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March 15

COPD 2

What is COPD 2? I've seen some references to it but no clear explanation of what it is, and the COPD article says nothing at all. There are several types of COPD (emphasyma, bronchitis etc.) but they don't seem to have numbers connected with them. There are also severity levels 1 thru 4, but no indication that it refers to that either. Is this standard terminology? Thanks. 2601:648:8202:96B0:54D9:2ABB:1EDB:CEE3 (talk) 01:41, 15 March 2020 (UTC)[reply]

Where did you see it? Some quick searching suggests that there are "2" main kinds of COPD, but it's not like they're numbered 1 and 2. Without further info, my guess would be that it's just short for COPD Stage 2 (i.e. "moderate" per this). Matt Deres (talk) 01:52, 15 March 2020 (UTC)[reply]

A friend of mine has it and was hospitalized for it, and then I did a web search and didn't find much. 2601:648:8202:96B0:54D9:2ABB:1EDB:CEE3 (talk) 03:00, 15 March 2020 (UTC)[reply]

Are you sure there wasn't some miscommunication and it's not actually the pandemic virus going around? The official name for the disease is COVID-19, but the virus itself has been named SARS-CoV-2, which could get clipped to "CoV-2". Otherwise, yeah, I'd go with the above respondent that it might be a COPD exacerbation, which can require hospitalization. --47.146.63.87 (talk) 06:32, 15 March 2020 (UTC)[reply]

Thanks, COPD exacerbation sounds like it exactly it. The idea of a sudden attack of something chronic otherwise seemed paradoxical. She has been out of the hospital for a few weeks, so if she had a covid-19 infection I think it would have incubated by now. That makes me feel relieved that she probably doesn't have it. She said she was not tested for it, but is staying at home. 2601:648:8202:96B0:386A:A40C:EBB1:ACC0 (talk) 08:05, 15 March 2020 (UTC)[reply]

How does COVID-19 kill?

What exactly does COVID-19 do to a person's body that causes them to die? Freeknowledgecreator (talk) 02:47, 15 March 2020 (UTC)[reply]

Basically blocks and damages the lungs til the person suffocates, thus the treatments with medical ventilators and ECMO. 2601:648:8202:96B0:54D9:2ABB:1EDB:CEE3 (talk) 02:59, 15 March 2020 (UTC)[reply]

This BBC website article from yesterday (14 March) is to the point, if fairly basic. {The poster formerly known as 87.81.230.195} 90.197.27.39 (talk) 07:39, 15 March 2020 (UTC)[reply]

COVID-19: typically death results from pneumonia, ARDS, and/or septic shock. It's likely in some cases coinfections like bacterial pneumonia contribute—this often occurs in influenza deaths—but we can't definitely say right now because of course everyone is running around trying to get a handle on things. Once things are more under control people will start reviewing case reports. Most deaths have occurred in people with preexisting conditions like hypertension. --47.146.63.87 (talk) 08:59, 15 March 2020 (UTC)[reply]

It's mostly collateral damage due to the body's own immune system. In older people the immune response tends to react more with an inflammatory response which then can cause damage to the lungs. This lung damage and the extra strain on the heart that now needs to pump more blood can then lead to acute respiratory distress syndrome (ARDS). A significant fraction of people over 80 already have heart failure to begin with. Even many "healthy" 80-year-olds who have not been diagnosed with heart failure, do have heart failure to some degree. They are at risk to get acute heart failure when the heart would be strained without any lung damage. In that case fluids builds up in the body, also in the lungs which then leads to ARDS without any lung damage. Also, people older than 80 often have impaired kidney function. If (additional) fluid builds up in the lungs due to heart failure, diuretics to remove all that excess fluids won't work well. And dialysis can at best only replace about 20% of a healthy kidney function. Count Iblis (talk) 21:13, 15 March 2020 (UTC)[reply]

There has been significant discussion/suggestion that the virus attacks heart muscle tissue as well. A number of patients have died of myocarditis after apparently weathering the respiratory storm, so to speak. NorthBySouthBaranof (talk) 21:43, 15 March 2020 (UTC)[reply]

G-force in a head-on collision

Suppose a 1-ton automobile runs head-on into a 2-ton automobile. Can you calculate the G-forces of people in each auto? It is an elastic collision, so the conservation of momentum doesn't apply. Does it depend on the details of the cars? Bubba73 ^{You talkin' to me?} 20:26, 15 March 2020 (UTC)[reply]

