I have long intended to write a paper entitled Why Elephants Are So Small, as a companion piece to Why Giraffes Have Short Necks (Taylor and Wedel 2013). I’ve often discussed this project with Matt, usually under the acronym WEASS, and its substance has come up in the previous post, and especially Mickey Mortimer’s comment:

I think it would be interesting to read a study on that — the order in which various factors restrict body size without transformative adaptations. Similarly, what the differences would be for an aquatic animal like a whale.

That is exactly what the WEASS project was supposed to consist of: a list of many candidate limitations on how big animals can get, some rough attempt to quantify their Big-O behaviour, some discussion of which factors seems to limit the sizes of modern terrestrial animals, and how dinosaurs (especially sauropods) worked around those limitations.

(Whales are different. I have in my mind a half-formed notion for a third paper, completing the trilogy, with a title along the lines of Why Whales Are Dirty Cheaters.)

What are those candidate limitations? Off the top of my head:

Biomechanical:

  • Bone strength
  • Cartilage strength
  • Cartilage thickness
  • Muscle strength
  • Nerve length and conduction time

Metabolic:

  • Blood pressure: column height and capillary length
  • Lung capacity
  • Tracheal dead space
  • Digestive efficiency
  • Metabolic overheating

Those are just some of the physical limits. There is anecdotal evidence that elephants are not very close to their mechanical limits in their usual behaviour: they could get bigger, and still work mechanically. (Follow the link at the start of this paragraph. You will thank me.)

There are plenty of other factors that potentially limit organism size, including:

Behavioural:

  • Feeding rate
  • Ability to navigate dense environments
  • Predator avoidance with limited athleticism
  • Difficulties in mating

Ecological:

  • Territory requirement
  • Time taken to reach reproductive maturity
  • Reproductive rate
  • Birth size
  • Lack of selection pressure: when there are no predators bigger than a lion, why would elephants need to evolve larger size?

I’m sure I am missing loads. Help me out!

I am haunted by something Matt wrote a while back when we were discussing this — talking about how alien sauropods are, and how easily we slip into assuming mammal-like paradigms.

We are badly hampered by the fact that all of the 250kg+ land animals are mammals. We only get to see one way of being big, and it’s obviously not the best way of being big. Our perceptions of how hard it is to be big are shaped by the animals that are bad at it.

So having written this blog post, I am wondering whether it’s time to breathe life back into this project, started in 2009 and repeatedly abandoned.

Small and large sauropods, with cross-sections through neck and leg. Bone shown in white, gullet in yellow. Modified from Twemoji12 1f995 (CC By 4.0) from the Twitter Emoji project. Downloaded from https://commons.wikimedia.org/wiki/File:Twemoji12_1f995.svg

Consider a small sauropod of length x, as shown on the left above. Its mass is proportional to x cubed, it stands on leg bones whose cross-sectional area is x squared, and it ingests food through a gullet whose cross-sectional area is x squared. Now consider a larger sauropod of length 2x, as shown on the right above. Its mass is proportional to 2x cubed = 8x, it stands on leg bones whose cross-sectional area is 2x squared = 4x, and it ingests food through a gullet whose cross-sectional area is 2x squared = 4x. The bigger sauropod has to carry proportionally twice as much mass on its leg bones, and ingest proportionally twice as much food through its gullet. (Similarly, a 104-foot tall gorilla, 20 times as tall as a real one, is only 400 times as strong but 8000 times as heavy — which is why we can’t have Skull Island.)

In practice, big animals tend to have adaptations such as thicker limb bones that mean the numbers aren’t quite as bad as this, but the principal holds: the bigger an animal gets, the worse the problems imposed by scaling. It’s not possible to “solve” this problem because so many biological properties scale this way. Something is always the limiting factor. Suppose it were leg-bones or gullet. If somehow a hypothetical ultra-sauropod evolved extra thick leg-bones and gullet, scaling of respiration would suffocate it, or scaling of digestion would starve it, or scaling of heat-loss through the skin would boil it. The fundamental reason that you can’t just scale an animal up is that some parts of its function scale with volume while most — respiration, digestion, etc. — scale with surface area.

Various Internet rumours have suggested that the Archbishop is a super-giant sauropod one third larger than the mounted Giraffatitan specimen MB.R.2181 (formerly HMN SII). This is incorrect.

Figure E. Skeletal inventory of NHMUK PV R5937, “The Archbishop”, showing which bones were excavated by Migeod’ expedition. Based on a skeletal reconstruction of Giraffatitan brancai kindly provided by Scott Hartman: note that this image does not illustrate the shapes or proportions of the Archbishop material. Bones prepared and available for study are shown in white; those still in jackets awaiting preparation in light grey; those excavated by Migeod but apparently lost or destroyed in dark grey.

Migeod’s assessment of the size of the animal was based on the vertebrae: “The [neck] vertebrae found give a 20-foot [6.10 m] length […] The length of the back including the sacral region was about 15 feet [4.57 m]. The eight or nine caudal vertebrae cover about 6 feet [1.83 m]” (Migeod 1931a:90). This gives the total preserved length of the skeleton as 41 feet (12.50 m). By comparison, Janensch (1950b:102) gives lengths of portions of the mounted skeleton of MB.R.2181 as 8.78m (neck), 3.92m (torso) and 1.07m (sacrum) for a torso-plus-sacrum length of 4.99m. On this basis, the preserved neck of NHMUK PV R5937 is only 69% as long as that of MB.R.2181, but since the first four vertebrae were missing and omitted from Migeod’s measurement, this factor cannot be taken at face value. More informative is the torso-plus-sacrum length, which in NHMUK PV R5937 is 92% the length of MB.R.2181.

