


confusing question about cars and their weight
Okay I'm at shaun's (chevy thunder) house right now and he has a 1:18 scale model car of a chevelle at his house. So we figure this thing is about 5 lbs or less and if you mutliply that by 18 that would mean the real car weighs 90 lbs. So okay yes there is alot of plastic in it so we figured if you made it all outta metal it would be like at the most 30 lbs. Anyhow considering I have 25 lbs weights at my house that could be squished to the size of this car I thought that would be right and the whole thing wuold be solid steel. Well, wrong then the car would be 540lbs. We figured out the model car would have to be 194lbs to be the the 3500 lbs that a chevelle would weigh. Well I'm not even sure this is possible because if you have a solid steel block that weighs 194 lbs it is gunna be alot bigger than the 1:18 scale model. It seems to me that the scale isn't right, but then I took the car and but it side by side 18 times and it is about the right width to be a real chevell. What are we missing here, other than brains.
Oh yeah shaun just notified me the chevelle is bluish green (tourquist) if it makes any difference. Also it is an ss with a big block 396.

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okay we have the ccworked out to 2050 cc approx. Now with the density of cast iron at 7.15g per cc
14714grams would be the weight of a cast iron block the size of the model. So 14.7kgs approx multiplied by 2.2 to get lbs. And you get 32lbs of a cast iron block the size of the 1:18 model. Now multiply that by 18 and you get a car sized cast iron blcok that is 582lbs. Now this isn't close the the weight of the car. WTF???????????????? The world makes no sense 


uhh the model is one of those steel bodied ones. Still that is going to be an insanly heavy model. Yes we were bored but dammit where does all that weight come from.



Okay I just worked out the real car by multiplying all sides by 18 if it were a solid block of steel. Now it works out to over 23 000 lbs. Maybe I'm just dumb



stone the scale does not apply to the weight of the model just size. It is like saying that 1/4 inch equals 1 foot.not a nickel bag equals a kilo.thought I would put it a way you might understand.
you need to lay off the wacky weed.you are reading way to much into a simple thing. 


Ok that dosent make any sence, we know the car is lighter than if it was real, but the scale always stays the same were just saying the model would have to be extreamly heavy to match a real car, and this is impossable because the most heavy material on earth isnt that heavy.



well what I did the second time was I took our dimensions and multiplied them by 18 each, then used that to find the size of the real car and then multiplied by the mass of steel. After that did the 2.2 to change to lbs from metric. :p I'm smart in a dumb way.



Do this take the 5 pounds of the model
divide 5 by 8.9 which is the spacific gravity of the metal on the model now take .5618 times 7.9 which is the specific gravity of iron take 4.438 by 16 length take 71 times 16 width take 1136 times 16 now you have 18176 take into consideration the seats, glass, plastic, rubber, and ect.. wont weight as much as iron, some parts on models are made thicker like body panels.plus i still think 5 pounds is to heavy I would it is probablt 1/2 that which brings you down to 9089 pounds. on top of that this formula only work exact on a cube. Not a irregular shape like a car, the closest way to determine that is to measure its volume by water emersion. I did it by 16 you need to do it by 18 [ March 27, 2003: Message edited by: cc69 ]</p> 


Hey stoned.... in order to convert from poundsmass to kilograms, you divide by 2.2 not multiply.
There are 2.2 lbm per kg 


In fact there's another question for you...are you converting from poundsmass or from pounds force??



[quote]Originally posted by stonedchihuahua:
<strong>Okay I just worked out the real car by multiplying all sides by 18 if it were a solid block of steel. Now it works out to over 23 000 lbs. Maybe I'm just dumb </strong><hr></blockquote> You are in the right track here. The model is 1/18 scale which means it is 18 times smaller than the prototype in three dimensions. The linear dimension conversion in any one direction is 18. However, the volume (weight too) conversion is 18x18x18 = 5832. The problem with you trying to scale the model up to full size comes with your assumption that the model is an exact scale copy of the real car. The model is actually far too heavy to be an exact replica. The body metal on the model is probably at least twice as thick as the real car's sheet metal when it should be 1/18th as thick (remember, the effect is cubed!). However, this is impossible to scale correctly since that would make the model body 0.12"/18 =0.0067", like extraheavy aluminum foil. Likewise, the solid tires are probably many times heavier in scale than the real pneumatic tires would be. Your assumption that the scale plastic parts on the model are lighter than the steel components is wrong also. Plastic is less dense than steel and thus lighter IF the two components have the same amount of material in them. The model has many times more plastic in scale terms in it's components than the full sized car has steel. You should be looking at the problem in reverse  ask, "how much should an exact scale model weigh?" Assuming your real car weighs 3000#, then a 1/18 scale model would weigh 3000#/5832 = 0.56#(half a pound)! A model of a real 50# tire should weigh 0.0085# or 3/8 ounce. The model tires probably weigh in at 5 or 6 ounces. Rather than being too heavy to lift, an EXACT scale model would seem feather light. There are professional engineers who make a career of finding ways to scale down real world problems to run physical experiments on them. Have you have seen engineers doing physical model expeirments on model boats in a wave pool on shows on the Discovery Channel? EVERYTHING in those experiments must be scaled down, not just the overall dimensions of the ship. Viscosity of the water, density of the water, strength of the materials in the model, weight of the model, frequency of the waves, roughness of the surface finish on the model's hull, velocity of the ship, etc., etc., etc., must me scaled correctly. That is why digital computers are taking over this task since it has become MUCH easier to account for all of the physical parameters there than in a physical model chamber. [ March 27, 2003: Message edited by: willys36@aol.com ]</p> 


alright... thanks, who woulda thought my bordness causes so much of a problem?


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