I appreciate the effort but I think you made a few mistakes.
1) You can't compute the average newton's of impact without knowing the length of time it took to decelerate [1]
2) The average Newtons of force during the impact is meaningless in this context. That missle won't completely stop until it reaches the river bottom (and even reaching terminal v in water will take many seconds) so it's going to be quite a low average because the divisor is so large.
3) You don't know the height of water spray is a function of momentum or energy or the maximum force during deceleration or something else.
4) If the height of water spray is a function of the total energy (plausible since I know at high speeds crater size is a function of energy) it should generate 16 times the spray of a similarly weighted diver entering the water at 1/4th the speed. I can't eyeball that to tell if it's within that range.
5) If instead the water height is a function of the momentum transfered over some critical time period you start having to solve some nasty diffeqs.
1: I suspect the calculator you used is assuming a car impacting an immovable object and using facts about the car crumple zone to compute the average force. Whatever it assumes won't be valid for a streamlined missle penetrating water.
I was pretty sleepy when I recorded that, but that's why I show all of my work. In hindsight, that does seem a bit high, but at the time I think I attributed it to parallax error. This is why I show all of my work.
I agree with you Peter that Ryan's calculation goes off the rails at the point where he used the impact force calculator. I would approach this by using something like Newton's impact depth formula: https://en.wikipedia.org/wiki/Impact_depth
I don't think that's going to help that much because once you get deep enough nothing that happens is going to kick up water on the surface.
Imagine you had an rod that was a mile long, it might splash really high on initial impact but after that it's going to just keep sliding into the water relatively smoothly (once the head gets deep enough under water it won't affect the surface much). What kicks up the water is the initial shock to water near the surface and the water closing back around the tail of the object.
I think the only real option is either numerical simulation or to get some scale models and try it.
There's another way to get the height of the splash without using the kinematic equation, trigonometry (look, I didn't say it was a fun way). We can see from the video that the top of the splash is roughly even with the top of the 7th floor of the red-and-gray building, and the bottom of the splash is roughly even with the top of the 4th floor. The "apparent height" of the splash is about 3 floors of the red-and-gray building, which is 9.9 m. That is not the height of the splash itself, but it will come in handy in the next step.
Using the rule of similar triangles, we know the ratio of the distance between the viewer and the red-and-gray building and the distance between the viewer and the splash must be the same as the ratio between the apparent height of the splash (9.9 m) and the actual height of the splash, x (I really wish I could post a diagram in the chat. That would make this so much easier to explain).
I estimate the distance between the two buildings to be 163.6 m. I got that answer by using my ruler to measure the distances between the red-and gray building, the viewer, and the splash on my computer screen at 2:43 in Ryan's video. I measured a distance of 10.25" between the viewer and the splash and a distance of 4.875" between the viewer and the front-left corner of the red-and-grey building. Then I converted from inches to meters using Ryan's value of 344 m between the viewer and the splash.
The ratio of distance between the viewer and the red-and-gray building and the viewer and the splash is 163.6 m / 344 m which should be equal to the apparent height, 9.9 m, divided by the actual height, x. Solving for x gives an actual height of 20.8 m.
I was only able to reach that answer thanks to the work you did. Thank you for showing your work. I really appreciate what you do to fight misinformation.
I think you have an error in your calculation on how high the water flew: if it were 58 meters, it would be above the level of the camera, which you calculated as around 38 meters.
The Kinzhal is 7.2 meters long, if intact. This would be approximately 2 floors, using the estimate of 3.3 meters per floor. The IRIS-T is 2.94 meters long, again intact, or almost 1 floor.
I’m just a retired EOD guy, but using freeze frame, how tall is the missile compared to the reference building across the street? We could also compare the width of the missile compared to one of the windows.
I’m not going to do math. We calculated explosives needed to destroy munitions using the rule of thumb of “make it green”. Anyone that has ever used C-4 (angry silly putty) will get the formula.
To me, as a simple moron, the item seems to be about 1 story tall, which excludes the Kinzhal. It does appear to have a greater diameter than the IRIS-T, comparing it to the windows of the building. Another option is that it is a Kinzhal that was intercepted and did not receive a catastrophic strike, instead, breaking up. The firing system may have been damaged which prevented the detonation upon hitting the water (how did we forget that?)… or it was the aft section/ propulsion system of the Kinzhal after said interception and break up.
Nit: You were all sciencey, but then you talked about the diver weighing kilograms. Kilograms is a unit of mass, not weight. Weight is the force of gravity, or to bring Newton into this, F = MA (2nd law) or weight of diver = diver's mass * 9.81 m/s^2 (acceleration of gravity at earth's surface).
If you try to calculate the splash height from the buildings (distance to the building infront is 145meters, splash went to the floor of the 7th floor which would be 23.1meters) gives you a height for the water spray of 2.94meters.
I tried using trigonometry for this too, but I got a height of 20.8 m. Can you please show your calculations for 2.94 m so we can compare? My math is in a post above.
I appreciate the effort but I think you made a few mistakes.
1) You can't compute the average newton's of impact without knowing the length of time it took to decelerate [1]
2) The average Newtons of force during the impact is meaningless in this context. That missle won't completely stop until it reaches the river bottom (and even reaching terminal v in water will take many seconds) so it's going to be quite a low average because the divisor is so large.
3) You don't know the height of water spray is a function of momentum or energy or the maximum force during deceleration or something else.
4) If the height of water spray is a function of the total energy (plausible since I know at high speeds crater size is a function of energy) it should generate 16 times the spray of a similarly weighted diver entering the water at 1/4th the speed. I can't eyeball that to tell if it's within that range.