This is consonant with measurements of individual elements, which compare as follows:

Table 4. Comparative measurements of Archbishop and Giraffatitan elements

ElementMeasurement (cm)ArchbishopGiraffatitanRatio
Torso plus sacrumtotal length4574990.916
C10 (mC4)centrum length991000.990
C11 (mC3)centrum length104100[1]1.040
D4 (mD3)centrum length27360.750
Longest riblength over curve2352630.894
Left scapulocoracoidlength over curve221238[2]0.929
Right humeruslength1462130.685
Right humeruswidth51590.864
Right iliumlength98123[3]0.797
Right iliumheight7996[4]0.823
Femurlength122196[5]0.622
Average0.846

Archbishop measurements taken from Migeod (1931a) and converted from imperial; Giraffatitan measurements are for MB.R.2181 except where noted, and are taken from Janensch (1950a:44) and Janensch (1961).
Notes.
[1] Janensch (1950a) did not report a total centrum length for C11, as its condyle had not been removed from the cotyle of C10; but since the length of its centrum omitting the condyle was, at 87 cm, identical to that of C10, it is reasonable to estimate its total length as also equal to that of C10.
[2] Janensch (1961:181) did not include measurements for the right scapula of MB.R.2181, which is incorporated into the mounted skeleton, but does give the proximodistal length of its right coracoid as 45 cm. Using the 193 cm length given for the similarly sized scapula Sa 9, we can deduce a reasonable total estimate of 238 cm for the scapulocoracoid.
[3] Estimated by Janensch (1950b:99) based on cross-scaling from the fibula and ilium of Find J from the Upper Saurian Marl.
[4] This is the measurement provided by Janensch (1961:199) for the ilium Ma 2, which is incorporated into the mounted skeleton, and which Janensch (1950b:99) considered to match MB.R.2181 very precisely.
[5] Based on a restoration of the midshaft which Janench (1950b:99) calcuated based on other finds.

Individual lines of this table should each be treated with caution: Migeod’s measurements may have been unreliable, and in any case are underspecified: for example, we do not know whether, when he gave a vertebra’s length, he included overhanging prezygapophyses or the condyle. Similarly, we know that Migeod (1931:96) wrote that a rib “was as much as 92.5 inches long”, but we do not know for certain that, like Janensch, he measured the length over the curve rather than the straight-line distance between the ends. And when Migeod says that the ilium “measured 38.5 by 31 inches” we do not know that the height was measured “at the public process”, as Janensch (1961:199) specified.

With those caveats in place, nevertheless, a picture emerges of a sauropod somewhat smaller than MB.R.2181, though by no means negligible. On average, the measurements come out about 15% smaller than those of Giraffatitan.

But this average conceals a great deal of variation. The cervical vertebrae are comparable in length to those of MB.R.2181 (The total of 203 cm for C10 and C11 in the Archbishop, only 1.5% longer than 200 cm for MB.R.2181, is a difference well within the margin of measurement error). The Archbishop’s scapulocoracoid may have been 93% as long as in MB.R.2181. But the limb bones are signficantly shorter (87% for the humerus and a scarcely credible 62% for the femur), and the humeri at least bseem to be have been proportionally more robust in the Archbishop: only 2.86 times as long as wide, whereas the ratio is 3.61 in MB.R.2181. If Migeod’s measurements can be trusted, we have here an animal whose neck is as long as that of Giraffatitan, but whose limbs are noticably shorter. These proportions corroborate the hypothesis that the Archbishop is not a specimen of Giraffatitan.

I have several small ordered sequences of data, each of about five to ten elements. For each of them, I want to calculate a metric which captures how much they vary along the sequence. I don’t want standard deviation, or anything like it, because that would consider the sequences 1 5 2 7 4 and 1 2 4 5 7 equally variable, whereas for my purposes the first of these is much more variable.

Here is a matric that I think does what I want, and will allow me to compare different sequences for variability-along-the-sequence.

For the n-1 pairs along the sequence of n elements, I take the difference (absolute value, so always positive) between elements i and i+1. Then I average all those differences. Then I divide the result by the average of the values themselves, to normalise for magnitude.

Some example calculations:

  • For the sequence 1 5 2 7 4, the differences are 4 3 5 3, for a total of 15 and an average of 3.75. The average of the values is 1+5+2+7+4 = 19/5 = 3.8, which gives me a metric of 3.75/3.8 = 0.987.
  • For the sequence 1 2 4 5 7, the differences are 1 2 1 2, for a total of 6 and an average of 1.5. The average of the values is again 3.8, which gives me a metric of 1.5/3.8 = 0.395.
  • So the first sequence is 0.987/0.395 = 2.5 times as sequentially variable as the second sequence.
  • And for the sequence 10 20 40 50 70 (which is the same as the previous one, but all values ten times greater), the differences are 10 20 10 20, for a total of 60 and an average of 15. The average of the values is 38, which gives me a metric of 15/38 = 0.395, the same as before — which is as it should be.

And now, my question! Does this metric, or something similar, already exist? If so, what is it called? Or if I should be using something else instead, what is it?

(It happens that my sequences are the aspect ratios of the cotyles of consecutive vertebrae, but that’s not important: whatever metric we land on should work for any sequences.)

Taylor 2015: Figure 8. Cervical vertebrae 4 (left) and 6 (right) of Giraffatitan brancai lectotype MB.R.2180 (previously HMN SI), in posterior view. Note the dramatically different aspect ratios of their cotyles, indicating that extensive and unpredictable crushing has taken place. Photographs by author.

And so the series continues: part 9, part 10 and part 11 were not numbered as such, but that’s what they were, so I am picking up the numbering here with #12.

If you’ve been following along, you’ll remember that Matt and I are convinced that BYU 9024, the big cervical vertebra that has been referred to Supersaurus, actually belongs to a giant Barosaurus. If we’re right about, then it means one of two things: either Supersaurus synonymous with Barosaurus, or there are two diplodocids mixed up together.

Jensen (1987:figure 8c). A rare — maybe unique? — photograph of the right side of the big “Supersaurus” cervical vertebra BYU 9024. We assume this was taken before the jacket was flipped and the presently visible side prepped out. We’d love to find a better reproduction of this image.

Which is it? Well, seventeen years ago Curtice and Stadtman (2002:39) concluded that “all exceptionally large sauropod elements from the Dry Mesa Quarry can be referred to one of two individuals, one a Supersaurus and one a Brachiosaurus […] further strengthening the suggestion that all of the large diplodocid elements belong to a single individual.” It is certainly suggestive that, of all the material that has been referred to Supersaurus, there are no duplicate elements, but there are nice left-right pairs of scapulocoracoids and ischia.

But do all those elements actually belong to the same animal? One way to address that question is to look at their relative sizes and ask whether they fit together.

Sadly, when Matt and I were at BYU we didn’t get to spend time with most of these bones, but there are published and other measurements for a few of them. Jensen (1985:701) gives the total lengths of the two scapulocoracoids BYU 9025 and BYU 12962 as 2440 and 2700 mm respectively. Curtice et al. (1996:94) give the total height of the last dorsal BYU 9044 as 1330 mm. We have measured the big cervical BYU 9024 (probably C9) ourselves and found it to measure 1370 mm in total length. Finally, while there is no published measurement for the right ischium BYU 12949 (BYU 5503 of Jensen’s usage), we can calculate it from the scalebar accompanying Jensen’s illustration (with all the usual caveats) as being 1235 mm long.

