- #1

Viky1147

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I need to scale down a human model by 1:20

i.e,

i need a human model of 9 cm height scaled down.

Help me calculate the weight of the 9 cm scaled down human model.

Regards,

Viky

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- Thread starter Viky1147
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- #1

Viky1147

- 17

- 7

I need to scale down a human model by 1:20

i.e,

i need a human model of 9 cm height scaled down.

Help me calculate the weight of the 9 cm scaled down human model.

Regards,

Viky

- #2

berkeman

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I would use the average density of a normal human to do the calculation. What is that average density? (You can probably find it fairly quickly using Google).Summary::Calculate scaled down human model weight for practical experiment

Hello,

I need to scale down a human model by 1:20

i.e,

i need a human model of 9 cm height scaled down.

Help me calculate the weight of the 9 cm scaled down human model.

Regards,

Viky

- #3

- #4

hutchphd

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- #5

sophiecentaur

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Why do you need to scale the weight`? If you want to know 'just for information' then you can use the same density and then the ratio would be proportional to the cube of the height.Summary::Calculate scaled down human model weight for practical experiment

Help me calculate the weight of the 9 cm scaled down human model.

BUT if you are trying to scale down and Engineer a model with appropriate materials then there is likely to be some other power law at work. Insects and other small animals use much 'weaker' structures than would be appropriate for humans and (of course) you couldn't just scale up a human body to be the height of an elephant and expect their skeleton would hold them up without modifying the skeleton along the lines of an elephant's.

This may not matter in your case but I had to point out possible problems that you could come across in really accurate down-scaling.

- #6

hutchphd

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Going bigger is usually lots harder than going smaller!

I am always amazed how people don't realize this.

- #7

sophiecentaur

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The OP's "Practical Experiment" may need to take this into account if there are any calculations of Forces or Powers or Velocities etc.. She needs to avoid a nonsense result if possible.

Going bigger is usually lots harder than going smaller!

I am always amazed how people don't realize this.

- #8

Richard R Richard

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Hello everyone, I have a different way of looking at it, If the **density is independent of the scale**….

$$P=mg=\frac{GmM}{R_E^2}\equiv\rho_bK_1L^3\frac{G\rho_EK_2L^3}{L^2}\equiv K\cdot L^4$$

If the densities remain constant and the universe shrinks any spatial dimension on a linear scale by 1:20 .the weight would vary with the fourth power of the scale….

$$P=mg=\frac{GmM}{R_E^2}\equiv\rho_bK_1L^3\frac{G\rho_EK_2L^3}{L^2}\equiv K\cdot L^4$$

If the densities remain constant and the universe shrinks any spatial dimension on a linear scale by 1:20 .the weight would vary with the fourth power of the scale….

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- #9

hutchphd

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Sorry, I did not understand any of that. Can you flesh it out a bit?

- #10

Richard R Richard

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The weight not only depends on the mass of the objects subjected to a gravitational field (density by the scale of length cubed), but also on the acceleration, if you develop the acceleration it is seen that it depends on the scale since it is length over time squared, the weight then depends on the fourth power of the scale ...

$$m = \rho_bV_b \equiv \rho_bK_1L^3$$

$$M = \rho_bV_E \equiv \rho_EK_2L^3$$

Weight by gravitational force between bodies of mass ##m## and ##M##

Sorry in Spanish (Peso = ##P##) but in English it is (##W## = weight)

$$F_G = P = W = \dfrac {mMG} {R^2} = \dfrac {mMGK_3}{L^2}$$

Uniting everything

$$W = \dfrac {\rho_bK_1L^3 \rho_EK_2L^3K_3G}{L^2} = KL^4$$

If the lengths are reduced in scale 1:20 the weight is reduced with the fourth power 1: 160000

It is not clear how the OP will use these results if the objective consisted of simulating a universe at 1:20 scale that would be the result, and if it is only how much a man of equal density weighs on the surface of the Earth and only 9 cm high then the result was previously given by 1: 8000 of the original weight.

$$m = \rho_bV_b \equiv \rho_bK_1L^3$$

$$M = \rho_bV_E \equiv \rho_EK_2L^3$$

Weight by gravitational force between bodies of mass ##m## and ##M##

Sorry in Spanish (Peso = ##P##) but in English it is (##W## = weight)

$$F_G = P = W = \dfrac {mMG} {R^2} = \dfrac {mMGK_3}{L^2}$$

Uniting everything

$$W = \dfrac {\rho_bK_1L^3 \rho_EK_2L^3K_3G}{L^2} = KL^4$$

If the lengths are reduced in scale 1:20 the weight is reduced with the fourth power 1: 160000

It is not clear how the OP will use these results if the objective consisted of simulating a universe at 1:20 scale that would be the result, and if it is only how much a man of equal density weighs on the surface of the Earth and only 9 cm high then the result was previously given by 1: 8000 of the original weight.

