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Smilodon populator

Guatemala GuateGojira Offline
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(05-17-2022, 10:31 PM)LonePredator Wrote: Yes, the 517kg weight must be treated with extreme caution because estimating the weight of Smilodon Populator is no easy task, we have no living animal which has the same or similar proportions as Smilodon Populators unlike in the case of prehistoric Lions and Tigers which are much more easier to estimate as animals with similar or almost same proportions as them still exist.

Now for this reason, the method used would make or break the whole estimation. And since the method of Christiansen and Harris (2005) is considered the most reliable, we could have used their 358.4kg specimen as surrogate to make a rough isometric estimation for these specimens but the problem is that the 358.4kg specimen was based on the humerus so isometric scaling can only be done on another humerus.

Unfortunately we only have skulls on our hands and it would not be possible to estimate their weight unless we get our hands on the original method of Christiansen and Harris and see how it works and how it could apply in this case if it really can.

I will like to read the study of Prevosti & Martin (2013) just to see what methods they used, because since I see "Anyonge" mentioned there, it smells at "exageration". But you are right, there is no modern animal with the same morphology, as Smilodon genus is like a cat that wanted to be a bear!


The issue with Christiansen & Harris (2005) is that they use a multifactorial aproach, as they apply several measurements in the same bone to get a "weighted" average. So, in order to apply this set of formulas to a new specimen, we will need to have a good set of measurements, if not, a single formula produce incorrect data. Check that in the set of measurements the "Total length" alone produce a very low value, so is not recomended to use a single measurement, with them at least.

Other issue, remember that we are using a single bone, but when a series of bones from the same specimen are used, the range of estimated weights is very large. For example the specimen CN52 has humerus/ulna/femur/tibia asociated, but check the average body mass "weighted", of the set of measurements, of each bone:

Humerus: 316.2 kg
Ulna: 231.2 kg
Femur: 246.8 kg
Tibia: 238.5 kg

As we can see, there is a big variation between the body masses of each bone, and also add the fact that the morphology is different so that may explain why the legs provide lower weights. The average of the 4 bone is 258.2 kg, so in theory that is the body mass taking in count all the bones, but as I said, the small size of the legs may affect this estimation.
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Guatemala GuateGojira Offline
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(05-18-2022, 02:20 AM)jrocks Wrote: i think when i was looking for the sources for the comment guate asked me on earlier, there was some study that said that the christiansen and harris 2005 formula did something that underestimated the smilodon specimens they estimated i think because of the bone width although im not sure, if i find the link to that study il edit it in here

Actually they did took in count the bone width in they study, the thing is that in order to use the formulas we need to be carefull, as they do not provide a straight value, but a series of processes. For example they separate the formulas "over all" and the formulas "excluding the lynx", as the value may change because of the lynx morphology. Second, we need to differentiate between the "unweighted" and the "weighted" values, and sadly many of the figures quoted in recent years, where the authors are "apparently" using they formulas, ignore that part.

Finaly, like I said to LonePredator, Christiansen & Harris (2005) propose a series of formulas for a set of measurements and latter, they average those figures to get the final weight of that particular specimen. Using only the "lenght" or the "wide" alone it produce incorrect values. For example, if we use the "total length" value "alone" for the largest femur of Panthera atrox and the Ngadong tiger, they produce incredible low weights of only about 270 - 295 kg, which is certainly incorrect.

Using Christiansen & Harris (2005) formulas is not easy, so we need to be carefull to use it and if an author says that they use them, check what formulas or methods they use, and if they include the full set of measurements, which probably is not the case.
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jrocks Offline
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( This post was last modified: 05-18-2022, 03:51 AM by jrocks )

(05-18-2022, 02:43 AM)GuateGojira Wrote:
(05-18-2022, 02:20 AM)jrocks Wrote: i think when i was looking for the sources for the comment guate asked me on earlier, there was some study that said that the christiansen and harris 2005 formula did something that underestimated the smilodon specimens they estimated i think because of the bone width although im not sure, if i find the link to that study il edit it in here

Actually they did took in count the bone width in they study, the thing is that in order to use the formulas we need to be carefull, as they do not provide a straight value, but a series of processes. For example they separate the formulas "over all" and the formulas "excluding the lynx", as the value may change because of the lynx morphology. Second, we need to differentiate between the "unweighted" and the "weighted" values, and sadly many of the figures quoted in recent years, where the authors are "apparently" using they formulas, ignore that part.