Jensen (1985:figure 7a). BYU 12946 (BYU 5503 of his usage), the right ischium assigned to Supersaurus. By measuring the bone and the scalebar, we can calculate the length as 1235 mm.

Do these measurements go together? Since we’re considering the possibility of Supersaurus being a big Barosaurus, the best way to test this is to compare the sizes of the elements with the corresponding measurements for AMNH 6341, the best known Barosaurus specimen.

For this specimen, McIntosh (2005) gives 685 mm total length for C9, 901 mm total height for D9 (the last dorsal) and 873 mm for the ischia (he only provides one measurement which I assume covers both left and right elements). The scapulocoracoids are more complex: McIntosh gives 1300 mm along the curve for the scapulae, and 297 mm for the length of the coracoids. Assuming we can add them in a straight line, that gives 1597 mm for the full scapulocoracoid.

I’ve given separate measurements, and calculated separate ratios, for the left and right Supersaurus scapulocoracoids. So here’s how it all works out:

Specimen Element Size (mm) Baro (mm) Ratio Relative
9024 Mid-cervical vertebra 1370 685 2.00 124%
9044 Last dorsal vertebra 1330 901 1.48 92%
9025 Left scapulocoracoid 2440 1597 1.53 95%
12962 Right scapulocoracoid 2700 1597 1.69 105%
12946 Right ischium 1235 873 1.41 88%

The first five columns should be self-explanatory. The sixth, “proportion”, is a little subtler. The geometric mean of the size ratios (i.e. the fifth root of their product) is 1.6091, so in some sense the Dry Mesa diplodocid — if it’s a single animal — is 1.6 times as big in linear dimension as the AMNH 6341 Barosaurus. The last column shows each element’s size ratio divided by that average ratio, expressed as a percentage: so it shows how big each element is relative to a hypothetical isometrically upsized AMNH Barosaurus.

As you can see, the cervical is big: nearly a quarter bigger than it should be in an upscaled Barosaurus. The two scaps straddle the expected size, one 5% bigger and the other 5% smaller. And the dorsal and ischium are both about 10% smaller than we’d expect.

Can these elements belong to the same animal? Maaaybe. We would expect the neck to grow with positive allometry (Parrish 2006), so it would be proportionally longer in a large individual — but 25% is a stretch (literally!). And it also seems as though the back end of the animal (as represented by the last dorsal and ischium) is growing with negative allometry.

A nice simple explanation would be that that all the elements are Supersaurus and that’s just what Supersaurus is like: super-long neck, forequarters proportionally larger than hindquarters, perhaps in a slightly more convergent-on-brachiosaurs way. That would work just fine were it were not that we’re convinced that big cervical is Barosaurus.

Here’s how that would look, if the BYU Supersaurus is a large Barosaurus with different proportions due to allometry. First, Scott Hartman’s Barosaurus reconstruction as he created it:

And here’s my crudely tweaked version with the neck enlarged 24% and the hindquarters (from mid-torso back) reduced 10%:

Does this look credible? Hmm. I’m not sure. Probably not.

So: what if we’re wrong?

We have to consider the possibility that Matt and I misinterpreted the serial position of BYU 9024. If instead of being C9 it were C14 (the longest cervical in Barosaurus) then the AMNH analogue would be 865 mm rather than 685 mm. That would make it “only” 1.58 times as long as the corresponding AMNH vertebra, which is only 3% longer than we’d expect based on a recalculated geometric mean scale of 1.5358 — easily within the bounds of allometry. We really really really don’t think BYU 9024 is a C14 — but it’s not impossible that its true position lies somewhere posterior of C9, which would mean that the allometric interpretation would become more tenable, and we could conclude that all these bones do belong to a single animal after all.

Of course, that would still leave the question of why the Supersaurus scapulocoracoids are 10% bigger than we’d expect relative to the last dorsal vertebra and the ischium. One possible explanation would be to do with preparation. As Dale McInnes explained, there’s some interpretation involved in preparing scaps: the thin, fragile distal ends shade into the cartilaginous suprascapula, and it’s at least possible that whoever prepped the AMNH 6341 scaps drew the line in a different place from Dale and his colleagues, so that the Barosaurus scaps as prepared are artificially short.

Putting it all together: it might easily be the case that all the elements really do belong to a single big diplodocid individual, provided that the big cervicals is more posterior than we thought and the AMNH scaps were over-enthusiastically prepped.

References

I got a wonderful surprise a couple of nights ago!

Supersaurus referred scapulocoracoid BYU 12962 back when it was still in the ground. Rough composite assembled from screenshots of the video below, from about 23m17s.

I found myself wondering where the widely quoted (and ludicrous) mass estimate of 180 tons for Ultrasauros came from, and went googling for it. That took me to a blog-post by Brian Switek, which linked to a Google Books scan of what turned out to be my own chapter on the history of sauropod research (Taylor 2010) in the Geological Society’s volume Dinosaurs and Other Extinct Saurians: a Historical Perspective. So it turns out that I once knew the answer to that question. My chapter references McGowan (1991:118), which says:

Jim Jensen’s (1985) Ultrasaurus (“beyond lizard”), found in Colorado in 1979, had an estimated length of more than ninety-eight feet (30 m), compared with seventy-four feet (22.5 m) for the Berlin specimen of Brachiosaurus. This is a length increase of 1.32, so the weight increase would be (1.32)^3 = 2.3, giving an estimated weight of almost 180 tons.

[As I noted in my 2010 chapter, that’s based on Colbert’s (1962) equally silly estimate of 78 tonnes for MB.R.2181 (formerly HMN S II), the Girafatitan brancai paralectotype.]

So that’s a funny story and a mystery solved, but where it gets really good is that as I was grubbing around in the search results that led me to that conclusion, I stumbled on Episode 21 of the I Know Dino podcast, which contains a glorious embedded video: The Great Dinosaur Discovery, a 1976 film by BYU about Jensen’s work at quarries including Dry Mesa, and heavily featuring bones of what would become Supersaurus!

It’s very well worth 25 minutes of your time, despite the horrible 1970s documentary music, and brings actual new information to the table.