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- #11

hutchphd

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I still don't follow your more general argument. What does it mean to "change lengths" by a factor of 20? Relative to what?

- #12

Richard R Richard

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I mean that any measurement in that space to scale is reduced 20 times with respect to the lengths in real scale. The lengths of man and those of the earth.What does it mean to "change lengths" by a factor of 20? Relative to what?

I agree that OP should only need the weight of an 8000 times smaller volume man placed in the gravitational field of the Earth if he maintains his density. He says it's for an experiment, but he doesn't say which one.

But it should be noted that if the dimensions of the Earth are reduced in scale 1:20 keeping its density constant then the weight varies with the fourth power.

- #13

hutchphd

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- #14

sophiecentaur

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This may be on purpose, of course but we could do with a bit of clarification before scaring off an OP with an avalanche of algebra.

There is more to scaling than just the one thing.

- #15

Dullard

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I would do it differently from @berkeman. I would find the BMI formula on the web, choose an appropriate BMI for the human I am scaling down and plug in the scaled height and the BMI to solve for the weight. (BMI = body mass index).

Actually, that would be a bad way to proceed. If you inspect the equation used to produce the BMI, you'll note that it assumes that the relationship between the weight is proportional to the square of the height. If 'proportions' are maintained (width, depth), the relationship is with the cube of the height. Empirically (for humans) the equation should use an exponent of approximately 2.5. The existing BMI was created for use in populations (not individuals) - it's utility is limited for the former and malpractice in the latter.

- #16

sophiecentaur

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iirc, the bending moment is what counts when considering structures and valid scaling could well involve a fifth power.

You pays your money and you takes your pick until the OP comes back to the thread.

- #17

Viky1147

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bit of a busy day and just now opened my laptop

i read all the above replies

the average density of a human is said to be 1kg/m3, scaling it down for a 9cm human model it gives 4kg.

i am amused to imagine a 9cm human model with 4KG!

then as per the reply above i came to conclusion the GRAVITATIONAL FORCE cannot be scaled down.

NOTE:

I want to use this scaled down human model for a scaled down study of my project. Using it in real time makes my study too costly.

Any guidance will do good. Hope i have redirected the intent of this discussion in right path

Thanks

Viky

- #18

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That's what I had in mind as described here. Adolphe Quetelet who came up with the original inverse square dependence on height was an astronomer, mathematician, statistician and sociologist. He must have had good reason to depart from the inverse cube dependence but I don't know what it is.Empirically (for humans) the equation should use an exponent of approximately 2.5.

- #19

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4 kg doesn't sound right. Can you post your calculation?the average density of a human is said to be 1kg/m3, scaling it down for a 9cm human model it gives 4kg.

- #20

Viky1147

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I was answering it for the scaling down the weight based on the ratio of 1:204 kg doesn't sound right. Can you post your calculation?

Height @ ratio 1:20 goes like 180cm = 9cm

Weight @ ratio 1:20 goes like 80Kg = 4Kg

Thanks,

Viky

- #21

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In addition to kuruman's comment, we're mostly water, which means our density can't be very different from 1,000kg/mthe average density of a human is said to be 1kg/m3,

- #22

Viky1147

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1:8000 of the original weightIt is not clear how the OP will use these results if the objective consisted of simulating a universe at 1:20 scale that would be the result, and if it is only how much a man of equal density weighs on the surface of the Earth and only 9 cm high then the result was previously given by 1: 8000 of the original weight.