Finaly, like I said to LonePredator, Christiansen & Harris (2005) propose a series of formulas for a set of measurements and latter, they average those figures to get the final weight of that particular specimen. Using only the "lenght" or the "wide" alone it produce incorrect values. For example, if we use the "total length" value "alone" for the largest femur of Panthera atrox and the Ngadong tiger, they produce incredible low weights of only about 270 - 295 kg, which is certainly incorrect.

Using Christiansen & Harris (2005) formulas is not easy, so we need to be carefull to use it and if an author says that they use them, check what formulas or methods they use, and if they include the full set of measurements, which probably is not the case.
found it were i saw it from, Sherani mentions it on page 15, for the 408.4 mm specimen does anyone know how much it weighed if the christiansen and harris formula is used on it, or it hasnt been done yet 

https://peerj.com/preprints/2327.pdf
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Guatemala GuateGojira Offline
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(05-18-2022, 03:05 AM)jrocks Wrote: found it Sherani mentions it on page 15, for the 408.4 mm specimen does anyone know how much it weighed if the christiansen and harris formula is used on it, or it hasnt been done yet 

https://peerj.com/preprints/2327.pdf

Thank you for redirecting me to that part of the document. I need to read it again.

About the large specimen of S. populator, the problem is that Christiansen & Harris (2005) use formulas for long bones, not for skull. The only one that use skull size, specifically Condylobasal length, is Van Valkenburg (1990) and while the formula is interesting, it has the problem that she used litterature values to create it, not real weights, and also that the formula ignore the variations between species: For example if we use that formula to check the weight of lions and tigers, the tigers will be underestimated as the formula will assume that they have the same body proportions with lions, which is not correct (tigers had proportionally smaller heads in relation with the body than lions).

So, while this formula has been used with the biggest skulls available of Smilodon populator and Machairodus horribilis, we must use those values carefully, as we don't know the proportions of those species and certainly will have the same issue as with the tiger-lion heads example.

In other paper, Christiansen & Harris (2009) use an isometrical approach and use a database of known animals with they own real weight to get the body mass of Panthera atrox, but the problem here is that while we can use this approach with animals of similar body morphology, Smilodon do not have such a thing and if we use the Panthera morphology we can underestimate (or overestimate, honestly we don't know....) the body mass of Smilodon specimens.

Body mass is a tricky business, and hard to found a reliable conclution or approach.
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jrocks Offline
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( This post was last modified: 05-18-2022, 04:27 AM by jrocks )

(05-18-2022, 04:23 AM)GuateGojira Wrote:
(05-18-2022, 03:05 AM)jrocks Wrote: found it Sherani mentions it on page 15, for the 408.4 mm specimen does anyone know how much it weighed if the christiansen and harris formula is used on it, or it hasnt been done yet 

https://peerj.com/preprints/2327.pdf

Thank you for redirecting me to that part of the document. I need to read it again.

About the large specimen of S. populator, the problem is that Christiansen & Harris (2005) use formulas for long bones, not for skull. The only one that use skull size, specifically Condylobasal length, is Van Valkenburg (1990) and while the formula is interesting, it has the problem that she used litterature values to create it, not real weights, and also that the formula ignore the variations between species: For example if we use that formula to check the weight of lions and tigers, the tigers will be underestimated as the formula will assume that they have the same body proportions with lions, which is not correct (tigers had proportionally smaller heads in relation with the body than lions).