Some of the highlights include:

— Right from the start, seeing Jensen himself: someone I’ve been sort of familiar with from the literature, but never really imagined as being an actual human being.

— From about two minutes in, Jensen seems be uncovering bones in dry sand, rather like kids in a palaeo pits at some museums. It takes about one minute to uncover a nice tibia. Is it ever really that easy? Is the Dry Mesa quarry that easy to work?

— Putting faces to the important names of Vivian and Eddie Jones, the uranium prospectors who first led Jensen to several of his important sites, and after whom the species Supersaurus vivianae and Dystylosaurus edwini were named.

Vivian “Supersaurus” Jones and Eddie “Dystylosaurus” Jones in the field [from about 4m41s in the video]

— From about 13m30s onwards, we see what I think must be the Supersaurus pelvis that’s now on display at the North American Museum of Ancient Life. (It doesn’t actually look all that big, in the scheme of things.)

— From 16m50s onwards, things start to get real, with the uncovering (real or re-enacted) of the first Supersaurus scapulocoracoid: that is, the one that Jensen referred to in his 1985 paper as “first specimen”, but which in the end he did not designate as the holtotype. This bone, once accessioned, became BYU 12962 (but Jensen refers to it in his papers as BYU 5501).

The first appearance in the film of the Supersaurus scap BYU 12962 fully unconvered [18m11s]. You can easily recognise it as the bone that Jensen posed with from the lobe-shaped acromion process.

— Within seconds of our seeing the scap, Jensen decides the best thing to do is illustrate how it’s “like a sidewalk” by walking up and down on it. Seriously.

Oh, Jim.

— At about 19m30s, we see what is probably the big Barosaurus vertebra BYU 9024 whose identity Jensen changed his mind about a couple of times. Unfortunately, the film quality is very poor here, and you can’t make much out.

— From 20 minutes in, the video shows comparative skeletal reconstructions of Brontosaurus (clearly from Marsh 1891), “Brachiosaurus” [i.e. Giraffatitan] (clearly from Janensch 1950) and Supersaurus. The fascinating thing is that the latter is restored as a brachiosaurid — in fact, as a scaled-up Janensch-1950 Giraffatitan with some tweaks only to the head and anterior neck. So it seems Jensen thought at this time that he’d found a giant brachiosaur, not a diplodocid. (Note that this film was made three years before the Ultrasaurus scapulocoracoid was discovered in 1979, so the presumed brachiosaurid identity cannot have rested in that.)

Brontosaurus (yellow), Brachiosaurus (blue), and Supersaurus (white) — which is restored as a brachiosaurid.

— During this section, a fascinating section of narration says “The animal found here is so much larger than anything ever dreamed of, the press, for lack of scientific name, called it a Supersaurus.” If this is legit, then it seems Jensen is not guilty of coining this dumb name. It’s the first I’ve heard of it: I wonder if anyone can corroborate?

— As 22m06s we are told: “It was an AP newsman who broke the story to the world. Time and Life followed. Reader’s Digest ran the story. And National Geographic, one of the quarry sponsors, began an article.” I would love to get hold of the AP, Time, Life and National Geographic articles. Can anyone help? It seems that all these organisations have archives online, but they all suffer from problems:

Here’s that scap again, in the process of being excavated. [22:05]

— As 22m40s, Jack McIntosh turns up to give an expert opinion. I don’t know how much film of him there is out there, but it’s nice that we have something here.

Everyone’s favourite avocational sauropod specialist, Jack McIntosh.

— At 23:17, we get our best look at the scap, with a long, slow pan that shows the whole thing. (That’s the sequence that I made the composite from, that we started this whole post with.)

All in all, it’s a facinating insight into a time when the Dry Mesa quarry was new and exciting, when it was thought to contain only a single giant sauropod, when that animal was known only informally as “Supersaurus” having been so nicknamed by the media, and when it was (it seems) thought to be brachiosaurid. Take 25 minutes, treat yourself, and watch it.

Update (the next day)

The Wikipedia entry on Jim Jensen says that “In 1973, Brigham Young University cooperated with producer Steve Linton and director John Linton in order to produce The Great Dinosaur Discovery, a 1-hour-long color documentary showing Jensen’s on-site finds in Dry Mesa. […] the full-length documentary was reduced to a 24-minute-long mini-film which started airing on American television channels throughout the USA as of 1976.”

Can anyone confirm that the original date was 1973, and not 1976 as given on the short version that’s linked above?

And, more important, does anyone have access to the full-hour version?

 

References

 

In part 2, we concluded that BYU 9024, the large cervical vertebra assigned by Jensen to the Supersaurus holotype individual, is in fact a perfectly well-behaved Barosaurus cervical — just a much, much bigger one than we’ve been used to seeing. Although we heavily disclaimered our size estimates, Andrea Cau quite rightly commented:

Thanks for the disclaimer: unfortunately, it is going to be ignored by the Internet.
[…]
So, my boring-conservative mind asks: what is the smallest size that is a valid alternative explanation? I mean, if we combine all possible factors (position misinterpretation, deformation effects, allometry and so on) what could be the smallest plausible size? Only the latter should be taken as “the size” of this animal, pending more material.

Andrea is right that we should take a moment to think a bit more about the possible size implications of BYU 9024.

BYU 9024, the huge cervical vertebra assigned to Supersaurus but which is actually Barosaurus, in left dorsolateral view, lying on its right side with anterior to the right. In front of it, for scale, a Diplodocus cervical from about the same serial position. (Note that the Diplodocus vertebra here appears proportionally bigger than it really is, due to being much closer to the camera.)

What we know for sure is that the vertebra is 1380 mm long (give or take a centimeter or two due to the difficulties of measuring big complex bones in an objective way, something we should write about separately some time.)

We are 99% certain that the bone is a Barosaurus cervical.

We are much less certain about the serial position of that bone. When we were at BYU, we concluded that it most resembled C9 of the AMNH specimen, but I honestly can’t remember the detail of our reasoning (can you, Matt?) and our scanned notebooks don’t offer much in the way of help. We know from McIntosh (2005) that the neural spine of C8 is unsplit and that C9 has the first hint of a cleft.  How does that compare with BYU 9024? Here’s a photo to help you decide:

BYU 9024, large cervical vertebra in left dorsolateral view, inverted (i.e. with dorsal towards us and anterior to the right). Note the shallow cleft between metapophyses at bottom left.

