Post by: jeffreyH on 05/02/2014 10:05:05
Post by: RD on 05/02/2014 10:41:18
e.g. n sin(1/n)
Quote
n n sin(1/n)http://en.wikipedia.org/wiki/Limit_of_a_sequence
1 0.841471
2 0.958851
...
10 0.998334
...
100 0.999983
n=1000000 , n sin(1/n)=0.999999999999833
[ the formula uses sin (http://en.wikipedia.org/wiki/Sine) in radians (http://en.wikipedia.org/wiki/Radians) not degrees ]
Post by: CliffordK on 05/02/2014 11:03:22
Did you say your equation has to equal anything other than 1?
Y = 1x
One could also have an equation like:
Y = 1 + |x| - x
If you want to shift the point where it becomes 1, then you just add in another factor:
Y = 1 + |x-n| - (x-n)
Post by: jeffreyH on 05/02/2014 11:16:54
Post by: evan_au on 05/02/2014 11:17:20
Depending on the computer language, it could be as simple as "(n>N)", where:
- n is an integer variable (or it could be a floating-point variable)
- N is an integer constant
- This expression returns 0 if n<=N
- This expression returns 1 if n>N
Post by: jeffreyH on 05/02/2014 11:30:44
Post by: CliffordK on 05/02/2014 12:01:39
My absolute value equation above will give you a nice linear progression that you could use, but it would just make all of your equations unnecessarily complicated.
As far as your escape velocity calculation, I think there is a problem.
Say you start 1000 miles below the surface of the Earth. The gravity at the starting point is less than at the surface, but to actually escape from the planet, one has to travel up the hole to the surface, and continue through the atmosphere and into space. Thus, starting at the bottom of the hole, the escape velocity must be greater than it would be starting at the surface.
Post by: jeffreyH on 05/02/2014 12:26:46
I think Evan is right, some kind of if-then-else statement would be easiest.
My absolute value equation above will give you a nice linear progression that you could use, but it would just make all of your equations unnecessarily complicated.
As far as your escape velocity calculation, I think there is a problem.
Say you start 1000 miles below the surface of the Earth. The gravity at the starting point is less than at the surface, but to actually escape from the planet, one has to travel up the hole to the surface, and continue through the atmosphere and into space. Thus, starting at the bottom of the hole, the escape velocity must be greater than it would be starting at the surface.
If gravity cancels at the centre then movement upwards is easiest surely and increases in an exponential manner from there. You have to take into account the partial gravitational cancellations all the way up to the surface. This is still a work in progress and once I have the proper spherical terms the profile may look rather different so any conclusions will have to wait. I have no idea how to test this but if true it would show that gravitation has the same properties as some of the atomic mechanisms. Of course I may also be talking rubbish.
Post by: jeffreyH on 05/02/2014 12:38:30
Post by: jeffreyH on 05/02/2014 12:42:47
I think Evan is right, some kind of if-then-else statement would be easiest.
My absolute value equation above will give you a nice linear progression that you could use, but it would just make all of your equations unnecessarily complicated.
As far as your escape velocity calculation, I think there is a problem.
Say you start 1000 miles below the surface of the Earth. The gravity at the starting point is less than at the surface, but to actually escape from the planet, one has to travel up the hole to the surface, and continue through the atmosphere and into space. Thus, starting at the bottom of the hole, the escape velocity must be greater than it would be starting at the surface.
Sorry I see your point. This is an escape velocity gradient. This is not saying that if you start at point P within the mass there will be a constant escape velocity. Internally this is an increasing vector function requiring more input of energy the further out one travels. This makes it HARDER to escape from an internal gravity well and not easier.
Post by: jeffreyH on 05/02/2014 12:54:14
Post by: JP on 05/02/2014 13:55:25
http://en.wikipedia.org/wiki/Heaviside_step_function
Post by: jeffreyH on 05/02/2014 16:33:27
The function you're looking for is the Heaviside step function H(x). Or to get the effect you want, 1-H(x).
http://en.wikipedia.org/wiki/Heaviside_step_function
I may yet need this so thanks for the link.