So, while this formula has been used with the biggest skulls available of Smilodon populator and Machairodus horribilis, we must use those values carefully, as we don't know the proportions of those species and certainly will have the same issue as with the tiger-lion heads example.

In other paper, Christiansen & Harris (2009) use an isometrical approach and use a database of known animals with they own real weight to get the body mass of Panthera atrox, but the problem here is that while we can use this approach with animals of similar body morphology, Smilodon do not have such a thing and if we use the Panthera morphology we can underestimate (or overestimate, honestly we don't know....) the body mass of Smilodon specimens.

Body mass is a tricky business, and hard to found a reliable conclution or approach.
oh ok thanks
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India LonePredator Offline
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( This post was last modified: 05-18-2022, 06:32 AM by LonePredator )

(05-18-2022, 02:33 AM)GuateGojira Wrote:
(05-17-2022, 10:31 PM)LonePredator Wrote: Yes, the 517kg weight must be treated with extreme caution because estimating the weight of Smilodon Populator is no easy task, we have no living animal which has the same or similar proportions as Smilodon Populators unlike in the case of prehistoric Lions and Tigers which are much more easier to estimate as animals with similar or almost same proportions as them still exist.

Now for this reason, the method used would make or break the whole estimation. And since the method of Christiansen and Harris (2005) is considered the most reliable, we could have used their 358.4kg specimen as surrogate to make a rough isometric estimation for these specimens but the problem is that the 358.4kg specimen was based on the humerus so isometric scaling can only be done on another humerus.

Unfortunately we only have skulls on our hands and it would not be possible to estimate their weight unless we get our hands on the original method of Christiansen and Harris and see how it works and how it could apply in this case if it really can.

I will like to read the study of Prevosti & Martin (2013) just to see what methods they used, because since I see "Anyonge" mentioned there, it smells at "exageration". But you are right, there is no modern animal with the same morphology, as Smilodon genus is like a cat that wanted to be a bear!


The issue with Christiansen & Harris (2005) is that they use a multifactorial aproach, as they apply several measurements in the same bone to get a "weighted" average. So, in order to apply this set of formulas to a new specimen, we will need to have a good set of measurements, if not, a single formula produce incorrect data. Check that in the set of measurements the "Total length" alone produce a very low value, so is not recomended to use a single measurement, with them at least.

Other issue, remember that we are using a single bone, but when a series of bones from the same specimen are used, the range of estimated weights is very large. For example the specimen CN52 has humerus/ulna/femur/tibia asociated, but check the average body mass "weighted", of the set of measurements, of each bone:

Humerus: 316.2 kg
Ulna: 231.2 kg
Femur: 246.8 kg
Tibia: 238.5 kg

As we can see, there is a big variation between the body masses of each bone, and also add the fact that the morphology is different so that may explain why the legs provide lower weights. The average of the 4 bone is 258.2 kg, so in theory that is the body mass taking in count all the bones, but as I said, the small size of the legs may affect this estimation.

Yes, I totally agree, any study that is based on or inspired by the Anyonge one needs to be looked at with skepticism.

And yes I just saw a glimpse of their method and I’ll read more of it later but I did not mean to say that I’ll redo the whole thing they did on this case, I meant to say that as they made different measurements of multiple parts of the same bone and then they calculated the weight possible for the specimen which that bone belonged to.

So what I meant to say is that as Christiansen and Harris had already estimated the possible weight for the specimen with a humerus of particular length has already been done (and they have already fixed the disparities by measuring different parts of the same bone)

So now I meant that as they have already estimated the weight of a Smilodon Populator which has a humerus of 390mm so now we can get a rough idea how much the Populator of 405mm will weight through isometry as all the other calculations have already been done.

And yes, this is a huge issue, especially the femur of the Populator would be very likely to produce an underestimation when estimated from extant specimens.