And here’s an anaglyph, to help you appreciate the 3D structure. (Don’t have any red-cyan glasses? GET SOME!)

BYU 9024, oriented similarly to the previous photograph.

The morphology around the crown of the neural spine is difficult to interpret, partly just because the fossil itself is a bit smashed up and partly because the bone, the (minimal) restoration and the matrix are such similar colours. But here’s my best attempt to draw out what’s happening, zoomed in from last non-anaglyph photo:

As you start at the prezygapophyses and work backwards, the SPRLs fade out some way before you reach the crown, and disappear at or before what appears to be an ossified midline ligament scar projecting anteriorly from very near the top of the vertebra. Posterior to that are two small, tab-like metapophyses that appear almost like separate osteological features.

Now this is a very strange arrangement. Nothing like it occurs in any of the cervicals of Diplodocus, where all the way from C3 back to the last cervical, the SPRLs run continuously all the way up from the prezygs to the metapophyses:

Hatcher (1901:plate V). Diplodocus carnegii holotype CM 84, cervical vertebra 2-15 in anterior view.

What we’d love to do of course is compare this morphology with a similar plate of the AMNH Barosaurus cervicals in anterior view, but no such plate exists and no such photos can be taken due to the ongoing entombment of the vertebrae. So we’re reduced to feeding on scraps. McIntosh (2005:47) says:

The neural spine of cervical 8 is flat across the top, and that of cervical 9 shows the first trace of a divided spine (Fig. 2.2A). This division increases gradually in sequential vertebrae, being moderately developed in cervicals 12 and 13, and as a deep V-shape in cervicals 15 and 16.

Sadly, McIntosh illustrates only cervicals 8 and 13 in anterior view: Fig 2.2A does not illustrate C9, as the text implies. And neither of the illustrated vertebrae much resembles what we see in BYU 9024.

So while in 2016 we interpreted BYU 9024 is having “the first trace of a divided spine”, we do hold open the possibility that what we’re seeing is a vertebra in which the spine bifurcation is a little more developed than we’d realised, but with strange morphology that does not correspond closely to any well-preserved vertebra we’ve seen of any sauropod. (Most Barosaurus cervicals are either crushed and damaged; the well preserved ones outside of the AMNH walkway tomb are from a more anterior part of the neck where there is no bifurcation of the spine.)

There is one more possibility. Here is a truly lovely (privately owned) Barosaurus cervical in the prep lab at the North American Museum of Ancient Life (NAMAL):

Uncrushed Barosaurus cervical vertebra, serial position uncertain, in the NAMAL prep lab.

In this blessedly undistorted vertebra, we can see that the summit of the neural spine is flared, with laterally projecting laminae that are likely homologous with metapophyses. (The vertebra is symmetrical in this respect.) Might it be possible that the tab-like metapophyses of BYU 9024 were like this in life, but have been folded upwards post-mortem?

All of this leaves the serial position of the vertebra far from certain. But what we can do is compare it with the lengths of all the known AMNH Barosaurus vertebrae. Columns 1 and 2 in the table below show the serial position and total length of the AMNH cervicals. Column 3 shows the factor by which the 1370 mm length of BYU 9024 exceeds the relevant cervical, and column 4 shows the corresponding estimate for total neck length, based on 8.5 m (Wedel 2007:206–207) for AMNH Barosaurus.

Cv# Length (mm) BYU 9224 ratio BYU 9024 neck length
8 618 2.217 18.84
9 685 2.000 17.00
10 737 1.859 15.80
11 775 1.768 15.03
12 813 1.685 14.32
13 850 1.612 13.70
14 865 1.584 13.46
15 840 1.631 13.86
16 750 1.827 15.53

So to finally answer Andrea’s question from waaay back at the start of this post, the smallest possible interpretation of the BYU 9024 animal gives it a neck 1.584 times as long as that of the AMNH individual, which comes out around 13.5 m (and implies a total length of maybe 43 m).

But I don’t at all think that’s right: I am confident that the serial position of BYU 9024 is some way anterior to C14, likely no further back than C11 — which gives us a neck at least 15 m long (and a total length of maybe 48 m and a mass of maybe 12 × 1.768^3 = 66 tonnes).

 

References

  • Hatcher, Jonathan B. 1901. Diplodocus (Marsh): its osteology, taxonomy and probable habits, with a restoration of the skeleton. Memoirs of the Carnegie Museum 1:1-63 and plates I-XIII.
  • McIntosh, John S. 2005. The genus Barosaurus Marsh (Sauropoda, Diplodocidae). pp. 38-77 in: Virginia Tidwell and Ken Carpenter (eds.), Thunder Lizards: the Sauropodomorph Dinosaurs. Indiana University Press, Bloomington, Indiana. 495 pp.
  • Wedel, Mathew J. 2007. Postcranial pneumaticity in dinosaurs and the origin of the avian lung. Ph.D dissertation, Integrative Biology, University of California, Berkeley, CA. Advisors: Kevin Padian and Bill Clemens. 290 pages.

Robin N. Kok asked an interesting question on Twitter:

For all the free money researchers throw at them, they might as well be shareholders. Maybe someone could model a scenario where all the APC money is spent on RELX shares instead, and see how long it takes until researchers own a majority share or RELX.

Well, Elsevier is part of the RELX group, which has a total market capitalisation of £33.5 billion. We can’t know directly how much of that value is in Elsevier, since it’s not traded independently. But according to page 124 their 2017 annual report (the most recent one available), the “Scientific, Technical and Medical” part of RELX (i.e. Elsevier) is responsible for £2,478M of the total £7,355M revenue (33.7%), and for £913M of the £2,284M profit (40.0%). On the basis that a company’s value is largely its ability to make a profit, let’s use the 40% figure, and estimate that Elsevier is worth £13.4 billion.

(Side-comment: ouch.)

According to the Wellcome Trust’s 2016/17 analysis of its open access spend, the average APC for Elsevier articles was £3,049 (average across pure-OA journals and hybrid articles).

On that basis, it would take 4,395,000 APCs to buy Elsevier. How long would that take to do? To work that out, we first need to know how many APC-funded articles they publish each year.

From page 14 of the same annual report as cited above. Elsevier published “over 430,000 articles” in a year. But most of those will have been in subscription journals. The same page says “Subscription sales generated 72% of revenue, transactional sales 26% and advertising 2%”, so assuming that transactional sales means APCs and that per-article revenue was roughly equal for subscription and open-access articles, that means 26% of their articles — a total of 111,800.

