
Post by therizinosaurus on Jun 28, 2010 16:15:21 GMT
There's a lot of people asking"What scale is xx figure in?" or "What other figures are in the same scale as xx figure?" Well, here's how to figure scale out very simply. 1. Take the size of the dinosaur toy. This can be in inches or centimeters. For our example, we will use the original Carnegie Dimetrodon, which is 3 inches long, or 7.6 cm. 2. Look up the actual size of the animal. Dimetrodon was 3 meters, or 11 feet. 3. If you're using feet, multiple the size by 12 to get it into inches (ex. 11x12=132 inches), or by 100 to get centimeters (3x100=300 cm). 4. Divide the number you calculated in 3 by the size of the toy itself (in the same scale). ex. 132/3=44 Therefore, the figure is 1/44 scale!
If you have any questions, feel free to ask
theri



Post by sbell on Jun 28, 2010 16:53:18 GMT
Thanks Theri, but there are 3 different situations where scale is used, and they can be described by three simple equations: where x=model dimension y=real world dimension n=scale factor scale is written 1:n or 1/n 1. To find the scale of an object: model dimension:actual dimension (using the same units). x:y=1:n or x/y=1:n so a 10cm model from a 1m real life object would be 10:100 or 1:10
a 6" model based on a 6' object would be 6:72 or 1:12
2. To scale up a model to real life, based on known scale: multiply model dimension by denominator of scale xn=y 10 cm model at 1:10 (also read as 1/10) would be 10*10=100 cm=1m real 6" model at 1:12 would be 6*12=72"=6' real 3. To scale real values down using a given scale: Divide real dimension value by scale denominator y/n=x 1m in real life at 1:10 would be 1/10=.1m=10cm model 6' in real life at 1:12 would be 6/12=.5'=6" model Using these, a person can quickly determine how big a model is supposed to be, what the scale is of a model, or what size of model to look for when trying to suit other models. All you need is two numbers, and you can find the third (side noteto be really sure, take measurements in two dimensionslike height and lengthand see how well they correlate...). Most important is to always have everything in the same units. Otherwise, the numbers come out nonsensical (1"/1'=1:1 scale!). And of course, there are systems for rounding scales to common onesdecimals, for example, are generally not allowed; and it is convention to round to 'normal' ones, so a 1:44 might be rounded to 1:45 or 1:40 or 1:50, depending on particular needs (as anyone that has tried scaling a company's figures can probably attest, their models rarely all come out to exactly their stated scale).



Post by wheezy on Jun 28, 2010 17:10:44 GMT
Thanks Theri, but there are 3 different situations where scale is used, and they can be described by three simple equations: where x=model dimension y=real world dimension n=scale factor scale is written 1:n or 1/n 1. To find the scale of an object: model dimension:actual dimension (using the same units). x:y=1:n or x/y=1:n so a 10cm model from a 1m real life object would be 10:100 or 1:10
a 6" model based on a 6' object would be 6:72 or 1:12
2. To scale up a model to real life, based on known scale: multiply model dimension by denominator of scale xn=y 10 cm model at 1:10 (also read as 1/10) would be 10*10=100 cm=1m real 6" model at 1:12 would be 6*12=72"=6' real 3. To scale real values down using a given scale: Divide real dimension value by scale denominator y/n=x 1m in real life at 1:10 would be 1/10=.1m=10cm model 6' in real life at 1:12 would be 6/12=.5'=6" model Using these, a person can quickly determine how big a model is supposed to be, what the scale is of a model, or what size of model to look for when trying to suit other models. All you need is two numbers, and you can find the third (side noteto be really sure, take measurements in two dimensionslike height and lengthand see how well they correlate...). Most important is to always have everything in the same units. Otherwise, the numbers come out nonsensical (1"/1'=1:1 scale!). And of course, there are systems for rounding scales to common onesdecimals, for example, are generally not allowed; and it is convention to round to 'normal' ones, so a 1:44 might be rounded to 1:45 or 1:40 or 1:50, depending on particular needs (as anyone that has tried scaling a company's figures can probably attest, their models rarely all come out to exactly their stated scale). see this is why i hated algebra you just overly complicated a simple process by adding in letters. therizinosaurus way is much easier.



Post by sbell on Jun 28, 2010 18:03:10 GMT
Mine is simply plugging in the two numbers you have to find the third, depending on what you need. I have even laid out where you put those two numbers.
Seeing as it is doubtful that the same numbers will be used every time.
And I did give examples. Simple ones, using the simplest possible examples from metric and insane (imperial).
And looking at it now, my second and third methods will work just fine.
However, I forgot about how decimals and fractions make bad bedfellows for method oneit's all about reciprocals, and who wants to look for the 1/x button on their calculator (or worse, the x^{1} button).
So really, to find the n value (the scale factor) just divide the real by the model (y/x) as Theri said.
Also, speaking of the example, 3 metres is closer to 10 feet (9.8'). So one other important thingif you are converting systems (or units) make sure the conversion is right, or none of it will matter!
meaning that in Theri's example, a 7.6cm figure, at 3m, would be 7.6/300 = 1:39.5 Due to the conversion error, Theri's imperial scale should have come to 3/9.8=3/117.6=39.2
Either one can be very handily rounded to 1:40, which would convert the figure to 3.04m or 10 feetan acceptable, negligible difference.