Post by: jeffreyH on 05/02/2014 18:53:55
Post by: RD on 05/02/2014 19:46:59
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2Fforum%2Findex.php%3Faction%3Ddlattach%3Btopic%3D50254.0%3Battach%3D18456%3Bimage&hash=77ae57461797f134d0a3c6ad731b9693)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2F5%2F50%2FEarthGravityPREM.svg&hash=819e98d66ce97de5d4533b0262610941)
http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth (http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth)
Post by: jeffreyH on 05/02/2014 21:07:54
Snap ?
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.thenakedscientists.com%2Fforum%2Findex.php%3Faction%3Ddlattach%3Btopic%3D50254.0%3Battach%3D18456%3Bimage&hash=77ae57461797f134d0a3c6ad731b9693)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2F5%2F50%2FEarthGravityPREM.svg&hash=819e98d66ce97de5d4533b0262610941)
http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth (http://en.wikipedia.org/wiki/Gravity_of_Earth#Depth)
The plateau profile is more rounded on the wiki graph as I haven't added the spherical geometry calculations to mine. As the wiki graph relates to density then my original assertion that mass-energy density affects gravitational feedback must have some credence.
Post by: JP on 06/02/2014 00:07:37
Post by: jeffreyH on 06/02/2014 00:36:49
What do you mean by gravitational feedback? The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).
I am not saying they are not negligible.
Post by: jeffreyH on 06/02/2014 00:38:12
Post by: JP on 06/02/2014 01:09:53
What do you mean by gravitational feedback? The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).
I am not saying they are not negligible.
Ah, but they are. Unless you're planning on replacing general relativity with your own theory.
Post by: jeffreyH on 06/02/2014 02:25:13
What do you mean by gravitational feedback? The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).
I am not saying they are not negligible.
Ah, but they are. Unless you're planning on replacing general relativity with your own theory.
Now that would be silly.
Post by: JP on 06/02/2014 02:34:17
What do you mean by gravitational feedback? The effects due to nonlinearity of Einstein's field equations is going to be negligible (and is ignored in these graphs).
I am not saying they are not negligible.
Ah, but they are. Unless you're planning on replacing general relativity with your own theory.
Now that would be silly.
Indeed, and it would belong in the New Theories section of the forum. :)
Post by: jeffreyH on 06/02/2014 12:31:06
Post by: jeffreyH on 06/02/2014 17:16:19
Post by: jeffreyH on 06/02/2014 17:35:36
Post by: jeffreyH on 06/02/2014 18:23:14
Post by: jeffreyH on 06/02/2014 18:44:18
Post by: jeffreyH on 06/02/2014 19:01:31
Post by: CliffordK on 06/02/2014 20:45:20
A(x) = acceleration due to gravity at distance X from the center of the Earth.
for a starting point d from the center of the earth.If you have two distinct equations, then you can simply add them together.
+ 
Are you using some kind of a numerical approximation of the area below the curve?
I'm not sure about the magnitude, but the shape of your curves are looking a bit more like what I would expect.
Post by: JP on 06/02/2014 21:47:11
Woo Hoo and there is the Earth's Scwarzschild gravity well.
It's great that you've gotten excel plots, but what exactly are you plotting? How did you calculate this? What equations did you use?
Post by: jeffreyH on 06/02/2014 21:50:49
Post by: JP on 06/02/2014 21:52:45
Post by: jeffreyH on 07/02/2014 05:22:14
Ok, Jeffrey. But if you're going to be publishing plots of your own calculations without telling us what you're doing, could you please keep it to the New Theories section? This part of the forum is for open discussion of the science behind "mainstream" ideas in physics, such as Einstein's field equations or Newtonian gravity.
OK then I'll post them. I suppose here is as good as anywhere. We want to find Ve both internally and externally to a mass. So first we have Ve = f(M)/r. At the surface and beyond f(M) will simply return M as we have no internal gravitational cancellations to take into account. Internally we have f(M) = M*s/v where s/v will give us our effective mass once we have cancelled equivalent opposing field strengths. v is the total mass volume so s = (v-vs)/v. vs is our cancelled mass. First we get the volume behind the direction of travel = v0 = (pi/6)h(3c^2+h^2).
c = SQRT(h(2r - h)). vs then equals v - 2(v - v0). I can post the spreadsheet somewhere for any interested parties. This gives the earth's Schwarzschild gravity well at just under 1 cm which agrees with the rs calculation. This shows the internal gravitation for a non compressed mass to follow a logistical curve rather than exponential. The curve will then go exponential when the rs compression limit is reached.