But overall I like the method of Christiansen and Harris, they calculated all the variations in the bone of Populator compared to a specimen from living species. Their calculation of different measurements of the same bone is arguably the best way to tell how much weight the current bone would produce with all those different variations in that same bone.
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Guatemala GuateGojira Offline
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(05-18-2022, 06:26 AM)LonePredator Wrote: So now I meant that as they have already estimated the weight of a Smilodon Populator which has a humerus of 390mm so now we can get a rough idea how much the Populator of 405mm will weight through isometry as all the other calculations have already been done.

Ohhh, I get it now. Yes, I think that we can make an estimation via isometry. In fact, you should use all the estimations of weight based in the humerus so your database will be larger.

Good idea, by the way. Like
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(05-18-2022, 10:29 PM)GuateGojira Wrote:
(05-18-2022, 06:26 AM)LonePredator Wrote: So now I meant that as they have already estimated the weight of a Smilodon Populator which has a humerus of 390mm so now we can get a rough idea how much the Populator of 405mm will weight through isometry as all the other calculations have already been done.

Ohhh, I get it now. Yes, I think that we can make an estimation via isometry. In fact, you should use all the estimations of weight based in the humerus so your database will be larger.

Good idea, by the way. Like

Thanks and I actually really wanted to get an idea of the weight of the supposedly giant specimen which you mentioned earlier but sadly only his skull is available
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jrocks Offline
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(05-18-2022, 10:29 PM)GuateGojira Wrote:
(05-18-2022, 06:26 AM)LonePredator Wrote: So now I meant that as they have already estimated the weight of a Smilodon Populator which has a humerus of 390mm so now we can get a rough idea how much the Populator of 405mm will weight through isometry as all the other calculations have already been done.

Ohhh, I get it now. Yes, I think that we can make an estimation via isometry. In fact, you should use all the estimations of weight based in the humerus so your database will be larger.

Good idea, by the way. Like

Hi Guate, i was just curious about if the 16.07 inch skull were to have its weight calculated the same way that 436 kg was calculated for 15.4 inch skull, how much would it have weighed

jrocks
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Guatemala GuateGojira Offline
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(05-28-2022, 04:09 AM)jrocks Wrote: Hi Guate, i was just curious about if the 16.07 inch skull were to have its weight calculated the same way that 436 kg was calculated for 15.4 inch skull, how much would it have weighed

jrocks

Apart from all the issues with that formula, the main problem is that the formula of Van Valkenburgn (1990) use the Condylobasal length, not the Greatest length (an error that many people made when they use it), and as you already probably know, we don't know the CBL of the giant skull. We can speculate based in the known specimens, but it will be a calculation over an asumption. 

This is the formula, by the way, using only felids: log mass = 3.11 * log CBL - 5.38
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( This post was last modified: 05-28-2022, 06:57 AM by LonePredator )

(05-28-2022, 06:43 AM)GuateGojira Wrote:
(05-28-2022, 04:09 AM)jrocks Wrote: Hi Guate, i was just curious about if the 16.07 inch skull were to have its weight calculated the same way that 436 kg was calculated for 15.4 inch skull, how much would it have weighed

jrocks

Apart from all the issues with that formula, the main problem is that the formula of Van Valkenburgn (1990) use the Condylobasal length, not the Greatest length (an error that many people made when they use it), and as you already probably know, we don't know the CBL of the giant skull. We can speculate based in the known specimens, but it will be a calculation over an asumption. 

This is the formula, by the way, using only felids: log mass = 3.11 * log CBL - 5.38

Do you mean to say that the 15.4 inch was the CBL while the 16.04 inch was the GSL?

And can you please tell me the source of the CBL regression equation which you mentioned?
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jrocks Offline
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(05-28-2022, 06:56 AM)LonePredator Wrote:
(05-28-2022, 06:43 AM)GuateGojira Wrote:
(05-28-2022, 04:09 AM)jrocks Wrote: Hi Guate, i was just curious about if the 16.07 inch skull were to have its weight calculated the same way that 436 kg was calculated for 15.4 inch skull, how much would it have weighed

jrocks

Apart from all the issues with that formula, the main problem is that the formula of Van Valkenburgn (1990) use the Condylobasal length, not the Greatest length (an error that many people made when they use it), and as you already probably know, we don't know the CBL of the giant skull. We can speculate based in the known specimens, but it will be a calculation over an asumption. 