At 111,800 APCs per year, it would take a little over 39 years to accumulate the 4,395,000 APCs we’d need to buy Elsevier outright.

That’s no good — it’s too slow.

What if we also cancelled all our subscriptions, and put those funds towards the buy-out, too? That’s actually a much simpler calculation. Total Elsevier revenue was £2,478M. Discard the 2% that’s due to advertising, and £2428M was from subscriptions and APCs. If we saved that much for just five and a half years, we’d have saved enough to buy the whole company.

That’s a surprisingly short time, isn’t it?

(In practice of course it would be much faster: the share-price would drop precipitously as we cancelled all subscription and stopped paying APCs, instantly cutting revenue to one fiftieth of what it was before. But we’ll ignore that effect for our present purposes.)

 

I got an email a couple of days ago from Maija Karala, asking me a question I’d not come across before (among several other questions): how much poop did Argentinosaurus produce in a day?

I don’t recall this question having been addressed in the literature, though if anyone knows different please shout. Having thought about it a little, I sent the following really really vague and hand-wavy response.

Suppose Argentinosaurus massed 73 tonnes (Mazzetta et al. 2004). In cattle, food intake varies roughly with body mass to the power 0.7 (Taylor et al. 1986), so let’s assume that the same is true of sauropods.

Let’s also assume that sauropods are like scaled-up elephants, in that both would have subsisted on low-quality forage. Wikipedia says elephants “can consume as much as 150 kg (330 lb) of food and 40 L (11 US gal) of water in a day.” Let’s assume that the “as much as” suggests we’re talking about a big elephant here, maybe 6 tonnes. So Argentinosaurus is 73/6 = 12 times as heavy, which means its food intake would be 12 ^ 0.7 = 5.7 times as much. That’s 850 kg per day.

Hummel et al. (2008, table 1) show that for a range of foods, the indigestible “neutral detergent fibre” makes up something around half of the mass, so let’s assume that’s the bulk of what gets pooped out, and halve the input to get about 400 kg of poop per day.

References

  • Hummel, Jürgen, Carole T. Gee, Karl-Heinz Südekum, P. Martin Sander, Gunther Nogge and Marcus Clauss. 2008. In vitro digestibility of fern and gymnosperm foliage: implications for sauropod feeding ecology and diet selection. Proceedings of the Royal Society B, 275:1015-1021. doi:10.1098/rspb.2007.1728
  • Mazzetta, Gerardo V., Per Christiansen and Richard A. Farina. 2004. Giants and Bizarres: Body Size of Some Southern South American Cretaceous Dinosaurs. Historical Biology 2004:1-13.
  • Taylor, C. S., A. J. Moore and R. B. Thiessen. 1986. Voluntary food intake in relation to body weight among British breeds of cattle. Animal Science 42(1):11-18.

You could drive several trucks through the holes in that reasoning, but it’s a start. Can anyone help to refine the reasoning, improve the references, and get a better estimate?

Norwescon 41 Guests of Honor: Ken Liu, Galen Dara, and, er, me. Mike would like to remind you that you can get your own ‘Kylo Stabbed First’ t-shirt here.

The week before last I was fortunate to be the Science Guest of Honor at Norwescon 41 in Seattle (as threatened back when). I had a fantastic time. I got to give talks on binocular stargazing and the sizes of the largest sauropods and whales (ahem), participate on panels on alien biology and creature drawing, and meet a ton of cool people, including my fellow Guests of Honor, multiple-award-winning author Ken Liu and multiple-award-winning artist Galen Dara, both of whom turned out to be humble, easygoing, regular folks (if frighteningly talented).

I also had a lot of great conversations with folks who were attending the con, which is exactly what I wanted. One of the most interesting was a hallway conversation with a fellow DM named Shawn Connor. He had a great question for me, which I liked so much I wanted to answer it here on the blog. Here’s his question, copied with permission from a follow-up email:

I run tabletop RPGs, and in my current game one of the characters is a caveman type who naturally grew up hunting dinosaurs. As one does. His weapon is a dinosaur bone, customized and used as a club. I have attached the picture that he came up with [below]. Now understanding the picture is obviously not of a real dinosaur bone – it’s probably a chicken bone or a cow bone or something – let’s assume for the sake of this exercise that it is and that it is four feet long stem to stern. Given that, two questions: discounting the extra bling attached how heavy would such a bone be, and what kind of dinosaur could it have come from?

I’m going to answer those questions out of order. Advance warning: this will be a loooong post that will go down several rabbit holes that are likely of more intense interest to me, personally, than to anyone else on the planet. Read on at your own risk.

Whose femur is in the image?

First, Shawn is correct in noting that the femur in the image provided by his player is not a dinosaur femur. The prominent trochanters and spherical head offset on a narrow neck clearly make it a mammal femur, and if it’s four feet long, it could only have come from an elephant or an indricothere. Or a giant humanoid, I suppose, which is what the anatomy of the bone in the image most closely resembles. (It also appears to be foreshortened to make the distal end look bigger, or deliberately distorted to enhance the clubby-ness.)

Mounted elephant at the Museum of Osteology in Oklahoma City, with Tyler Hunt for scale.

But let’s play along and assume it’s from a non-human mammal. How big? Back in 2016 I was fortunate to get to measure most of the mounted large mammal skeletons at the Museum of Osteology in Oklahoma City, along with Tyler Hunt, then a University of Oklahoma undergrad and now finishing up his MS thesis under my mentor, Rich Cifelli.* The mounted elephant at the Museum of Osteology has a shoulder height of 254 cm (8 ft, 4 in) and a femur length of 102 cm (3 ft, 4 in). Assuming isometric scaling, a world record elephant with a shoulder height of 366 cm (12 ft) would have a femur length of 147 cm (4 ft, 10 in). So a four-foot (122 cm) femur would belong to an elephant roughly in the middle of that range, about ten feet (3 m) tall at the shoulder. That’s the size of the big bull elephant mounted at the Field Museum in Chicago.

The big mounted bull elephant at the Field Museum is 10 feet tall at the shoulder and weighed 6 tons in life. Note Mike for scale on the lower right. He and the elephant are about equidistant from the camera, so he should make a roughly accurate scale bar. Photo from our visit in 2005!