Post by brontodocus on Jun 28, 2010 18:36:32 GMT
Good topic! However, I think that correct use of the rule of proportion is a prerequisite but it is only one part of the problem. In scientific descriptions of dinosaur taxa the total length refers to the length of skull plus the length of the vertebral column measured along the vertebral centra, so as if the vertebra would be a completely straight line. One example is Madsen's 1976 Osteology of Allosaurus. So all curves in the vertebra of an animal in a natural position would make it shorter in direct line. Of course, vinyl figures don't have a skeleton but it is mostly possible (yet hard) to figure out where the centra would be if this was a real animal. I know this would be subject to some error but you can come nearer to the "real" length of a figure if you try it. The often vivid poses with curved tails etc. can lead to confusion, one example is the new WS Apatosaurus which is 328 mm if measured in direct line but 489 mm if one attempts to measure along the position where the vertebral column would be. I know, it's an extreme example but I found most figures to be actually a good deal longer than stated by the manufacturer. It is never a bad idea to rely on your own measurements anyway. But it even gets more complicated, we all know that not all figures are that accurate regarding proportions. Often the tails seem too short and of course that should be considered if you're really picky on scale. One possibility would be to consider only the snoutvent length (but there's much less data on that than on total length) or, perhaps better, compare the length of the figure's head with published data on skull length. However, the latter would make little sense for dinosaurs with proportionally minute heads like sauropods or figures with exagerratedly large heads which is also quite common. But I calculate scales of my theropod, ceratopsian and quite a lot of other figures by comparing their head length with published data on skull lengths and then compare the total length of the figure with data on total length. Often enough both results are not consistent with each other. Many dinosaurs are known from good skull material but there are only a few where the complete vertebra is known so often the skull is more reliable. And it's a question of personal taste, I don't like Triceratops with a 4 m head or Tyrannosaurus with one that's close to 3 m and in these cases I'd rather reduce their total length. And, yes, I did this with almost every dinosaur figure I have. I feel a little embarassed about it but I couldn't resist... Anyway, regardless how good or accurate a figure is, one can only achieve an approximation. But if you're picky on scale I'd say one shouldn't neglect the overall proportions or curvature of a figure. Sorry, I don't want to make things overly complicated but I just don't think it's just a math problem. Edit: I just saw that Madsen 1976 relied more on femoral length to calculate total length, a better example is Paul's Predatory Dinosaurs of the World 1988.



Post by wheezy on Jun 28, 2010 20:44:57 GMT
Mine is simply plugging in the two numbers you have to find the third, depending on what you need. I have even laid out where you put those two numbers. Seeing as it is doubtful that the same numbers will be used every time. And I did give examples. Simple ones, using the simplest possible examples from metric and insane (imperial). And looking at it now, my second and third methods will work just fine. However, I forgot about how decimals and fractions make bad bedfellows for method oneit's all about reciprocals, and who wants to look for the 1/x button on their calculator (or worse, the x ^{1} button). So really, to find the n value (the scale factor) just divide the real by the model (y/x) as Theri said. Also, speaking of the example, 3 metres is closer to 10 feet (9.8'). So one other important thingif you are converting systems (or units) make sure the conversion is right, or none of it will matter! meaning that in Theri's example, a 7.6cm figure, at 3m, would be 7.6/300 = 1:39.5 Due to the conversion error, Theri's imperial scale should have come to 3/9.8=3/117.6=39.2 Either one can be very handily rounded to 1:40, which would convert the figure to 3.04m or 10 feetan acceptable, negligible difference. its not the equation is the way it is explained with letters not numbers that makes it hard to follow.



Post by Godzillasaurus on Jun 28, 2010 20:46:18 GMT
Oh my this topic makes my head hurt already but very helpful!



Post by wheezy on Jun 28, 2010 21:03:19 GMT
for example to find out the scale of the an animal that is supposed to be 30 ft. and the figure is 9 in. you would take 30 x 12 = 360 take 360 divide by 9. you get 40. which means its a 1/40 scale.
Another quick tip all figures in a 1/40 scale, the animals actual meter length and the figures length in inches should be the same number. for example stegosaurus was a 9 meter long dinosaur so the figure would be 9 inches inches in the 1/40 scale. of corse you have to take into account of what brontodocus just posted. If the figure has a a curved tail you would have to use a flexible ruler to get the actual length of the figure.



