Understanding the Coriolis Effect: Paris Gun Shells and Deviation Measurements

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Discussion Overview

The discussion centers around the Coriolis effect as it relates to the deviation of artillery shells, specifically those fired from the Paris Gun, over long distances. Participants explore whether the drift caused by the Coriolis effect can be scaled linearly with distance or if it exhibits non-linear behavior, while also considering the implications of Earth's rotation and the shell's trajectory.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant questions if the Coriolis effect's drift can be scaled down linearly, suggesting that deviations could be calculated proportionally over shorter distances.
  • Another participant asserts that the drift is non-linear and only applicable under specific conditions, such as treating the Earth as a spinning cone.
  • A participant introduces the idea that aerodynamic effects and the shell's trajectory could influence the landing point, suggesting that the shell would follow a sub-orbital path if not for the Coriolis effect.
  • Another participant emphasizes that the Coriolis force is significant due to the eastward motion of the cannon and the shell, explaining that the impact point shifts westward when fired southward, and that this shift is not linear due to the relationship between latitude and Earth's rotation.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the Coriolis effect's impact on projectile motion, with some arguing for linear scaling and others asserting non-linear behavior. The discussion remains unresolved regarding the exact nature of the Coriolis effect in this context.

Contextual Notes

Participants note that the relationship between the Coriolis effect and distance may depend on various factors, including the latitude of the firing location and the speed of the shell, which introduces complexity into the analysis.

restfull
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According to Wikipedia, the shells of the Paris Gun fired over 120 km landed "1,343 meters (4,406 ft) to the right of where it would have hit if there were no Coriolis effect"...

Is it then correct to say that it would have deviated by 134.3 metres over 12 km, and 13.43 metres over 1.2 km etc etc?

i.e. is it correct that the drift due to the coriolis effect can be scaled down in this way, or is it non linear with respect to distance
 
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restfull said:
According to Wikipedia, the shells of the Paris Gun fired over 120 km landed "1,343 meters (4,406 ft) to the right of where it would have hit if there were no Coriolis effect"...

Is it then correct to say that it would have deviated by 134.3 metres over 12 km, and 13.43 metres over 1.2 km etc etc?

No.

i.e. is it correct that the drift due to the coriolis effect can be scaled down in this way, or is it non linear with respect to distance

Only if the Earth is a spinning cone. Of course it's non-linear.
 
Ignoring aerodynamic effects on the shell, I would assume it would follow a sub-orbital path and land exactly where it should. It would only land to the "right" if the Earth were treated as flat, and the shell was fired in a certain direction at a certain latitude. Sort of like the difference between rhumb line (consant headin) versus GPS (great circle) navigation.

Another issue would be the speed of the Earth's rotation and the speed, direction, latitude of the shell. Assuming the shell travels at a very high speed, than I assume that the fact the Earth moves while the shell travels through the air isn't going to create that much difference in the flight.

or am I missing the point and there's some other factor in play here?
 
Yes, you are missing the point: the "coriolis force" as cited in the original post. The cannon and shell in it have a certain speed eastward because they are attached to the Earth and the Earth is rotating to the east. If the cannon is fired southward, the impact area is also moving eastward but, in the northern hemisphere, faster- thus, the shell hits slightly to the west of "due south" of the cannon. It may not "create that much difference" but enough: 1346 meters in 120 km.

The amount of "rotation" at any latitude is proportional to the circumference and therefore the radius of the great circle around the Earth at that latitude. That itself is radius of the Earth times the sine of the "co-latitude" (angle measured from the north pole). The distance between the cannon and the point of impact is proportional to the difference in co-latitudes and so the rotation involves a sine function and is not linear.