How do I calculate the scale weight of an object??

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ButchAZ

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Let’s say a real airplane weighs 150,000 lbs. If I have a 1/72 scale model of it, what would the “scale” weight of the model be? I’m completely flummoxed.
 
Hi Butch. An object's weight depends on the material(s) of which it's made. With an imaginary model (i.e., an exact replica, with each of the thousands of modeled parts being made from the same material as the corresponding parts in the actual airplane), then the weight would be 150000÷72.

The weight of your actual model may be found by weighing it, of course. Maybe that'll be close enough for your purpose. By the way, what is the purpose? :)
[imath]\;[/imath]
 
I don’t think that’s correct. If the real thing weighed 150,000 lbs, a model with a wing span of about 18” would weigh over a ton. Impossible. I’m not interested in the weight of the plastic parts that make up the model. There has to be a way to calculate what a 1/72 replica would weigh if it were made of the same materials and parts as the real thing, only 1/72 the size. It’s not a linear relationship, that much I know.
 
ButchAZ said:
Let’s say a real airplane weighs 150,000 lbs. If I have a 1/72 scale model of it, what would the “scale” weight of the model be? I’m completely flummoxed.
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If made of the same materials (with the same density), the model's weight would be proportional to the volume of the original. Do you know how that relates to the scale?

But if your model is made of, say, balsa or plastic, it will be a lot lighter than that!
 
ButchAZ said:
Let’s say a real airplane weighs 150,000 lbs. If I have a 1/72 scale model of it, what would the “scale” weight of the model be? I’m completely flummoxed.
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Further to the above responses: It is highly unlikely that a 1/72 "model" of an aeroplane is going to be constructed of exactly the same materials (throughout) as the aircraft it emulates or it would weigh 2,083⅓ lbs (150,000 ÷ 72), therefore, it is probably made from typical modelling materials which means it would weigh much less and it would, therefore, be impossible to find a "scaled" weight for it.
It is also likely to be (well?) under a metre long and (probably) quite easily lifted by you (assuming you're a healthy, able-bodied adult) so the simplest way to find it's actual weight is to step onto a set of (bathroom?) scales holding the model then repeat without the model and subtract the second weight (your own) from the first. ?
 
Dr.Peterson said:
If made of the same materials (with the same density), the model's weight would be proportional to the volume of the original. Do you know how that relates to the scale?

But if your model is made of, say, balsa or plastic, it will be a lot lighter than that!
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It looks like I need to elaborate on this.

Although I've had trouble finding the phrase "scale weight" used in this sense in order to prove my understanding, I believe "scale weight" refers not to the actual expected weight of a model, but to the "scaled weight", that is, what it would weigh if it were exactly the same but scaled down. That is, we can ignore the reality of the model in this calculation.

(I find the term being used here, for example, with no definition but with the comment "Models generally make no attempt to replicate scale weight, only size.")

The important thing is that you will not be dividing by 72, as everyone else is reflexively doing. As I said, weight will be proportion to the volume, not length, so you need to divide by the cube of the scale number. The answer will be much smaller than you might think.

@ButchAZ, can you tell us your definition of "scale weight"? And show us in what way you are flummoxed? And why you are asking?
 
The Highlander said:
It is highly unlikely … to be constructed of exactly the same materials (throughout)
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I'd say it's impossible. That's why my model is completely imaginary.

I'm waiting for Butch to explain why he's interested. He might be a science-fiction or game author, such that laws of physics don't matter.
[imath]\;[/imath]
 
ButchAZ said:
I’m not interested in the weight of the plastic parts that make up the model. There has to be a way to calculate what a 1/72 replica would weigh if it were made of the same materials and parts as the real thing, only 1/72 the size.
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Ah, that's a good clarification, Butch. (I'd missed post#3 until today.) Why are you interested in that weight? Would an estimate be acceptable? :)
[imath]\;[/imath]
 
ButchAZ said:
I don’t think that’s correct. If the real thing weighed 150,000 lbs, a model with a wing span of about 18” would weigh over a ton. Impossible. I’m not interested in the weight of the plastic parts that make up the model. There has to be a way to calculate what a 1/72 replica would weigh if it were made of the same materials and parts as the real thing, only 1/72 the size. It’s not a linear relationship, that much I know.
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It appears that this reply was sitting in moderation during some or all of the subsequent discussion.

Did you see post #6, where I explained that mass (given a fixed density) is proportional to volume, and therefore to the cube of lengths? So we find the "scale weight" by dividing the real-world weight by the cube of the scale factor, 72: [math]150,000\times\frac{1}{72^3}=0.4\text{ lb}[/math]
That seems low to me, but I'm not sure of the dimensions of such a plane, and therefore of the model.
 
Dr.Peterson said:
It appears that this reply was sitting in moderation during some or all of the subsequent discussion.

Did you see post #6, where I explained that mass (given a fixed density) is proportional to volume, and therefore to the cube of lengths? So we find the "scale weight" by dividing the real-world weight by the cube of the scale factor, 72: [math]150,000\times\frac{1}{72^3}=0.4\text{ lb}[/math]
That seems low to me, but I'm not sure of the dimensions of such a plane, and therefore of the model.
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Obviously, you are correct that a real reduction in linear scale by [imath]1/72[/imath] would result in a reduction in volume of [imath]1/72^3[/imath]. I suspect, however, that such a model is reduced by [imath]1/72[/imath] only in two dimensions. A model with the thickness of each layer of skin reduced by a factor of 72 may very well be so fragile as to be impractical (if even physically possible). As a practical matter, the volume probably needs to be reduced by a factor equal to [imath](1/72)^2 * (1/x)[/imath], where x might only be 10 or 15. Indeed, I suspect most model builders focus only on length and height in terms of scaling. The information on the thickness of the skin may not even be available.
 
I recently wondered how tall I would have to be, scaled up equally in 3 dimensions, to weigh as much as the heaviest elephant.

I took the 24000 lbs. given by Google, divided it by my 185 lbs, took the cube root of this ratio and multiplied it by my height of 6 ft. I got roughly 30.5 ft. tall.
 
MarkFL said:
I recently wondered how tall I would have to be, scaled up equally in 3 dimensions, to weigh as much as the heaviest elephant.

I took the 24000 lbs. given by Google, divided it by my 185 lbs, took the cube root of this ratio and multiplied it by my height of 6 ft. I got roughly 30.5 ft. tall.
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which would limit the number of venues indoors that you could visit.
 
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