The G-forces on the people is independent from the relative mass of the cars involved. It depends on the time vs. distance for the people to go from speed-on-impact to zero -- which might differ for each vehicle, depending on a variety of factors (including mass). 107.15.157.44 (talk) 21:01, 15 March 2020 (UTC)[reply]

Indeed, this is the whole purpose of crumple zones. Someguy1221 (talk) 21:52, 15 March 2020 (UTC)[reply]

The G-loading cannot be calculated for any collision, using the basic principles of physics. (Contrary to the original question, momentum IS conserved in any collision but that doesn’t help with calculating the G-loading.) The reason is that the distance involved in decelerating the bodies varies from one body to the next. In a vehicle collision this distance is closely related to the crumple zone of each vehicle. Dolphin (t) 21:59, 15 March 2020 (UTC)[reply]

Some of the energy goes into deforming the automobiles. Bubba73 ^{You talkin' to me?} 23:54, 15 March 2020 (UTC)[reply]

It is not an elastic collission. Momentum is conserved. No you can't calculate the G forces on the occupants, that's what supercomputers are for. The accelerations in general in the 1 tonne car will be around twice those in the two tonne car, independent of the distribution of structural properties of the two cars, because of Newton's laws F1=-F2 and F=ma.Greglocock (talk) 00:21, 16 March 2020 (UTC)[reply]

To make it a little bit clearer: Momentum is a vector property (it has size and direction). The total momentum is the (vector) sum of the two individual momentums of the cars, both before and after the crash. Momentum is always conserved in a closed system. See Momentum#Conservation. What is not conserved in an inelastic collision is kinetic energy (which is a scalar property, not a vector property. --Stephan Schulz (talk) 02:13, 16 March 2020 (UTC)[reply]

If the vehicles involved are traveling with a speed of one inch a year, the g-force will be low. Supposing the movement of the passengers is suitably constrained (safety belt; air bag) there are three things that can be used for a back-of-the-envelope calculation, assuming a constant deceleration from the moment of contact to a complete halt: (1) the speed v of the vehicle; (2) the time t it takes to come to a halt; (3) the distance s traveled by the passenger from the moment of collision to a halt. Any two out of these three will suffice, since they are connected by the relationship s = ¹/₂vt. {See Kinematics#Particle trajectories under constant acceleration.) Then the experienced deceleration a is given by a = v/t = ¹/₂v²/s = 2s/t². Since the passengers are also subject to the force of gravity g, the combined force is the square root of a² + g². To convert to the correct magnitude when using g-force as the unit, divide by g. --Lambiam 04:06, 16 March 2020 (UTC)[reply]

Of course it varies based on vehicle and crash conditions, but realistic deceleration curves tend to be rather pointy. From the examples here, especially for what looks like car vs. concrete wall, peak head acceleration is about double the average value.[1] Someguy1221 (talk) 05:49, 16 March 2020 (UTC)[reply]

One bit of technical jargon that will help is the phrase "crash pulse." This term refers to the waveform that describes the velocity and/or the acceleration during impact.

If you use this phrase to search the web archives of vehicle safety research at a site like the National Highway Traffic Safety Administration, you'll find tons of research on physics-based computational models, and tons of real-world data from tests and simulations.

The short answer, as has been described above, is that you can't use simple physics to describe the pulse - because the actual duration, intensity, and shape all depend on the complicated details of the materials and structures that are involved in the collision. Even for a simple case - something much less complicated than a modern motor vehicle - the reality is that an inelastic collision is governed by material deformation - and the math and physics that govern material deformation constitute one of the most complicated areas of computational physics.

Nimur (talk) 17:40, 17 March 2020 (UTC)[reply]

White scar removal question

We are not supposed to answer questions calling for medical advice. Talk to a doctor.

The following discussion has been closed. Please do not modify it.

Is there any way to permanently get rid of a decade-old white scar that is about 1.0-1.5 inches long and 0.05-0.10 inches wide? For instance, with some kind of laser? Futurist110 (talk) 21:41, 15 March 2020 (UTC)[reply]

Wait another 10 years? Or go to a really good plastic surgeon. Greglocock (talk) 00:22, 16 March 2020 (UTC)[reply]

First talk to your family doctor, and see who they can refer you to - maybe a dermatologist. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:24, 16 March 2020 (UTC)[reply]