Post by wheezy on Jun 28, 2010 21:08:54 GMT
Good topic! However, I think that correct use of the rule of proportion is a prerequisite but it is only one part of the problem. In scientific descriptions of dinosaur taxa the total length refers to the length of skull plus the length of the vertebral column measured along the vertebral centra, so as if the vertebra would be a completely straight line. One example is Madsen's 1976 Osteology of Allosaurus. So all curves in the vertebra of an animal in a natural position would make it shorter in direct line. Of course, vinyl figures don't have a skeleton but it is mostly possible (yet hard) to figure out where the centra would be if this was a real animal. I know this would be subject to some error but you can come nearer to the "real" length of a figure if you try it. The often vivid poses with curved tails etc. can lead to confusion, one example is the new WS Apatosaurus which is 328 mm if measured in direct line but 489 mm if one attempts to measure along the position where the vertebral column would be. I know, it's an extreme example but I found most figures to be actually a good deal longer than stated by the manufacturer. It is never a bad idea to rely on your own measurements anyway. But it even gets more complicated, we all know that not all figures are that accurate regarding proportions. Often the tails seem too short and of course that should be considered if you're really picky on scale. One possibility would be to consider only the snoutvent length (but there's much less data on that than on total length) or, perhaps better, compare the length of the figure's head with published data on skull length. However, the latter would make little sense for dinosaurs with proportionally minute heads like sauropods or figures with exagerratedly large heads which is also quite common. But I calculate scales of my theropod, ceratopsian and quite a lot of other figures by comparing their head length with published data on skull lengths and then compare the total length of the figure with data on total length. Often enough both results are not consistent with each other. Many dinosaurs are known from good skull material but there are only a few where the complete vertebra is known so often the skull is more reliable. And it's a question of personal taste, I don't like Triceratops with a 4 m head or Tyrannosaurus with one that's close to 3 m and in these cases I'd rather reduce their total length. And, yes, I did this with almost every dinosaur figure I have. I feel a little embarassed about it but I couldn't resist... Anyway, regardless how good or accurate a figure is, one can only achieve an approximation. But if you're picky on scale I'd say one shouldn't neglect the overall proportions or curvature of a figure. Sorry, I don't want to make things overly complicated but I just don't think it's just a math problem. Edit: I just saw that Madsen 1976 relied more on femoral length to calculate total length, a better example is Paul's Predatory Dinosaurs of the World 1988. one ? does this include animals that have an actual raising back for example hadrosaurs spines get slope up from the neck which kind of makes a hump. would they measure the skull down the neck up the hump then straight down to the tip of the tail. As far as i know they are measured from tip of the nose the the neck then straight to the tip tail as in this is the natural way the vertebra would have occured and the hump would never actually straitened out.



Post by brontodocus on Jun 28, 2010 21:35:33 GMT
one ? does this include animals that have an actual raising back for example hadrosaurs spines get slope up from the neck which kind of makes a hump. would they measure the skull down the neck up the hump then straight down to the tip of the tail. As far as i know they are measured from tip of the nose the the neck then straight to the tip tail as in this is the natural way the vertebra would have occured and the hump would never actually straitened out. Good point, wheezy. yes, that's the tricky part: There's a difference between length along the vertebral centra and length along the outer curve (more close to length along distal tips of neural spines), e.g. for Ouranosaurus you'd get two very different results. So for all these figures of dinosaurs with high neural spines it's better to measure along the base of the sail or hump. It's not easy and subject to some error of course. No, I didn't mean to measure along the edge of humps or sails. Most of the time it takes me about a quarter of an hour to have an opinion on the scale of a newly acquired figure. I always try to measure head length, total length as described above, snoutvent length and hip height. And then check for good skeletal drawings and publications, but often enough I rely on sources like Paul 1988 or Weishampel et al. "The Dinosauria" or else.



Post by brontodocus on Jun 28, 2010 22:04:38 GMT
I thought it'd be better to show an illustration of what I mean: The red line goes from the snout to the tip of the tail following the vertebral centra. Obviously, if this line was stretched out it would not be far longer than the direct line but I guess about one head length longer. And it's also why I think I won't be able to measure without some error and why I prefer to have skeletal drawings at hand. And: In this case it is also obvious that I'm not talking about huge differences between direct line and length along vertebral centra. There are probably many who find it neglectable, since often it's only like 10% more than the direct line but I won't neglect that.