This is the formula, by the way, using only felids: log mass = 3.11 * log CBL - 5.38

Do you mean to say that the 15.4 inch was the CBL while the 16.04 inch was the GSL?

And can you please tell me the source of the CBL regression equation which you mentioned?

no they are 2 separate skulls, one is the one found uruguay which had a GSL of 15.4 inches and its CBL was 14.9 inches, the other is called smilodon bonaeresis 10-1 and its GSL is 16.07 inches, however its CBL isnt known
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( This post was last modified: 05-29-2022, 09:44 PM by LonePredator )

(05-28-2022, 09:34 AM)jrocks Wrote:
(05-28-2022, 06:56 AM)LonePredator Wrote:
(05-28-2022, 06:43 AM)GuateGojira Wrote:
(05-28-2022, 04:09 AM)jrocks Wrote: Hi Guate, i was just curious about if the 16.07 inch skull were to have its weight calculated the same way that 436 kg was calculated for 15.4 inch skull, how much would it have weighed

jrocks

Apart from all the issues with that formula, the main problem is that the formula of Van Valkenburgn (1990) use the Condylobasal length, not the Greatest length (an error that many people made when they use it), and as you already probably know, we don't know the CBL of the giant skull. We can speculate based in the known specimens, but it will be a calculation over an asumption. 

This is the formula, by the way, using only felids: log mass = 3.11 * log CBL - 5.38

Do you mean to say that the 15.4 inch was the CBL while the 16.04 inch was the GSL?

And can you please tell me the source of the CBL regression equation which you mentioned?

no they are 2 separate skulls, one is the one found uruguay which had a GSL of 15.4 inches and its CBL was 14.9 inches, the other is called smilodon bonaeresis 10-1 and its GSL is 16.07 inches, however its CBL isnt known

Okay, then you can use isometry but the result will be unreliable because of using only a single isometric calculation.

Nevertheless, if you isometrically scale a Populator specimen with a skull of 15.4 inches which weighs 436kg then a specimen with a skull of 16.04 inches would be 495kg assuming isometry within both specimens.

BUT this result is *not* reliable because we just used a single calculation and making calculation using only a single dimension or single calculation often gives wrong weight estimates.

One more reason this is unreliable is because as @GuateGojira stated, the 436kg estimate itself may be inaccurate.

If you can find or derive a regression equation between weights and GSL then you may be able to get an actually reliable estimate (possibly an equation specifically made for Smilodon Populators)

You can use the same relation between GSL and CBL as you have in the smaller skull and then use that to estimate the weight from Guate’s formula but like Guate said, that would be all guesswork and that will also become unreliable in that sense.
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( This post was last modified: 05-29-2022, 02:36 AM by jrocks )

(05-28-2022, 09:44 AM)LonePredator Wrote:
(05-28-2022, 09:34 AM)jrocks Wrote:
(05-28-2022, 06:56 AM)LonePredator Wrote:
(05-28-2022, 06:43 AM)GuateGojira Wrote:
(05-28-2022, 04:09 AM)jrocks Wrote: Hi Guate, i was just curious about if the 16.07 inch skull were to have its weight calculated the same way that 436 kg was calculated for 15.4 inch skull, how much would it have weighed

jrocks

Apart from all the issues with that formula, the main problem is that the formula of Van Valkenburgn (1990) use the Condylobasal length, not the Greatest length (an error that many people made when they use it), and as you already probably know, we don't know the CBL of the giant skull. We can speculate based in the known specimens, but it will be a calculation over an asumption. 

This is the formula, by the way, using only felids: log mass = 3.11 * log CBL - 5.38

Do you mean to say that the 15.4 inch was the CBL while the 16.04 inch was the GSL?

