 81900 g
Fiber: 17640 - 19404 g
Linoleic acid: 7560 - 10080 g
Linolenic acid: 693 - 1008 g
Calcium: 630000 - 819000 mg
Iron : 6930 - 11340 mg
Magnesium: 195300 - 258300 mg
Phosphorus: 441000 - 787500 mg
Potassium: 2961000 mg
Sodium: 1449000 mg (This is an upper limit - not a recommended allowance)
Zinc: 5040 - 6930 mg
Copper: 560700 - 567000 mcg
Manganese : 1134 - 1386 mg
Selenium: 34650 mcg
Vitamin A: 441000 - 567000 mg
Vitamin E: 9450 mg
Vitamin C: 47250 mg
Thiamin: 693 - 756 mg
Riboflavin: 693 - 819 mg
Niacin: 8820 - 10080 mg
B6: 819 mg
B12: 1512 mcg
Choline: 267750 - 346500 g
Vitamin K: 47250 - 56700 mcg
Folate: 252000 mcg

So our giant will be eating a whole lot, to say the least. For reference, an entire cow, properly butchered, yields approximately 513,713 calories. That would be a suitable entrée for our giant, with a side salad of an entire lettuce patch.

  • 6
    $\begingroup$ You are assuming that the dietary requirements scale linearly with mass. If this was true, elephants would starve. $\endgroup$ – BobTheAverage Apr 26 '17 at 15:32
  • $\begingroup$ Do you have a reference for your calories for an entire cow? I looked and couldn't find one. $\endgroup$ – BobTheAverage Apr 26 '17 at 15:36
  • $\begingroup$ The second answer on this Yahoo answers post, which includes calculations and sources: answers.yahoo.com/question/index?qid=20081220234733AAfe5zj $\endgroup$ – Cameron Apr 26 '17 at 15:39
  • 1
    $\begingroup$ 3kg of Potassium and 1.4kg of Sodium. Wow. In pure form that would make for quite a reaction. $\endgroup$ – Tom Carpenter Apr 26 '17 at 22:10
  • 1
    $\begingroup$ You ought to express as more suitable units, e.g. kg not mg. $\endgroup$ – JDługosz Apr 27 '17 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.