* Two further notes: first, I have roughly a zillion awesome photos from that 2016 visit to the Museum of Osteology, both of the specimens and of Tyler and me measuring them – not having posted them yet is one of the things I was whingeing about in the post that kicked off our return-to-weekly-posting thing this year. And second, I owe a belated and public thanks to the folks at the Museum of Osteology for accommodating Tyler and me. They helped us with ladders and so on and basically gave us free rein to play with collect data from their mounted skeletons, which was incredibly generous and helpful, and fortunately reflects the pro-research and pro-researcher attitude of most museums.

Which dinos had four-foot femora?

As for what kind of dinosaur a four-foot femur could have come from, we can rapidly narrow it down to a handful of clades: sauropods, ornithopods, theropods, and stegosaurs.

  • Sauropods. The longest complete femora of Patagotitan are 238 cm (7 ft, 10 in; Carballido et al. 2017), and an incomplete femur of Argentinosaurus has an estimated complete length of 250 cm (8 ft, 2 in; Mazzetta et al. 2004). So a four-foot femur would not be from a particularly large sauropod – something about elephant-sized, as you might expect from the elephant comparison above. Our old friend Haplocanthosaurus will fit the bill, as we’ll see in a bit.
  • Ornithopods. Femora of 172 cm (5 ft, 8 in) are known for the hadrosaurs Shantungosaurus (Hone et al. 2014) and Huaxiaosaurus (Zhao and Li 2009), and Zhao et al. (2007) reported a 170 cm (5 ft, 7 in) femur for Zhuchengosaurus (Huaxiaosaurus and Zhuchengosaurus may be junior synonyms of Shantungosaurus). But those are all monsters, well over 10 metric tons in estimated mass. So a four-foot femur would be from a large but not insanely large hadrosaur.

Mmmmmm…suffering. OM NOM NOM NOM!!

  • Theropods. Among the largest theropods, the holotype of Giganotosaurus has a femur length of 143 cm (4 ft, 8 in; Coria and Salgado 1995), and ‘Sue’ the T. rex (a.k.a. FMNH PR2081) has a right femur 132 cm long (4 ft, 4 in; Brochu 2003). So a four-foot femur from a theropod would definitely be from one of the monsters. The femur of Saurophaganax was 113.5 cm long (Chure 1995), just under four feet, which I only note as an excuse to use the above photo, which I adore.
  • Stegosaurs. I don’t know the longest femur that has been recovered from a stegosaur, but getting in the ballpark is easy. NHMUK PV R36730 has a femur 87 cm long, and the whole animal was approximately 6 m long (Maidment et al. 2015). Partial bits and bobs of the largest stegosaurs suggest animals about 9 m long, implying a femur length of about 130 cm (4 ft, 3 in), or just over the line.

I think that’s it. I don’t know of any ceratopsians or ankylosaurs with femora long enough to qualify – I assume someone will let me know in the comments if I’ve forgotten any.

How much would a four-foot femur weigh?

There are a couple of ways to get to the answer here. One is to use Graphic Double Integration, which is explained in this post.

Limb bones are not solid – in terrestrial tetrapods there is virtually always a marrow cavity of some sort, and in marine tetrapods the limb bones tend to be cancellous all the way through. Estimating the mass of a limb bone is a lot like estimating the mass of a pneumatic bone: figure out the cross-sectional areas of the cortex and marrow cavity (or air space if the bone is pneumatic), multiply by the length of the element to get volumes, and multiply those volumes by the density of the materials to get masses. I piled up all the relevant numbers and formulas in Tutorial 24, a move that has frequently made me grateful to my former self (instead of cussing his lazy ass, which is my more usual attitude toward Past Matt).

Currey and Alexander (1985: fig. 1)

Sauropod limb bones are pretty darned dense, with extremely thick cortices and smallish marrow spaces that are not actually hollow (tubular) but are instead filled with trabecular bone. My gut feeling is that even a four-foot sauropod femur would be almost too heavy to lift, let alone wield as a club, so in the coming calculations I will err in the direction of underestimating the mass, to give our hypothetical caveman the best possible chance of realizing his dream.

Some of the proportionally thinnest cortices I’ve seen in sauropod limb bones are those of the macronarian Haestasaurus becklesii NHMUK R1870, which Mike conveniently showed in cross-section in this post. I could look up the actual dimensions of the bones (in Upchurch et al 2015: table 1 – they passed the MYDD test, as expected), but for these calculations I don’t need them. All I need are relative areas, for which pixels are good enough.

First, I took Mike’s photo into GIMP and drew two diameters across each bone, one maximum diameter and a second at right angles. Then I drew tick marks about where I think the boundaries lie between the cortex and the trabecular marrow cavity. Next, I used those lines as guides to determine the outer diameters (D) and inner diameters (d) in pixels, as noted in the image.

For the radius, on the left, the mean diameters are D = 891 and d = 648. I could divide those by 2 to get radii and then plug them into the formula for the area of a circle, etc., but there’s an easier way still. For a tubular bone, the proportional area of the inner circle or ellipse is equal to k^2, where k = r/R. Or d/D. (See Wedel 2005 and Tutorial 24 for the derivation of that.) For the Haestasaurus radius (the bone, not the geometric dimension), d/D = 0.727, and that number squared is 0.529. So the marrow cavity occupies 53% of the cross-sectional area, and the cortex occupies the other 47%.

For the ulna, on the right, the mean diameters are D = 896 and d = 606, d/D = 0.676, and that number squared is 0.457. So in this element, the marrow cavity occupies 46% of the cross-sectional area, and the cortex occupies the other 54%.

(For this quick-and-dirty calculation, I am going to ignore the fact that limb bones are more complex than tubes and that their cross-sectional properties change along their lengths – what I am doing here is closer to Fermi estimation than to anything I would publish. And we’ll ground-truth it before the end anyway.)

Left: rat humerus, right: mole humerus. The mole humerus spits upon my simple geometric models, with extreme prejudice. From this post.

You can see from the photo (the Haestasaurus photo, not the mole photo) that neither bone has a completely hollow marrow cavity – both marrow cavities are filled with trabecular bone. By cutting out good-looking chunks in GIMP and thresholding them, I estimate that these trabecular areas are about 30% bone and 70% marrow (actual marrow space with no bone tissue) by cross-sectional area. According to Currey and Alexader (1985: 455), the specific gravities of fatty marrow and bone tissue are 0.93 and 2.1, respectively. The density of the trabecular area is then (0.3*2.1)+(0.7*0.93) =  1.28 kg/L, or about one quarter more dense than water.