The values for h and c can be found at this page.
http://keisan.casio.com/exec/system/1223382199
Post by: jeffreyH on 07/02/2014 06:47:39
Correction there are 8 microseconds difference.
Post by: evan_au on 07/02/2014 08:10:30
Assuming a spherically symmetric Earth, the gravitational force would be the attraction of a 10m radius sphere (possibly with a significant content of gold, osmium or other dense materials). You would be effectively weightless (if you weren't instantly crushed by thousands of kilometers of iron and rock).
This 10m radius sphere does not have enough density to form a black hole.
I think the graphed reduction of escape velocity near the center of the Earth is not [edit] a real effect - I think it might be some sort of underflow error.
Post by: jeffreyH on 07/02/2014 08:40:08
The latter graph seems to show the escape velocity decreasing when you reach 10m from the center of the Earth.
Assuming a spherically symmetric Earth, the gravitational force would be the attraction of a 10m radius sphere (possibly with a significant content of gold, osmium or other dense materials). You would be effectively weightless (if you weren't instantly crushed by thousands of kilometers of iron and rock).
This 10m radius sphere does not have enough density to form a black hole.
I think the graphed reduction of escape velocity near the center of the Earth is a real effect - I think it might be some sort of underflow error.
Hi Evan This is a gradient equation. What it says is at any point P you can find the velocity needed to escape the gravitational field. It shouldn't be viewed as a representation of a 10m mass. This is 10m from the centre of the earth. The well at the centre is not an error and matches the event horizon if and only if the mass of the earth were concentrated at that point which it isn't in this equation. This iss a view zoomed in to the core. The mass is at normal density.
Post by: jeffreyH on 07/02/2014 18:38:20
Post by: jeffreyH on 07/02/2014 18:43:49
Post by: jeffreyH on 07/02/2014 19:05:37
Post by: jeffreyH on 07/02/2014 19:48:05
Post by: jeffreyH on 07/02/2014 20:04:38
Post by: alancalverd on 07/02/2014 21:39:43
se= sqrt (2gr)
where g is the gravitational acceleration at r. At the center of a sphere both g and r are zero, and increase smoothly (if the sphere is homogeneous) to the surface. For the nonhomogeneities of the earth, see the plot of g versus r given in the Wikipedia reference, which takes account of the depth/density profile of the planet.
Post by: jeffreyH on 07/02/2014 22:48:48
Escape speed at a distance r from the center of mass of a sphere is
se= sqrt (2gr)
where g is the gravitational acceleration at r. At the center of a sphere both g and r are zero, and increase smoothly (if the sphere is homogeneous) to the surface. For the nonhomogeneities of the earth, see the plot of g versus r given in the Wikipedia reference, which takes account of the depth/density profile of the planet.
Have you a link for the wikipedia reference. I can't find it?
Post by: CliffordK on 07/02/2014 22:54:46
Were you looking at theoretical singularities with all the mass concentrated at a single point? That is a completely different beast than a planet with the mass surrounding a person as one bores towards the center.
Your escape velocity should always increase as you move towards the center of the body. I.E. no jagged lines as you have on some of your graphs.
I threw this chart together last night.
Acceleration Due to Gravity.
I drew a linear line from the center of Earth to the surface, which I believe is indicative of a constant density of Earth. Current theories indicate an iron core, and greater density in the center of Earth, but for the purposes of my estimate, this was good enough.
[ Invalid Attachment ]
Equations: (d is the distance to the center of Earth).
From center to surface:
Acceleration due to gravity is: 9.8*d/(6371 km)
From the surface into space,
Acceleration due to gravity is:
I'll try to convert my "acceleration" to a "velocity" soon.
