How high do they get on the Bay of Funday (at most)?

That depends on what they're smoking.

__________________
"Ignorance more frequently begets confidence than does knowledge" -- Charles Darwin

Too small to measure - completely swamped by background!

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Any day you wake up on "the right side of the dirt" is a good day.

T. Anderson

A decent rule of thumb is that the tidal effect is proportional to the area in the sky. (It's also proportional to density, which is why the moon's effect is twice the sun's, even though they are approx. the same area in the sky.) At its closest, Venus subtends about a minute of angle, about a thirtieth of that of the moon--so its area is 1/900. The Bay of Fundy tide is about 17 meters, so the Venus tide, if it were proportional, would be 17//900 meters, or about 2 centimeters. Of course, that would be when it was lined up with the sun, too.

Interesting. You could actually visualize the change, although it would be lost in the waves. Thank you...I read the tides on the Great Lakes are of an order of an inch or so, so this is a similar amount.

Sounds way too large, hhEb09'1.

The tidal force goes as ....(ignoring higher order terms)....

F = GMmr/d^3......where r is earth's radius in this case.

Making a ratio of Venus vs moon, the only thing that will change in the eqn. is (the M) for mass of moon vs. mass of venus, and the distance, d, of each from earth.

Mass of Venus is 71 times that of the moon, but its closest distance is 107 times that of the moon.

Thus; since tidal force goes as the inverse 3rd power of distance.....

the tidal force of Venus = 71 X 1/(107)^3 = 58 X 10^ - 6 that of the moon.

If I did it right ....that's 58 millionth of the moon's tidal force.

If you think about it that rule of thumb can't be correct since the 'area' changes as inverse square of distance, whereas tidal force changes as inverse cube of distance.
G^2

I see the booby trap here, and I fell into it. My initial response to the aforementioned rule of thumb was "no way". Then I checked out the case of a more distant object with the same density and angular diameter, and found that its tidal force was indeed equal to that of the nearby one. In my haste I concluded that the rule of thumb was correct. Now I see that it does not hold as a general rule. At a given distance the mass, and thus the tidal force, varies as the 1.5 power of the angular area.

I did indeed forget about that factor. So, instead of squaring 30, I should have cubed it. 1/27000 is 37 millionths, which is (in the) same (ballpark) as Gsquare's calculation. So, instead of a couple centimeters, it would be less than a millimeter.

Sorry to get your hopes up, Tom

And Fundy enjoys its exceptional tides because it happens to resonate with the 12h50m period of the lunar tides. It's presumably not going to amplify the effect of the ~12h Venusian signal to the same extent.

Grant Hutchison

I was going to mention this, and then realized I didn't really know what the resonance was--after all, the solar tide would be at nearly the same timing as Venus--so I just went with the premise of the OP. And made other mistakes.

I'll have to look into it.

PS: On a google, the first page returned led me to a Fisheries and Ocean Canada website. Looking over the tides for the next week, it looks to me that the tide is around ten meters (instead of the 17 I used before) and it has a period of near 12h 20-25m! There doesn't seem to be much variance over the week, either. So, I don't know how this works.

The resonant period should be 12h25m, I think, not 12h50m as I stated above.
I remembered to halve the 24 hours, but not the extra 50 minutes of the lunar cycle.

Grant Hutchison

Oops. Yep, me too.