It is not clear to me what triangle and included angle you are considering when you say

“The geometry of it is such that the included angle between W and L in the right triangle with L as a hypotenuse ON THE GROUND and the W as the leg that makes the crater width …”

My understanding is that the included angle between W and L is 90°, since W and L are twice the semi-major and major axes of the ellipses, as shown in the figures in the paper. That is, L is the maximum dimension of the ellipse, but it seems like you are viewing L differently. A diagram would be helpful. I have no doubt that you are very knowledgeable about ellipses and impact craters, but I am not understanding your interpretation here.

To check Zamora’s result, I did a calculation based on a very simple model. I assumed that a spherical object sweeps out a cylinder as it approaches and impacts the surface of the Earth, which is represented by a plane, where the cylinder is inclined at angle θ to the plane. I simply calculated the intersection of the cylinder and the plane, to represent the shape of the impact crater. The result is that the intersection is an ellipse with eccentricity cos(θ), and sin(θ) = W/L, which agrees with Zamora.

I have not yet closely read Zamora’s paper, but I notice two things. (1) He gives an incorrect definition of ellipticity, and (2) the ballistic equation he cites is not the correct equation in this case. It assumes that gravity is constant over the trajectory and the trajectory is over a flat surface, neither of which is true. At best is is an approximation, which may be good enough for his argument, but it is not accurate. The ballistic equation describes a parabola, but the suborbital trajectory of the ice boulders is elliptical.

]]>A width-to-length ratio of 0.58 corresponds to a cone inclined at 35° using the relationship sin(θ) = W/L. The proposed conical cavities could have been made by impacts of material ejected at approximately 35° in ballistic trajectories from the point of convergence in the Great Lakes Region.

I am interested in where Antonio got that cone inclination from the Davias data. The way I look at it, he got his math wrong. I derived a very different values for the angle and the ratio. I come up with 43.652°, measured vis-a-vis the horizontal.

Michael Davias long ago shared his data with me, slightly before he completed the count of CBs. Davias ended up with 45,000, and what I have includes 43,900 of them (97.55% of the total). I’ve averaged their widths and lengths, and they came out to average W = 0.21743 km wide and L = 0.37514 km long – a ratio of 0.7235:1.000.

**EXCEPT: ** The sine is not the right trig function to be using.

How to explain this in 1,000 words or less? . . . *Technical stuff coming up:*

The geometry of it is such that the included angle between W and L in the right triangle with L as a hypotenuse ON THE GROUND and the W as the leg that makes the crater width*** – and because the impactor is assumed to be spherical, we can draw it as SQUARE to the angle made by the target surface and the incoming path, the angle θ. Thus we use NOT the sine value but the COSINE value. **Basic trig, it is a case of the near side over the hypotenuse** – and that is COSINE, not SINE. The lower this W/L ratio, the steeper the impact angle.

This is easily visualized by looking at a hypothetical vertical impact and a hypothetical (basically) horizontal impact. The former would have a W/L = 1.000, and the latter would have W/L of essentially zero. The lower the angle relative to the ground, the greater L becomes. This is what the COSINE does, not the sine.

(*** This is NOT the bolide diameter, but the crater width.)

(I could be wrong, but, folks, I don’t think so. My main gig in my early engineering years was solving scads and scads of triangles – probably upwards of 100,000 – *LONGHAND*. And I HAD to get them right. And I’ve checked and double-checked this before writing this comment.)

Given that we are given W and L, I determined that the direct method is that W/L gives the COSINE, not the sine.

The Sine would work if the angle was taken OFF THE VERTICAL, but to GET to the vertical, you have to go round about, if you started with W and L. But W/L itself does NOT give you the angle off the vertical. W/L can only lead directly to the angle measured off the surface. Because L is measured ON the surface, not on the vertical. And W cannot HELP but be related geometrically to the WIDTH of the bolide. Again, because L is measured on the surface, not on the vertical, working off the vertical is WRONG.

The cosine should be used, and then the arccosine angle should be derived, and then that angle subtracted from a 90° vertical impact angle. This fits my trig sketch and fits the way the impactors come in and are discussed.

It ALSO agrees with the 20° downward angle always referred to in, say, Chelyabinsk. The reference 0° position is not the vertical, but the horizontal (the ground). The 20° was measured versus a horizontal path (a 0° path parallel to the ground).

Thus Zamora’s angle has been gotten wrong. He also got a different value than my 43,900 CBs gave. He got And I don’t think the 1,000 other craters Davias has identified and measured is going to change my angle much. My average angle is 43.652° measured off the horizontal. This is based on arccos (0.27143km/.37514km). (I feel comfortable going to 5 decimal places on such a large population in the database, even though Davias only took the L and W values to 2 places.)

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(Can I be wrong because this is an ellipse and maybe there is something magical about ellipses? An ellipse is defined as “*the locus of all points P such that the sum of the distances from P to two fixed points F, F1 (foci) are constant.*”

Yeah, and in the real world, every one of those points has the same coordinates (relative to the intersection of the major and minor axes, A and B) as if you took a circle and stretched everyone of P’s Y coordinates in one direction by the ratio of A/B. I learned this from experience. Note that this intersection point of A and B is NOT the intersection of the right angle cone’s axis and the plane cutting across it.)

It’s not all harsh critiquing, though… I will post at least one positive comment, too. . . .

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