5) If instead the water height is a function of the momentum transfered over some critical time period you start having to solve some nasty diffeqs.
1: I suspect the calculator you used is assuming a car impacting an immovable object and using facts about the car crumple zone to compute the average force. Whatever it assumes won't be valid for a streamlined missle penetrating water.
I was pretty sleepy when I recorded that, but that's why I show all of my work. In hindsight, that does seem a bit high, but at the time I think I attributed it to parallax error. This is why I show all of my work.
Looking on physics stack exchange shows this is a super complicated
But if we use total energy as an estimate an IRIS-T should spash 16x and the Russian missle 450x.
This looks closer to 16 but I don't think this is accurate enough to conclude anything.
I agree with you Peter that Ryan's calculation goes off the rails at the point where he used the impact force calculator. I would approach this by using something like Newton's impact depth formula: https://en.wikipedia.org/wiki/Impact_depth
I don't think that's going to help that much because once you get deep enough nothing that happens is going to kick up water on the surface.
Imagine you had an rod that was a mile long, it might splash really high on initial impact but after that it's going to just keep sliding into the water relatively smoothly (once the head gets deep enough under water it won't affect the surface much). What kicks up the water is the initial shock to water near the surface and the water closing back around the tail of the object.
I think the only real option is either numerical simulation or to get some scale models and try it.
There's another way to get the height of the splash without using the kinematic equation, trigonometry (look, I didn't say it was a fun way). We can see from the video that the top of the splash is roughly even with the top of the 7th floor of the red-and-gray building, and the bottom of the splash is roughly even with the top of the 4th floor. The "apparent height" of the splash is about 3 floors of the red-and-gray building, which is 9.9 m. That is not the height of the splash itself, but it will come in handy in the next step.
Using the rule of similar triangles, we know the ratio of the distance between the viewer and the red-and-gray building and the distance between the viewer and the splash must be the same as the ratio between the apparent height of the splash (9.9 m) and the actual height of the splash, x (I really wish I could post a diagram in the chat. That would make this so much easier to explain).
I estimate the distance between the two buildings to be 163.6 m. I got that answer by using my ruler to measure the distances between the red-and gray building, the viewer, and the splash on my computer screen at 2:43 in Ryan's video. I measured a distance of 10.25" between the viewer and the splash and a distance of 4.875" between the viewer and the front-left corner of the red-and-grey building. Then I converted from inches to meters using Ryan's value of 344 m between the viewer and the splash.
The ratio of distance between the viewer and the red-and-gray building and the viewer and the splash is 163.6 m / 344 m which should be equal to the apparent height, 9.9 m, divided by the actual height, x. Solving for x gives an actual height of 20.8 m.
This could be a better answer
I was only able to reach that answer thanks to the work you did. Thank you for showing your work. I really appreciate what you do to fight misinformation.
I think you have an error in your calculation on how high the water flew: if it were 58 meters, it would be above the level of the camera, which you calculated as around 38 meters.
He might have meant 58 feet or around 17,5 meters ?
I'd be interested in your take on the corrections proposed by readers.
I’m on a plane right now, but I think that some of them have merit.
The guy has totally moved out of his house by now 🤣
Not to try to out nerd all the nerds, but…
The Kinzhal is 7.2 meters long, if intact. This would be approximately 2 floors, using the estimate of 3.3 meters per floor. The IRIS-T is 2.94 meters long, again intact, or almost 1 floor.
I’m just a retired EOD guy, but using freeze frame, how tall is the missile compared to the reference building across the street? We could also compare the width of the missile compared to one of the windows.
I’m not going to do math. We calculated explosives needed to destroy munitions using the rule of thumb of “make it green”. Anyone that has ever used C-4 (angry silly putty) will get the formula.
To me, as a simple moron, the item seems to be about 1 story tall, which excludes the Kinzhal. It does appear to have a greater diameter than the IRIS-T, comparing it to the windows of the building. Another option is that it is a Kinzhal that was intercepted and did not receive a catastrophic strike, instead, breaking up. The firing system may have been damaged which prevented the detonation upon hitting the water (how did we forget that?)… or it was the aft section/ propulsion system of the Kinzhal after said interception and break up.
Ryan, you’re right! Science is fun.
Nit: You were all sciencey, but then you talked about the diver weighing kilograms. Kilograms is a unit of mass, not weight. Weight is the force of gravity, or to bring Newton into this, F = MA (2nd law) or weight of diver = diver's mass * 9.81 m/s^2 (acceleration of gravity at earth's surface).
This kind of math makes my head hurt. Glad you’re here to do it for me.
If you try to calculate the splash height from the buildings (distance to the building infront is 145meters, splash went to the floor of the 7th floor which would be 23.1meters) gives you a height for the water spray of 2.94meters.
I tried using trigonometry for this too, but I got a height of 20.8 m. Can you please show your calculations for 2.94 m so we can compare? My math is in a post above.
I just scretched it in CAD, i made a mistake with the "," (whoops) My actual result is 29.4meters )
https://cdn.discordapp.com/attachments/838512433035083796/1192131825225703494/image.png?ex=65a7f647&is=65958147&hm=e370b97e61bb76f72ea0810c13f6e525d20d9692d94164d83081d1c9be60f543&
I messured the distance on google earth from roof to roof which gave me 145m, and 163.6m from 7 floors *3.3m , rest are the mesurements from ryans
I feel like 20~30m is close enough that im guessing it is just tolerances
Yeah, I agree 20-30 m is a good estimate. Thanks for the reply.