Post by: jeffreyH on 07/02/2014 23:09:41
RD posted a couple of models for the internal gravity of the Earth (http://www.thenakedscientists.com/forum/index.php?topic=50254.msg430176#msg430176). The gravity should go to zero at the middle of the body.
Were you looking at theoretical singularities with all the mass concentrated at a single point? That is a completely different beast than a planet with the mass surrounding a person as one bores towards the center.
Your escape velocity should always increase as you move towards the center of the body. I.E. no jagged lines as you have on some of your graphs.
I threw this chart together last night.
Acceleration Due to Gravity.
I drew a linear line from the center of Earth to the surface, which I believe is indicative of a constant density of Earth. Current theories indicate an iron core, and greater density in the center of Earth, but for the purposes of my estimate, this was good enough.
[ Invalid Attachment ]
Equations: (d is the distance to the center of Earth).
From center to surface:
Acceleration due to gravity is: 9.8*d/(6371 km)
From the surface into space,
Acceleration due to gravity is:
I'll try to convert my "acceleration" to a "velocity" soon.
I appreciate that. My view is that gravitational cancellation is a gradient. If it all cancels out at the centre then this cancellation should decrease going outward but it is not a straight line. It can't be. The straight line approach makes no sense when the external field is inverse square. The cancellation must describe a curve with a particular functional description.
Moreover I believe that the current view of internal forces is inadequate. The only way to say which position is true is via experimentation. This is a thorny problem.
Post by: CliffordK on 07/02/2014 23:44:34
According to RD's Notes (http://www.thenakedscientists.com/forum/index.php?topic=50254.msg430176#msg430176), the linear line represents the acceleration due to gravity if the planet or body has a constant density.
Anyway, it should be good enough for a crude estimate. Otherwise one would need well defined acceleration or density functions.
Post by: jeffreyH on 07/02/2014 23:54:00
Technically you can divide your sphere into cubes, and calculate the acceleration due to gravity to each sub-cube, or perhaps one could do it with concentric rings or shells.
According to RD's Notes (http://www.thenakedscientists.com/forum/index.php?topic=50254.msg430176#msg430176), the linear line represents the acceleration due to gravity if the planet or body has a constant density.
Anyway, it should be good enough for a crude estimate. Otherwise one would need well defined acceleration or density functions.
I'm just thinking how many times in physics things have been counter-intuitive. Heisenberg's uncertainty principle, general relativity etc. I think g will increase with depth and fall sharply at the very core. I know this runs counter to common sense but this is exactly the point. An experiment to determine g in a deep mine shaft would resolve the issue. If it decreases linearly then my position is wrong. If it increases, however, we have a whole new view of gravity that may aid in a quantum theory of gravitation. Isn't that worth determining. I was as surprised as anyone when the figures went the wrong way but it makes sense when you think about partial field cancellations.
Post by: jeffreyH on 08/02/2014 00:16:29
http://davidpratt.info/gravity.htm
" In 1981 a paper was published showing that measurements of G in deep mines, boreholes, and under the sea gave values about 1% higher than that currently accepted.4 Furthermore, the deeper the experiment, the greater the discrepancy. However, no one took much notice of these results until 1986, when E. Fischbach and his colleagues reanalyzed the data from a series of experiments by Eötvös in the 1920s, which were supposed to have shown that gravitational acceleration is independent of the mass or composition of the attracted body. Fischbach et al. found that there was a consistent anomaly hidden in the data that had been dismissed as random error. On the basis of these laboratory results and the observations from mines, they announced that they had found evidence of a short-range, composition-dependent ‘fifth force’. Their paper caused a great deal of controversy and generated a flurry of experimental activity in physics laboratories around the world.5"
Now does this 1% increase agree with my model? That is the interesting question.
