Smilodon populator

[LonePredator]

(05-04-2022, 11:04 AM)jrocks Wrote:

(04-30-2022, 08:58 AM)tigerluver Wrote:

(04-30-2022, 08:25 AM)LonePredator Wrote:

(04-30-2022, 03:09 AM)tigerluver Wrote:

(04-30-2022, 02:18 AM)LonePredator Wrote:

(04-30-2022, 02:10 AM)tigerluver Wrote:

(04-29-2022, 02:41 AM)LonePredator Wrote:

(04-29-2022, 02:33 AM)tigerluver Wrote:

(04-28-2022, 10:05 PM)LonePredator Wrote: @tigerluver How did they estimate the size of Smilodon Populator when it was so different from currently living felids? Like it’s morphology was entirely different from the living cats.

Can you tell me the equations which has been used for it and are all these allometric equations similar in different kinds of uses? For example, is an allometric equation to find the weight of smilodon populator similar to say Cave Bear? Can you tell me a little bit about this?

You make an excellent observation and truly, no publication has really been able to address the limitation.

The oldest estimates from Anyonge used circumferences of the long bones. Christiansen and Harris (2005) did similar but also added many other dimensions of the bone. Both used extant felids as the source of the regression equations and cannot actually account for the unique proportions of Smilodon. The hope is that circumference has minimal interspecific different in estimating body mass in these. While Campione et al. (2012) did show circumference is best correlated with body mass across taxa, the percent error is still actually quite significant. Another example of the problem of Smilodon is that it is front heavy, therefore there is a comparatively disproportionate amount of weight supported by the forelimbs as compared to hindlimbs (while in extant cats the distribution is more even). If we think about this theoretically, this would result in the humerus robusticity overestimating (a greater percentage of the body weight is stressing the forelimbs, so they will be proportionately thicker for the total weight) while femur robusticity (a lesser percentage of the body weight is stressing the hindlimbs, so they would be proportionately thinner for the body weight) would underestimate.

The allometric equations were based on felids. Separate equations have been made for bears. When we compare the estimates from the equations, the results are different. This means an allometric equation applied to a felid cannot really apply accurately to a bear.

The giant Smilodon skull was estimated using a regression developed from felid skulls. The issue here is that Smilodon had a smaller skull for its body as compared to extant felids. Therefore, this estimate is probably a significant underestimate.

So there is no perfect formula for Smilodon. Volumetric estimation can be attempted but it gets exposed to subjectivity as the model is essentially art subject to the artist's thoughts.

Thankyou very much. A very well detailed explanation. I will check out those sources. As far as I remember, I think that the oldest estimates which came from Anyonge were gross overestimations.

Just one more question, do you think that the isometric scaling of extant felids by scaling the the mass cubically in relation to scaling the length linearly give an accurate enough bodymass estimate? I’m aware it’s just a law of physics which only works on objects of similar proportions but...

What if I do the same for the shoulder height, chest girth and head body length. I linearly scale all these dimensions to the dimensions of the big Tiger and then cubically scale the mass and then average out all the three mass estimates then will it practically hold up?

Do you think that will give a good enough estimate of the mass of the big Tiger and can you roughly predict a percentage error that I could get in my result?

And once again, thanks a lot for your detailed explaination.

Isometry is the most parsimonious when are comparing animals of the same species so I prefer it.

Isometry works with 3 dimensions (length, width, height). Therefore, if you use three measurements that are surrogates for these dimensions, cubically scale the mass, then average them together, you should get an estimate that ideally accurately estimates the mass/volume. There is no real way to calculate percent error unfortunately. 

Another way to do it is to multiple the difference ratio of length by the difference ratio of width and the difference ratio of height. This is based on the formula for volume (length * width * height). For instance, if a tiger is 1.2x longer, 1.1x width, and 1.4x taller than another tiger, you would do:

Mass bigger tiger = Mass smaller tiger * (1.2 * 1.1 * 1.4) = Mass smaller tiger * 2.002

This way is mathematically more accurate than averages the cubically scaled masses together.

This makes it much more simpler. Thanks a lot! I was calculating with all three dimensions separately and then averaging them out and even though it was working well, I’ll try this out now.

Just one more question. I’ve seen in some studies which show that the bite force of Jaguar is 3/4th that of a Bengal Tiger while the bodymass of the Jaguar is only half of the Tiger.

But can we really say that Jaguar has pound for pound stronger bite? Because when we scale the Jaguar to the same weight as the Tiger (by 2x) then the bite force would only increase by 1.777 times, right? Because the cross sectional area would only increase by an exponent of 2 while the mass increases by an exponent of 3.

And I calculated this and found that when mass of the Jaguar is scaled to 200kg from 100 then the force would still be about 12% weaker. Don’t you also think that the Jaguar would still have a weaker bite force than the Tiger if we scale a Jaguar by 2 times to make it equal in weight as the Tiger??

In the literal sense your math would be right. However, as skulls get bigger there are allometric effects such as increase in the length of the masseteric fossa which would increase the bite force. 

Therefore, regressions are made to account for the allometric changes. Wroe et al. finds the jaguar's bite force is above the regression line:

*This image is copyright of its original author

The positioning of where the point is as compared to the regression line is measured by the bite force quotient (BFQ). In the study, the jaguar's BFQ is 1.37 and the tiger's 1.27. A weakness of the study is that actual weights of animals are not used, just estimates. Therefore, the BFQ may not be accurate.

A more fruitful measurement would probably be bite force over skull length. I ran the regression quickly (can't figure out how to label datapoints, if I have time later I will update the post):

*This image is copyright of its original author

Based on the second graph, the jaguar has significantly stronger bite than predicted for its skull length.

Do you happen to have a table for the regression graph of logarithm of bite force against the log of bodymass? It’s hard to get the log cb to log bm ratio just from the graph so could you share a regression table of it if you have it?

The table is in the supplement here: https://royalsocietypublishing.org/doi/s....2004.2986

@tigerluver @GuateGojira

so are these weights right for the largest specimens of each prehistoric cat

smilodon populator 16.07 inch skull at around 490 kg if isometric scaling is used to scale up the 15.4 inch skull specimen to the 16.07 inch one

american lion 18.4 inch skull at 375 kg

natadomeri lion 19.08 inch skull at 390 kg

ngandong tiger 480 mm femur at 368 kg, i think i saw somewhere that there was a huge mandible found of a bornean tiger is it known much does that specimen is

also how did m. giganteus compare to these cats as i cant find much on them

Not much was found of the Bornean Tiger but based on it there was an isometric size estimate with extant Tigers as surrogates and it suggests that this Tiger could have been extremely large. See the peer reviewed paper (Sherani, 2019). The weight was estimated at ~480kg but since not much was available, the estimate is likely to have high deviation from the actual weight.

The weight was estimated at 480kg but with a ±60kg deviation so 420-540kg according to the the paper below. Check it out.[/color]

https://sci.bban.top/pdf/10.1080/08912963.2019.1625348.pdf?download=true Reply