Upper limits on the mass of land animals estimated through the articular area of limb-bone cartilage Michael P. Taylor School of Earth and Environmental.

Upper limits on the mass of land animalsestimated through the articular area of limb-bone cartilage

Michael P. Taylor

School of Earth and Environmental SciencesUniversity of Portsmouth

Portsmouth PO1 3QL

[email protected]

Upper limits on the mass of land animalsestimated through the articular area of limb-bone cartilage

Michael P. Taylor

School of Earth and Environmental SciencesUniversity of Portsmouth

Portsmouth PO1 3QL

[email protected]

(featuring BIG SAUROPODS)

Michael P. Taylor: Upper limits on the mass of land animals estimated through the articular area of limb-bone cartilage

How big can land animals get?



Palaeontologist: 80 kg



Palaeontologist: 80 kg100



Palaeontologist: 100 kgRhino: 1000 kg




Elephant: 10000 kg




Elephant: 10000 kgSauropod: 100000 kg





????: 1000000 kg





???: 1000000 kgGodzilla: 10000000 kg


What limits the size of land animals?



* Bone strength (Hokkanen 1985)




* Muscle mass (Hokkanen 1985)





* Metabolic scaling (Seymour and Lillywhite 2000)






* Metabolic overheating (Alexander 1998)







* Limits on limb-bone allometry (Christiansen 2002)







* Limits on limb-bone allometry (Christiansen 2002)

* Strength of articular cartilage (THIS STUDY!)


Were sauropods terrestrial?

... Orthodoxy has changed over time ...

Zallinger's mural(1947)

BBC's WalkingWith Dinosaurs(1999)


Aquatic/amphibious sauropods

Most early workers considered sauropods to be aquatic,or at least “amphibious”.




* Owen (1859) thought that Cetiosaurus was a marine crocodile.





* Colbert (1961) argued that the dorsally positioned nares of Diplodocus indicated an aquatic lifestyle.






* Hatcher (1901), Hay (1910) and others felt that the cartilaginous joints of sauropod limbs would not support their weight on land.






* Hatcher (1901), Hay (1910) and others felt that the cartilaginous joints of sauropod limbs would not support their weight on land.

* Burian (1957) restored Brachiosaurus walking on the bottom of a lake, “snorkelling” with its long neck and high nostrils.


Terrestrial sauropods

Many lines of evidence show that sauropods were primarilyterrestrial.




* Extreme lightening of vertebrae (skeletal pneumaticity) is an adaptation for terrestrial life (Wedel 2003)





* Sauropod feet were too compact for walking in swamps: individuals have been found mired (Russell et al. 1980)






* Tall, relatively narrow torsos characterise terrestrial animals and are biomechanically adapted for heavy loads (Coombs 1975)






* Tall, relatively narrow torsos characterise terrestrial animals and are biomechanically adapted for heavy loads (Coombs 1975)

* Many sauropods found in seasonally dry environments, e.g. Morrison Formation (Dodson et al. 1980)


Limbs of dinosaurs and mammals compared

Jensen 1988 compared dinosaur limb-bones unfavourably with thoseof mammals.

“The limb and foot joints in the most agile dinosaur,large or small, are structurally and functionally inferiorto those of proboscidians and, in large measure, to allmammals [because mammals have] compact, bone-to-bone join geometry that includes ball-and-socket jointsand curvilinear flanged joints mating perfectly withmatching incurvate forms”



Jensen 1988 compared dinosaur limb-bones unfavourably with thoseof mammals.

“The limb and foot joints in the most agile dinosaur,large or small, are structurally and functionally inferiorto those of proboscidians and, in large measure, to allmammals [because mammals have] compact, bone-to-bone join geometry that includes ball-and-socket jointsand curvilinear flanged joints mating perfectly withmatching incurvate forms”

... which is a bit rude in a paper that named a new sauropod(Cathetosaurus, currently considered a species of Camarasaurus).



Humeri of Camarsaurus and Brontops



But sauropods, like extant dinosaurs (birds), would have had large caps of hyaline cartilage on each articular surface.

These would achieve the close fitting that is otherwise not possible.


Digression: how thick were cartilage caps?

Brachiosaurus brancai (HMN S II) mount in the Humbold Musuem, Berlin.

Cartilage must have filled the large part of the acetabulum not filled by the head of the femur.

So pretty darned thick!


Were the cartilage caps strong enough?

The method is simple:

1. Choose a big dinosaur.





2. Find the mass of the dinosaur.






3. Calculate the articular area of its limb-bone cartilage.







4. Divide mass by area to find the stress acting of the cartilage.








5. Compare this with the known compressive strength of cartilage.



The method is simple IN PRINCIPLE:





5. Compare this with the known compressive strength of cartilage.

BUT EVERY STEP EXCEPT #4 IS A MINEFIELD.


1. Choose a big dinosaur

We need a dinosaur that:* is huge* is well enough represented to estimate its mass* is known from material including femur and humerus




Amphicoelias fragillimus and Bruhathkayosaurus are truly huge, butare known only from scraps.