Post by sbell on Jun 28, 2010 22:05:29 GMT
Mine is simply plugging in the two numbers you have to find the third, depending on what you need. I have even laid out where you put those two numbers. Seeing as it is doubtful that the same numbers will be used every time. And I did give examples. Simple ones, using the simplest possible examples from metric and insane (imperial). And looking at it now, my second and third methods will work just fine. However, I forgot about how decimals and fractions make bad bedfellows for method oneit's all about reciprocals, and who wants to look for the 1/x button on their calculator (or worse, the x ^{1} button). So really, to find the n value (the scale factor) just divide the real by the model (y/x) as Theri said. Also, speaking of the example, 3 metres is closer to 10 feet (9.8'). So one other important thingif you are converting systems (or units) make sure the conversion is right, or none of it will matter! meaning that in Theri's example, a 7.6cm figure, at 3m, would be 7.6/300 = 1:39.5 Due to the conversion error, Theri's imperial scale should have come to 3/9.8=3/117.6=39.2 Either one can be very handily rounded to 1:40, which would convert the figure to 3.04m or 10 feetan acceptable, negligible difference. its not the equation is the way it is explained with letters not numbers that makes it hard to follow. Then that's the fault of the reader! Seriously, how do people work out scales without algebra? Even if you don't use letters (actually, representations of values as determined by your situation), it's not like you are recreating the equation each time. As for what to measurethey are toys. At best, you will get approximate values. Even the skeletons on which the toys are based have imperfect lengths, because of the potential for soft tissue, including in between the vertebrae, which is often not considered (until recently). This is one of the reasons that rounding the scale gives a more conservative value.



Post by Horridus on Jun 28, 2010 22:09:55 GMT
I was going to say  most of the models we're looking at are far from perfect. An approximation is the best you'll achieve in most cases.



Post by wheezy on Jun 29, 2010 1:08:01 GMT
I thought it'd be better to show an illustration of what I mean: The red line goes from the snout to the tip of the tail following the vertebral centra. Obviously, if this line was stretched out it would not be far longer than the direct line but I guess about one head length longer. And it's also why I think I won't be able to measure without some error and why I prefer to have skeletal drawings at hand. And: In this case it is also obvious that I'm not talking about huge differences between direct line and length along vertebral centra. There are probably many who find it neglectable, since often it's only like 10% more than the direct line but I won't neglect that. so then the natural arc in the backs are measured in the total lengths given. the thing its imposible for these animal to have a straight spine so y the arch is included doesn't make sense. for example the ouanosaurus here is estimated at 24 ft in length in reality though it would appear more like 20ft. Because the arch in neck into the back would be about 4 ft. to get the 24 ft you would have to stretch out the red line. A ? about the pic the black shadow is supposed to represent the actual animal. Y is the neck between the head and sail so large on top? This would have made it impossible for the animal to tilt its head down to drink.



Post by brontodocus on Jun 29, 2010 20:01:02 GMT
so then the natural arc in the backs are measured in the total lengths given. the thing its imposible for these animal to have a straight spine so y the arch is included doesn't make sense. for example the ouanosaurus here is estimated at 24 ft in length in reality though it would appear more like 20ft. Because the arch in neck into the back would be about 4 ft. to get the 24 ft you would have to stretch out the red line. A ? about the pic the black shadow is supposed to represent the actual animal. Y is the neck between the head and sail so large on top? This would have made it impossible for the animal to tilt its head down to drink. I agree that it's problematic. So in this case total length would mean a length that could not be achieved in reality as a straight line. On the other hand, if only the length along the direct line was regarded there would still be the problem with the neck that could be hold out more straightly or be flexed upward which would make the length variable in a single individual and that's not what you want to measure. Even if the tissue between the vertebral centra has to be taken into account, the length along the vertebral column is something that can be measured and that would not be affected by individual posture of neck, trunk or tail. So it provides length data that can be reproduced by others. And I agree about the black shadow outline. I just didn't care too much about it in this case as it was just the largest skeletal drawing of Ouranosaurus I could find doing a quick google search, others were smaller. And my scanner has been out of order for quite some time now...



Post by dinosaurmoe22 on Jul 6, 2010 19:17:38 GMT
MY THREAD ;D



Post by foxilized on Sept 1, 2010 23:51:06 GMT
Anybody could help with the scale of the Carnegie Spinosaurus? Would 1:40 be correct? What size was the real animal supposed to be?



Post by brontodocus on Sept 2, 2010 0:15:17 GMT
Anybody could help with the scale of the Carnegie Spinosaurus? Would 1:40 be correct? What size was the real animal supposed to be? I guess 1:40 would fit really well. Wasn't the skull estimated to be 1.75 m long? The Carnegie's head is approx. 44 mm which would represent a 1.76 m long head. Along the curve the length of mine is 421 mm which would result in an animal of 16.84 m length and that's within the estimates of 16 to 18 m total length.



Post by therizinosaurus on Sept 2, 2010 0:24:01 GMT
It's 1/40 scale, the old one is 1/65



Post by foxilized on Sept 2, 2010 0:51:18 GMT
Thanks.