80Kg is the weight

1:8000 of 80 KG is 0.01KG (10 grams)

Something doesn't fit right

Thanks,

Viky

- #23

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No - your height, breadth and depth all go down by a factor of 20, so your volume goes down by a factor of ##20^3=8000##. So for constant density your mass also decreases by 8000, giving about 10g, as @Richard R Richard estimated.I was answering it for the scaling down the weight based on the ratio of 1:20

Height @ ratio 1:20 goes like 180cm = 9cm

Weight @ ratio 1:20 goes like 80Kg = 4Kg

Thanks,

Viky

- #24

Viky1147

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Agreed,In addition to kuruman's comment, we're mostly water, which means our density can't be very different from 1,000kg/m^{3}.

density varies in different parts of the human body

- #25

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It does but if you homogenize the innards of a 1.80 m human in a blender (don't try this at home) keeping the outside shape intact, you will get a fluid of density 985 kg/mAgreed,

density varies in different parts of the human body

- #26

sophiecentaur

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Guidance can only be useful for you if youAny guidance will do good. Hope i have redirected the intent of this discussion in right path

Your 9cm model person will weigh around the same amount as a large mouse or a new born puppy. You're talking (1/20)X(1/20)X(1/20) of your weight (1/8000) for a model of the same proportions as your body.

- #27

berkeman

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In your other thread about advanced granular convection flow (marked "Advanced" in the Classical Physics forum), you mention that you are a Mechanical Engineer. Why are you having trouble with this simple problem of scaling down a model of a person?Summary::Calculate scaled down human model weight for practical experiment

i need a human model of 9 cm height scaled down.

Help me calculate the weight of the 9 cm scaled down human model.

I am a mechanical engineer

- #28

Viky1147

- 17

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How do you propose to scale down human model to 1:20 ratio.In your other thread about advanced granular convection flow (marked "Advanced" in the Classical Physics forum), you mention that you are a Mechanical Engineer. Why are you having trouble with this simple problem of scaling down a model of a person?

Did you see my replies in this same thread?

Weight doesn't work that way.

If its simple problem for you please let me know the solution let's see.

- #29

Viky1147

- 17

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I see you fail to understand and provide me lots of replies.Guidance can only be useful for you if youdescribe what you want to do in this project. If you want just to stand your little person on a shelf then the weight (do you mean mass?) will not be important. If you want to put this person on a structure then the structure would also need to be scaled properly but a 1/20 scale bridge, using normal bridge materials would be far 'stronger' in terms of how many model people it will support so what do you want your project to do?

Your 9cm model person will weigh around the same amount as a large mouse or a new born puppy. You're talking (1/20)X(1/20)X(1/20) of your weight (1/8000) for a model of the same proportions as your body.

My question is this

I need to Scale down a human model (height & weight) for a study.

I couldn't experiment it on a 1:1 model, due to cost.

So, I need a human model of 9cm height.

Now don't say data is insufficient. This is what the requirement is.

And by guidance means, i meant if you couldn't answer you could direct me towards the answer is what i meant.

- #30

Motore

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And by guidance means, i meant if you couldn't answer you could direct me towards the answer is what i meant.

No - your height, breadth and depth all go down by a factor of 20, so your volume goes down by a factor of 20^{3}= 8000 . So for constant density your mass also decreases by 8000, giving about 10g, as @Richard R Richard estimated.

- #31

sophiecentaur

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I have counted several simple answers in this thread - including one from me. There are also the reasonings behind them. Your 'figurine' could be made of any material that's just a bit less dense than water and you could test this by checking it floats with only a small proportion above water.And by guidance means, i meant if you couldn't answer you could direct me towards the answer is what i meant.

It seems that you have ignored the questions that we have asked but many of them could be very relevant if you plan to do anything more than just stand this figure on your 'apparatus'. Did you consider re-stating your question to make it clearer what you really mean? It appears that you want to keep this project a secret and so the information you have taken from the thread may let you down.

- #32

jbriggs444

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The data is insufficient. We need to know what particular aspect you are attempting to model.Now don't say data is insufficient. This is what the requirement is.

If you were trying to model pressure on shoe material then scaling the weight down by a factor of 20

- #33

Viky1147

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My question is thisThe data is insufficient. We need to know what particular aspect you are attempting to model.

If you were trying to model pressure on shoe material then scaling the weight down by a factor of 20^{2}= 400 could be appropriate. That would preserve the pressure per square inch on the shoe material.

I need to Scale down a human model (height & weight) for a study.

I couldn't experiment it on a 1:1 model, due to cost.

So, I need a human model of 9cm height.

What data is missing here in your proposed calculation?

- #34

jbriggs444

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The purpose and methods of the study. If you want the scale model to behave similarly to an unscaled human, those details matter.My question is this

I need to Scale down a human model (height & weight) for a study.

I couldn't experiment it on a 1:1 model, due to cost.

So, I need a human model of 9cm height.

What data is missing here in your proposed calculation?

- #35

berkeman

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