Argentinosaurus and Paralititan are huge and their masses can bemeaningfully estimated, but the humerus of the former and the femurof the latter are unknown.





Argentinosaurus and Paralititan are huge and their masses can bemeaningfully estimated, but the humerus of the former and the femurof the latter are unknown.

Brachiosaurus is well-known; its mass can be estimated from the typespecimen of B. brancai and both humerus and femur are well preservedin the type specimen of B. altithorax ... which is close enough.


2. Find the mass of the dinosaur

Mass estimates for Brachiosaurus have varied wildly:



Mass estimates for Brachiosaurus have varied wildly:Colbert 1962: 78 tonnes (volume of model)Russell et al. 1980: 15 tonnes (limb-bone allometry)Alexander 1989: 47 tonnes (model)Anderson et al. 1985: 29 tonnes (allometry)Paul 1988: 32 for B. brancai, 35 for B. altithorax (model)Gunga et al. 1995: 74 tonnes (model)Christiansen 1997: 37 tonnes (model)Henderson 2003: 26 tonnes (model)




Estimates based on limb-bone allometry are not measurements: ignore.




Estimates based on limb-bone allometry are not measurements: ignore.Colbert's model was grotesquely fat – probably on steroids.




Estimates based on limb-bone allometry are not measurements: ignore.Colbert's model was grotesquely fat – probably on steroids.Gunga et al.'s model is made from round (not elliptical) sections.




Estimates based on limb-bone allometry are not measurements: ignore.Colbert's model was grotesquely fat – probably on steroids.Gunga et al.'s model is made from round (not elliptical) sections.

Average of Alexander, Paul, Christiansen and Henderson is 36 tonnes.


3. Calculate the articular area of limb-bone cartilage

Proximal surfaces of humerus and femur(Reconstruction from Janensch 1950)



I scanned plate LXXIV (limb bones) of Riggs 1904 on Brachiosaurus.



I scanned plate LXXIV (limb bones) of Riggs 1904 on Brachiosaurus.I threw away the anterior views and just kept the proximal views.

Femur Humerus



I scanned plate LXXIV (limb bones) of Riggs 1904 on Brachiosaurus.I threw away the anterior views and just kept the proximal views.I mapped all the bone to black and background to white.

Femur Humerus



I scanned plate LXXIV (limb bones) of Riggs 1904 on Brachiosaurus.I threw away the anterior views and just kept the proximal views.I mapped all the bone to black and background to white.I counted the black pixels.

Femur96447 pixels

Humerus96023 pixels



I scanned plate LXXIV (limb bones) of Riggs 1904 on Brachiosaurus.I threw away the anterior views and just kept the proximal views.I mapped all the bone to black and background to white.I counted the black pixels.From the 204cm length of humerus, I measured 97 pixels per 10cm

Femur96447 pixels= 0.1025 m2

Humerus96023 pixels= 0.1021 m2


4. Stress on cartilage

Proximal articular areas of Brachiosaurus are:0.1025 m2 (femur) and 0.1021 m2 (humerus)




=> Total area for two femora and two humeri:2 x 0.1025 + 2 x 0.1021 = 0.4092 m2





For Brachiosaurus we assume even distribution of mass (Alexander 1989)





For Brachiosaurus we assume even distribution of mass (Alexander 1989)(Not true for all sauropods: Diplodocus carried 80% mass on hindlimbs.)






Mass estimated at 36 metric tonnes = 36000 kg






Mass estimated at 36 metric tonnes = 36000 kgAcceleration due to gravity is 9.8 m2







=> weight = 9.8 x 36000 = 352800 Newtons







=> weight = 9.8 x 36000 = 352800 Newtons=> compressive stress = 352800 / 0.4092 = 862 KPascals







=> weight = 9.8 x 36000 = 352800 Newtons=> compressive stress = 352800 / 0.4092 = 862 KPascals

Is that a lot?


5. Strength of cartilage



The compressive strength of cartilageis a most holy and sacred secret

that cannot and must not be divulged!



“Fresh compact bone loaded parallel to its grain has a compressive strengthof 1330 to 2100 kg/cm^2 (19,000 to 30,000 lb/in^2) ... Values for cartilagevary, but are lower than those for bone.”

– Hildebrand 1988.





“The strength of cartilage is considerably less than that of bone.” – McGowan 1999.






“ ”– Alexander 1989.






“ ”– Alexander 1989.

“Compression strength is a mechanical property that has meaning withrespect to the hardness of rigid materials. Applying this concept toresilient materials is not something I'm comfortable with.”

– pers. comm., permission to cite not sought.



In the end I only found one paper containing hard numbers:

“Axial load up to 5 MPa produces an almost elasticdeformation, an increasing axial load results in aplastic deformation [...] An axial load of 25.8 +/-5.2 MPa (sigma max) causes a break of cartilage.

– Spahn and Wittig 2003.



In the end I only found one paper containing hard numbers:

“Axial load up to 5 MPa produces an almost elasticdeformation, an increasing axial load results in aplastic deformation [...] An axial load of 25.8 +/-5.2 MPa (sigma max) causes a break of cartilage.

– Spahn and Wittig 2003.

