What is the weight of a model of the Eiffel Tower that is 0.5 metres high, constructed with exactly the same materials as the original?
I will share with you some data (not every fact is necessary to solve the problem): between 1887 and 1889 three hundred workers joined together 18,038 pieces of puddled iron using two and a half million rivets. Its actual weight is 10.100 tons. The tower, including the tv antenna, is 324 metres high. The walk to the highest level is 1.652 steps. The height varies 15 cm due to temperature oscillations.
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15.59 kg assuming that 10.1 tons is metric since everything else is.
Assuming the given means that materials shrinks proportionally in all 3 dimensions:
volume shrinkage factor = (0.5/324)^3
model weight = 10,100 x (0.5/324)^3 = 37.12 grams
37 grammes
I think you mean “exactly the same TYPES of materials.” A model constructed with exactly the same materials would have exactly the same weight.
I would guess that a 0.5m replica would probably weigh 14-15kg.
ok i guess it it’s weight is…less that 5.500 tons
The ratio of
x/0.5 = 10.1/324
would give a close approximation of an “exact” model.
Solving for X = 0.0155864 tons.
You could also estimate how much the model varies in height due to temperature.
X/0.5 = 0.15/324
X= 0.0002314
Realistically, I like to see a 0.5m model with 2 and a half million rivets!
If the model is made with literally the same materials as the original then it would weigh the same. But if you mean the same type of material, then a model that is 648 times shorter than the original has (1/648)^3 the volume. The densities are the same so the mass is also (1/648)^3 smaller. The weight would be about 37.1 grams, which is less than 7 US quarters.
OK, Bilbao. You’ve really gone waaay off the deep end this time. I don’t even have the slightest clue how to approach this. The first issue to cross in my mind is the bounding of the constraints of the puzzle.
1) “exactly the same materials”, one flaw. If a beam happens to be say 10 feet long. You’ve already blown the exact same materials. Because no matter how you stack it it’s a lot taller than 0,5 m. Oh, you say, “exact same materials” in smaller dimensions. OK, then assuming that the pieces are scaled down to be much smaller with exacting precision on the length and thickness of each piece. Do we also need extremely small bolts and hammers to generate rivets? Let’s assume that we aren’t being so precise or literal.
2) Taking the object down to smaller dimensions is not a linear fit is it? In other words, just because all the parts are identical in reduced scale does not mean the end result is identical in assembled, scaled dimensions. Does it?
3) If there is a formula to solve something like this where would one find it, because I wouldn’t know how to solve it from scratch. I am neither a civil engineer nor a theoretical mathematician.
4) Am I waaay over thinking this?
weight = density x volume
same material – so same density
for similar volume, proportionate to length cube
therefore weight of model = (0.5/324)^3 x 10,100tons = 37.1g
I almost fell for living in the US and got confused by those metric tons.
If the height is scaled down to 1/648 size, then so must be the width, and length.
The tower occupies some percentage of a box around it with equal height, length and width, so say:
V = k * l * w * h
where k is some constant of proportionality. The model will occupy the same percentage of space in a box around it, so:
V(model) = k * (1/648)l * (1/648)w * (1/648)h
So V(model) = [(1/648)^3] * V.
Since it is made out of the same things, the average density of the model will be the same as the original, so:
m / V = m(model) / V(model).
This proportion implies m(model) = [(1/648)^3] * m.
Expressed in grams, the mass of the Eiffel tower is:
10,100 tons * 1000 kg/ton * 1000 g/kg = 1.01 * 10^10 g
So, m(model) = [(1/648)^3] * 1.01 * 10^10 g,
= 37.12 g.
.5 meters is .154% of the original. So .154% of 10.1 tons is .0155 tons or 31.173 pounds.
Correction, the weight is 10,000 tons. (you need a comma instead of a period)Going with 10,100 tons the weight would be, 15.58 tons.
Which seems huge, so I might be missing something.
Ok, I’m back. Heres what I’m thinking:
A 1% scale model of a 1 cubic meter is a cubic centimeter. But simply taking 1% of a cubic meter is actually 10000 cubic centimeters. So to get a scale representation of the original cube, which would be 1 centimeter (1% of the height)you need to take 1% of 1% of 1% of the original cube.
So, .5 meters is 0.154% of 324 meters. Which means you need to take .154% of .154% of .154% of the original to get an accurate scale model (at .154%)
.154%^3 of 10100 tons is .000037 tons or 0.0742 pounds. Which seems very light, but taking into account that the structure is basically a skeleton using the least amount of material as possible, it then seems feasible.
Hi Hendy, take this puzzle as a ‘thought experiment’. Although I was not clear about this, please assume you take the original tower and you magically scale it down until it is 0.5 m high. What would be its weight?
To go from 324m height to 0.5m, you would need to divide by 648. So 10100 tons becomes 15586kg. But that would just make the tower smaller in one dimension. Then you need to divide it 2 times again by 648 to make it smaller in the other two dimensions. So, the small tower would weight approx. 37,12g.
Take a cube for example of 1x1x1m of water. It would weight 1000kg. A cube of 0,1×0,1×0,1m of water would weight only 1 kg. So you would need to divide each dimension of the three by 10. So the total weight would be divided by 10x10x10 = 1000.
Alright fourth time is a charm. Assuming the weight is in metric tons (tonnes) then it is still 0.000037 tons or tonnes, but if it is US tons (and why would France report in US conversions?) then it is about 0.074 pounds or 0.0335 Kg, if it is in tonnes then it is 0.081 pounds or 0.037 Kg.
Thanks for understanding beyond my english…:-)
Correct answer is 37.12g, as explained in above comments
Bilbao strikes again this time with scaled models (fluid mechanics)
Ooops! I totally goofed on this one!