"As mentioned above, measurements of gravity below the earth’s surface are consistently higher than predicted on the basis of Newton’s theory.13 Sceptics simply assume that hidden rocks of unusually high density must be present. However, measurements in mines where densities are very well known have given the same anomalous results, as have measurements to a depth of 1673 metres in a homogenous ice sheet in Greenland, well above the underlying rock. Harold Aspden points out that in some of these experiments Faraday cage-type enclosures are placed around the two metal spheres for electrical screening purposes. He argues that this could result in electric charge being induced and held on the spheres, which in turn could induce ‘vacuum’ (or rather ether) spin, producing an influx of ether energy that is shed as excess heat, resulting in errors of 1 or 2% in measurements of G.14"
Post by: JP on 08/02/2014 00:59:43
I'd like to ask that you keep this thread to asking questions about Newtonian gravity, general relativity and perhaps some questions about peer-reviewed theories of quantum gravity. Please keep posts discussing your own models and their results to a thread in the New Theories section.
Thanks!
Post by: jeffreyH on 08/02/2014 01:14:38
Jeffrey, while I appreciate that you're discussing gravity and making calculations using your model, we have a strict policy here that we don't allow new models to be proposed and discussed outside of the New Theories sub-forum. A major reason for this is that this is mostly a science Q&A board and users who come here seeking answers (without any scientific background) will find the forum very confusing if new theories are mingled in with accepted theories.
I'd like to ask that you keep this thread to asking questions about Newtonian gravity, general relativity and perhaps some questions about peer-reviewed theories of quantum gravity. Please keep posts discussing your own models and their results to a thread in the New Theories section.
Thanks!
Is it possible for this thread to be moved into new theories?
Post by: jeffreyH on 08/02/2014 01:17:33
Post by: jeffreyH on 08/02/2014 01:41:43
http://earthdynamics.org/papers-ED/2010/2010-Steinberger-etal-Icarus.pdf
This discusses investigations of gravity anomalies with respect to density variations so is not new theory material but I will be looking into this further.
Post by: alancalverd on 08/02/2014 02:30:25
Have you a link for the wikipedia reference. I can't find it?
http://en.wikipedia.org/wiki/Escape_velocity (http://en.wikipedia.org/wiki/Escape_velocity)
The equation I quoted is about 2/3 of the way down the article
Post by: jeffreyH on 08/02/2014 03:17:44
Have you a link for the wikipedia reference. I can't find it?
http://en.wikipedia.org/wiki/Escape_velocity (http://en.wikipedia.org/wiki/Escape_velocity)
The equation I quoted is about 2/3 of the way down the article
I'd already sorted the Ve calculations I was more interested in a plot of Ve from the centre of gravity. I am transferring posts on this subject to here if anyone is interested in carrying on the debate.
http://www.thenakedscientists.com/forum/index.php?topic=50260.0
Post by: jeffreyH on 08/02/2014 03:25:39
If we consider a spherical mass as being made up of a gradient of internal spheres going down to the core we can treat each one as a separate gravitational source working out from the centre. At the very centre all gravitation cancels. At any point P out from the radius we will have a sphere at which P will be coincident with a spherical surface. We could treat our source of gravitation as being concentrated within each of these spheres as we move outwards and calculate Ve using this method. I will show the problems with this approach over the following posts.
Post by: jeffreyH on 08/02/2014 03:34:51
Post by: CliffordK on 08/02/2014 03:47:28
Your new topic appears to be related to atoms and not planets, so I didn't merge the two topics.
As RD had mentioned quite a while ago, if the core is denser than the crust, then the acceleration due to gravity does increase slightly as one descends through the crust, or deeper into the ocean. We currently do not have any technology that would allow drilling into the mantle, or core of the planet.
See:
https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth
Post by: jeffreyH on 08/02/2014 07:01:16
I went ahead and moved this discussion to New Theories.
Your new topic appears to be related to atoms and not planets, so I didn't merge the two topics.
As RD had mentioned quite a while ago, if the core is denser than the crust, then the acceleration due to gravity does increase slightly as one descends through the crust, or deeper into the ocean. We currently do not have any technology that would allow drilling into the mantle, or core of the planet.
See:
https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth
Sorry forget the other thread and we'll stick with this one. Well there must be another way other than drilling down that far. If I'm wrong then I'd rather know that sooner than later.
I'm not sure I agree with the method of gravitational cancellation used in the wikipedia entry. I'll have to put together an illustration of exactly what I mean by that. I can see how it makes sense with the internal spheres idea but I think it is slightly different than that.