But that’s just the trabecular area, which accounts for about one half of the cross-sectional area of each bone. The other half is cortex, which is probably close to 2.1 kg/L throughout. The estimated whole-element densities are then:

Radius: (0.53*1.28)+(0.47*2.1) = 1.67 kg/L

Ulna: (0.46*1.28)+(0.54*2.1) = 1.72 kg/L

Do those numbers pass the sniff test? Well, any skeletal elements that are composed of bone tissue (SG = 2.1) and marrow (SG = 0.93) are constrained to have densities somewhere between those extremes (some animals beat this by building parts of their skeletons out of [bone tissue + air] instead of [bone tissue + marrow]). We know that sauropod limb bones tend to have thick cortices and small marrow cavities, and that the marrow cavities are themselves a combination of trabecular bone and actual marrow space, so we’d expect the overall density to be closer to the 2.1 kg/L end of the scale than the 0.93 kg/L end. And our rough estimates of ~1.7 kg/L fall about where we’d expect.

Femur of Haplocanthosaurus priscus, CM 572, modified from Hatcher (1903: fig. 14).

To convert to masses, we need to know volumes. We can use Haplocanthosaurus here – the femur of the holotype of H. priscus, CM 572, is 1275 mm long (Hatcher 1903), which is just a hair over four feet (4 ft, 2.2 in to be exact). The midshaft width is 207 mm, and the proximal and distal max widths are 353 and 309 mm, respectively. I could do a for-real GDI, but I’m lazy and approximate numbers are good enough here. Just eyeballing it, the width of the femur is about the same over most of its length, so I’m guessing the average width is about 23 cm. The average width:length ratio for the femora of non-titanosaur sauropods is 3:2 (Wilson and Carrano 1999: table 1), which would give an anteroposterior diameter of about 15 cm and an average diameter over the whole length of 19 cm. The volume would then be the average cross-section area, 3.14*9.5*9.5, multiplied by the length, 128 cm, or 36,273 cm^3, or 36.3 L. Multiplied by the ~1.7 kg/L density we estimated above, that gives an estimated mass of 62 kg, or about 137 lbs. A femur that was exactly four feet long would be a little lighter – 86.6% as massive, to be exact, or 53.4 kg (118 lbs).

I know that the PCs in RPGs are supposed to be heroes, but that seems a little extreme.

But wait! Bones dry out and they lose mass as they do so. Lawes and Gilbert (1859) reported that the dry weight of bones of healthy sheep and cattle was only 74% of the wet mass. Cows and sheep have thinner bone cortices than sauropods or elephants, but it doesn’t seem unreasonable that a dry sauropod femur might only weigh 80% as much as a fresh one. That gets us down to 43 kg – about 95 lbs – which is still well beyond what anyone is probably going to be wielding, even if they’re Conan the Cimmerian.

Picture is unrelated.

I mentioned at the top of this section that there are a couple of ways to get here. The second way is to simply see what actual elephant femora weigh, and then scale up to dinosaur size. According to Tefera (2012: table 1), a 110-cm elephant femur has a mass of 21.5 kg (47 lbs). I reckon that’s a dry mass, since the femur in question had sat in a shed for 50 years before being weighed (Tefera 2012: p. 17). Assuming isometry, a four-foot (122 cm) elephant femur would have a dry mass of 29.4 kg (65 lbs). That’s a lot lighter than the estimated mass of the sauropod femur – can we explain the discrepancy?

 

Femora of a horse, a cow, and an elephant (from left to right in each set), from Tefera (2012: plate 1).

I think so. Elephant femora are more slender than Haplocanthosaurus femora. Tefera (2012) reported a circumference of 44 cm for a 110-cm elephant femur. Scaling up from 110 cm to 122 cm would increase that femur circumference to 49 cm, implying a mean diameter of 15.6 cm, compared to 19 cm for the Haplo femur. That might not seem like a big difference, but it means a cross-sectional area only 2/3 as great, and hence a volume about 2/3 that of a sauropod femur of the same length. And that lines up almost eerily well with our estimated masses of 29 and 43 kg (ratio 2:3) for the four-foot elephant and sauropod femora.

A Better Weapon?

Could our hypothetical caveman do better by choosing a different dinosaur’s femur? Doubtful – the femora of ‘Sue’ are roughly the same length as the Haplo femur mentioned above, and have similar cross-sectional dimensions. Hadrosaur and stegosaur femora don’t look any better. Even if the theropod femur was somewhat lighter because of thinner cortices, how are you going to effectively grip and wield something 15-19 cm in diameter?

I note that the largest axes and sledgehammers sold by Forestry Suppliers, Inc., are about 3 feet long. Could we get our large-animal-femur-based-clubs into the realm of believability by shrinking them to 3 feet instead of 4? Possibly – 0.75 to the third power is 0.42. That brings the elephant femur club down to 12.3 kg (27 lbs) and the sauropod femur club down to 18 kg (40 lbs), only 2-3 times the mass of the largest commonly-available sledgehammers. I sure as heck wouldn’t want to lug such a thing around, much less swing it, but I can just about imagine a mighty hero doing so.

Yes, there were longer historical weapons. Among swing-able weapons (as opposed to spears, etc.), Scottish claymores could be more than four feet long, but crucially they were quite light compared to the clubs we’ve been discussing, maxing out under 3 kg, at least according to Wikipedia.

T. rex FMNH PR2081 right fibula in lateral (top) and medial (bottom) views. Scale is 30 cm. From Brochu (2003: fig. 97).

If one is looking for a good dinosaur bone to wield as a club, may I suggest the fibula of a large theropod? The right (non-pathologic) fibula of ‘Sue’ is 103 cm long (3 ft, 4.5 in), has a max shaft diameter just under 3 inches – so it could plausibly be held by (large) human hands, and it probably massed something like 8-9 kg (17-20 lbs) in life, based on some quick-and-dirty calculations like those I did above. The proximal end is even expanded like the head of a war club. The length and mass are both in the realm of possibility for large, fit, non-supernaturally-boosted humans. Half-orc barbarians will love them.

And that’s my ‘expert’ recommendation as a dice-slinging paleontologist. Thanks for reading – you have Conan-level stamina if you got this far – and thanks to Shawn for letting me use his question to freewheel on some of my favorite geeky topics.

References