THANK YOU, SPAHN! THANK YOU, WITTIG!


Conclusions

* Assume a mass of 36 tonnes for Brachiosaurus* Assume even distribution of bodyweight on fore and hind limbs

The total area of proximal articular facets is 0.409 m2

When standing still, compressive stress on cartilage is 862 KPa


Conclusions

* Assume a mass of 36 tonnes for Brachiosaurus* Assume even distribution of bodyweight on fore and hind limbs

The total area of proximal articular facets is 0.409 m2

When standing still, compressive stress on cartilage is 862 KPa

This is about 1/6 of Spahn and Wittig's figure of 5 MPa before plastic deformation of cartilage occurs.

So a stationary Brachiosaurus would be comfortable on land.


Conclusions

Locomotory stress is about twice that of standing (Jayes and Alexander1978). For Brachiosaurus, this is about 1.7 MPa, which is still safe.


Conclusions

Godzilla's mass is approximately10000000 kg.

Assuming his leg bones scale isometricallyfrom those of Brachiosaurus, and that hemaintains bipedal posture, he will suffer11 times the stress on his cartilage.

18.7 MPa should suffice to crush hisarticular cartilage caps like over-ripewater-melons.

So the world is safe ... for now.


Conclusions

Crap.Godzilla's mass is approximately10000000 kg.

Assuming his leg bones scale isometricallyfrom those of Brachiosaurus, and that hemaintains bipedal posture, he will suffer11 times the stress on his cartilage.

18.7 MPa should suffice to crush hisarticular cartilage caps like over-ripewater-melons.

So the world is safe ... for now.


Afterword: why you shouldn't trust these figures

The findings of this study should be regarded as a first step, withcorrections and refinements hopefully to follow.

Some areas where refinement is needed:

* How much of the articular cartilage is in contact at once?

* What is the strength of articular cartilage?

* How is mass distributed between fore and hind limbs?

* How does joint reaction force compare with weight?


How much of the articular surface is in contact at once?

Nocartilage


Nocartilage Thin

cartilage



Nocartilage Thin

cartilage

Bird-likecartilage


The whole articulararea is in contact.


What is the strength of articular cartilage?

Although I am using Spahn and Wittig's figure of 5MPa, the issueis not resolved.

“It can be estimated from Lucas and Bresler's (1961)analysis of weightlifting that stresses of up to 6 MPaare liable to occur in human intervertebral discs.”

– Alexander 1985

30 MPa of force can be experienced in the human knee (Grodzinskyet al. 2000) – six times Spahn and Wittig's elastic limit!

Cartilage is a complex material, so its compressive strength dependson how quickly the force is applied, how much shear acts, the watercontent of the cartilage, etc.


How is mass distributed between fore and hind limbs?

For Brachiosaurus, mass appears to be evenly distributed betweenfore and hind limbs (Alexander 1989), and the articular areas of thehumerus and femur are similar. This does not apply to all sauropods.

For example, diplodocids carry about 84% of their mass on their hindlegs (Alexander 1985's figure for Diplodocus) – but the articular areasof their humerus and femur are similar (measurement of Apatosaurus).

Calculations for diplodocids must take uneven distribution into account.

Why does Apatosaurus have such disproportionately large articular areasin its forelimbs?

(Guess: to absorb shock when descending abruptly from bipedal rearing.)


How does joint reaction force compare with weight?

It is “obvious” that the force acting at a joint is equal to weight.

Obvious – yes. True – no.

Hip joint reaction forces in stationary humans is 4.2 times weight!

Why? I don't know but I intend to find out.

If this were also true in Brachiosaurus, locomotion would seem to beimpossible.


Help! Too much uncertainty!

We don't confidently know:

* The masses of dinosaurs (factor of perhaps 5)* The relationship between mass and joint reaction force (4.2)* The extent of articulation between limb bones (?2)* How locomotory forces exceeds static forces (?3)* The strength of articular cartilage (6!)

So my figures are correct within a factor of 756.

So are these results worth anything?


Help! Too much uncertainty!

We don't confidently know:

* The masses of dinosaurs (factor of perhaps 5)* The relationship between mass and joint reaction force (4.2)* The extent of articulation between limb bones (?2)* How locomotory forces exceeds static forces (?3)* The strength of articular cartilage (6!)

So my figures are correct within a factor of 756.

So are these results worth anything?

“The best way to get information [on the Internet]isn't to ask a question, but to post the wrong information.”

– [email protected]


Acknowledgements

Thanks to my supervisor David M. Martill.

Thanks to John R. Hutchinson, Adam P. Summers and H. ToddWheeler for email discussions concerning the properties ofcartilage.

Thanks to Mathew J. Wedel for supplying literature andenthusiasm.

Thanks to Spahn and Wittig for actually writing down a numberfor the compressive strength of hyaline cartilage!

Upper limits on the mass of land animals estimated through the articular area of limb-bone cartilage Michael P. Taylor School of Earth and Environmental.

Documents

mass of land animals

size of land animals

upper limits

bone cartilage michael

articular area of limb

bone strength hokkanen

muscle mass hokkanen

taylor school of earth