Post by: alancalverd on 08/02/2014 09:33:36
I'd already sorted the Ve calculations I was more interested in a plot of Ve from the centre of gravity.
Just take the square root of the g/r plot. The PREM graph (see reply #14) is probably the best estimate for this planet. The shell integrals and presumed densities are all pretty much as I recall from undergraduate days.
Post by: jeffreyH on 08/02/2014 18:23:24
As an additional note before I finish this post. I will set out the method of proof in later posts.
Post by: jeffreyH on 08/02/2014 21:57:20
Post by: jeffreyH on 08/02/2014 22:19:43
Post by: jeffreyH on 08/02/2014 22:56:30
Post by: jeffreyH on 09/02/2014 00:48:21
In figure 1 we see the situation where our mass is at the centre of gravity where all forces cancel. The vector direction of travel is shown by V. The mass marked as s as can be thought of as subtractive as its gravitation is in the direction of travel and can be cancelled by f = a - s. In figure 2 our mass is moving towards the surface and s, still needing to be cancelled, no longer equals a. We now need to cancel the equivalent force. Inverting this spherical dome gives c, where all mass distances are equivalent and so all forces equalize. Cancelling this way gives us a new portion of the mass at a. a now contains the only mass that can be considered as acting against the momentum of our moving mass. However a can now be thought of as having a different centre of gravity to be used in our Ve calculation. If this is not adjusted correctly then the plot of internal gravitation will not meet with the inverse square plot at the surface. The line L is the demarcation of the sum of positive gravitational vectors in s and negative gravitational vectors in a. Figure 3 shows that L is coincident with the spherical surface indicating that no cancellations exist as s is now equivalent with free space.
Post by: jeffreyH on 09/02/2014 14:31:04
Post by: jeffreyH on 09/02/2014 18:19:09
Post by: jeffreyH on 09/02/2014 22:14:38
The calculation of a modified centre of gravity and the gravitational strength contributions are non-trivial functions.
Post by: jeffreyH on 09/02/2014 23:10:05
Post by: jeffreyH on 10/02/2014 01:59:36
Post by: jeffreyH on 11/02/2014 21:47:32
Post by: jeffreyH on 20/02/2014 03:57:46
Post by: jeffreyH on 22/02/2014 21:58:18
In section 2.2 I have found that I should be able to use zonal spherical harmonics with assumptions of latitude perpendicular to the direction of Ve. From what I have read so far the contributions should be greater nearer the surface than at the centre of gravity as the distance from the non-cancelled mass declines as the escaping mass nears the surface. Calculations nearer the centre of gravity may as well consider the mass to be spherical and uniform and ignore mass distribution as it can be considered to be a greater distance away than the non-cancelled mass when approaching the surface.
Post by: jeffreyH on 23/02/2014 07:20:03
V = 1/6pi * h(3a^2 + 3b^2 + h^2)
I will post results once done.
Post by: jeffreyH on 25/02/2014 02:16:28
http://www.cscamm.umd.edu/tiglio/GR2012/Syllabus_files/EinsteinSchwarzschild.pdf
Of particular interest is the statement on page numbered 928.
"Equation (14) represents a complicated relation between the particle density
n and the function a representing the gravitational field. The limiting case,
however, in which the gravitating particles are concentrated within an infinitely
thin spherical shell, between r = ro- A and r = ro, is comparatively simple.
Of course, this case could only be realized if the individual particles had the
rest-volume zero, which cannot be the case. This idealization, however, still is
of interest as a limiting case for the radial distribution of the particles"
Post by: jeffreyH on 25/02/2014 03:12:30
https://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&ved=0CFIQFjAE&url=http%3A%2F%2Fthescipub.com%2Fpdf%2F10.3844%2Fpisp.2012.50.57&ei=3QcMU_uxKoir7Aayr4DoDg&usg=AFQjCNFVWsjr4uP2vgxds8d_eqMI_Sg_5Q&sig2=r_8jC_XdyGl1Ex9Ei73cxw&bvm=bv.61725948,d.ZGU