And can you please tell me the source of the CBL regression equation which you mentioned?

no they are 2 separate skulls, one is the one found uruguay which had a GSL of 15.4 inches and its CBL was 14.9 inches, the other is called smilodon bonaeresis 10-1 and its GSL is 16.07 inches, however its CBL isnt known

Okay, then you can use isometry but the result will be unreliable because of using only a single isometric calculation.

Nevertheless, if you isometrically scale a Populator specimen of 15.4 inches which weight 436kg then a specimen of 16.04 inches would be 495kg assuming isometry within both specimens.

BUT this result is *not* reliable  because we just used a single calculation and making calculation using only a single dimension or single calculation often gives wrong weight estimates.

One more reason this is unreliable is because as @GuateGojira stated, the 436kg estimate itself may be inaccurate.

If you can find a regression equation between weights and GSL then you may be able to get an actually reliable estimate (possibly an equation specifically made for Smilodon Populators)

You can use the same relation between GSL and CBL as you have in the smaller skull and then use that to estimate the weight from Guate’s formula but like Guate said, that would be all guesswork and that will also become unreliable in that sense.
dang with that isometry formula it can weigh a crazy amount, as for the 15.4 inch skull does guate think 436 kg is an overestimate or an underestimate, also is it known how much smilodon bonaerensis 46 weighs as its full skeleton is completely preserved
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(05-28-2022, 09:33 PM)jrocks Wrote:
(05-28-2022, 09:44 AM)LonePredator Wrote:
(05-28-2022, 09:34 AM)jrocks Wrote:
(05-28-2022, 06:56 AM)LonePredator Wrote:
(05-28-2022, 06:43 AM)GuateGojira Wrote:
(05-28-2022, 04:09 AM)jrocks Wrote: Hi Guate, i was just curious about if the 16.07 inch skull were to have its weight calculated the same way that 436 kg was calculated for 15.4 inch skull, how much would it have weighed

jrocks

Apart from all the issues with that formula, the main problem is that the formula of Van Valkenburgn (1990) use the Condylobasal length, not the Greatest length (an error that many people made when they use it), and as you already probably know, we don't know the CBL of the giant skull. We can speculate based in the known specimens, but it will be a calculation over an asumption. 

This is the formula, by the way, using only felids: log mass = 3.11 * log CBL - 5.38

Do you mean to say that the 15.4 inch was the CBL while the 16.04 inch was the GSL?

And can you please tell me the source of the CBL regression equation which you mentioned?

no they are 2 separate skulls, one is the one found uruguay which had a GSL of 15.4 inches and its CBL was 14.9 inches, the other is called smilodon bonaeresis 10-1 and its GSL is 16.07 inches, however its CBL isnt known

Okay, then you can use isometry but the result will be unreliable because of using only a single isometric calculation.

Nevertheless, if you isometrically scale a Populator specimen of 15.4 inches which weight 436kg then a specimen of 16.04 inches would be 495kg assuming isometry within both specimens.

BUT this result is *not* reliable  because we just used a single calculation and making calculation using only a single dimension or single calculation often gives wrong weight estimates.

One more reason this is unreliable is because as @GuateGojira stated, the 436kg estimate itself may be inaccurate.

If you can find a regression equation between weights and GSL then you may be able to get an actually reliable estimate (possibly an equation specifically made for Smilodon Populators)

You can use the same relation between GSL and CBL as you have in the smaller skull and then use that to estimate the weight from Guate’s formula but like Guate said, that would be all guesswork and that will also become unreliable in that sense.
dang with that isometry formula it can weigh a crazy amount, as for the 15.4 inch skull does guat think 436 kg is an overestimate or an underestimate, also is it known how much smilodon bonaerensis 10-1 weighs as its full skeleton is completely preserved

I’ll have to check to make sure but probably an overestimate if you estimate this one with the Christiansen & Harris estimate as reference.

And as far as I remember I think I had seen a weight estimate for that particular fully preserved specimen but I don’t remember an exact number so if I see it, I’ll tell
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