Impact of air resistance at varying angles

[eddywalrus]

I recently read that the impact of air resistance on the horizontal range increases as the launch angle increases (http://moodle.davidson.edu/moodle2/pluginfile.php/121168/mod_resource/content/2/Brancozio%20fly-ball%20paper.pdf). A graph depicting this is attached. Is there a reason for why this is? I'm not a very good mathematician so it would be great if the explanations didn't involve too much calculus / are mainly qualitatitive (although I do understand a bit of differentiation and integration)

 

[A.T.]

I recently read that the impact of air resistance on the horizontal range increases as the launch angle increases

Only up to a certain angle. At 90° launch angle air resistance doesn't change the horizontal range. Same for 0°. It can only affect non-zero ranges, which are between 0 and 90°.

 

[eddywalrus]

Thanks for the response and clarification, but would you be able to provide an explanation for why this is the case (if the angle x is restricted to 0 < x < 90)?

 

[A.T.]

Thanks for the response and clarification, but would you be able to provide an explanation for why this is the case (if the angle x is restricted to 0 < x < 90)?

Why air resistance reduces non-zero ranges? Well, it slows down the projectile.

 

[eddywalrus]

Nope, as in why the impact of air resistance on range increases as the angle increases; why is it that, at higher angles, the range is reduced by so much more than it is at lower angles?

Thanks for your help!

 

[A.T.]

Nope, as in why the impact of air resistance on range increases as the angle increases;

The impact is zero at 0° and 90° and non-zero in between. So it must increase and then get back to zero as you go from 0° to 90°.

 

[jerromyjon]

You have to consider the ratio of linear projectile velocity which is contributing to work in the "x" direction.

 

[eddywalrus]

The impact is zero at 0° and 90° and non-zero in between. So it must increase and then get back to zero as you go from 0° to 90°.

But the fact that the impact is zero at 0 and 90 degrees, and non-zero in between, only suggests that, as you said, the impact of air resistance on the range must increase and get back to zero -- it doesn't explain why, for example, the impact of air resistance on range is greater for 80 degrees than for 60 degrees (or am I missing something here?)

 

[eddywalrus]

You have to consider the ratio of linear projectile velocity which is contributing to work in the "x" direction.

Could you explain this a bit further please?

 

[A.T.]

But the fact that the impact is zero at 0 and 90 degrees, and non-zero in between, only suggests that, as you said, the impact of air resistance on the range must increase...

Right, so it must be increasing with increasing angle, for certain angles, which is what you wanted explained.

it doesn't explain why, for example, the impact of air resistance on range is greater for 80 degrees than for 60 degrees (or am I missing something here?)

Right, it doesn't. You wanted a qualitative explanation. Justifying particular quantitative results is more complex. Also, if you look at your graph, you will find that the impact at 80° is actually less than at 60°.

 

[eddywalrus]

Right, so it must be increasing with increasing angle, for certain angles, which is what you wanted explained.

Right, it doesn't. You wanted a qualitative explanation. Justifying particular quantitative results is more complex. Also, if you look at your graph, you will find that the impact at 80° is actually less than at 60°.

But is there an explanation for why it is the way it is? The fact that the impact is zero at 0 and 90 degrees only indicates that the impact increases with increasing angle, but doesn't explain it per say. Thanks for your help anyway!

 

[jerromyjon]

It boils down to x velocity. At 10 degrees most of the velocity is in y, very little in x, so very little air resistance contributes to deceleration of x velocity. As x velocity increases at 20 degrees y velocity decreases so more air resistance to decelerate x, less for y.

 

[eddywalrus]

It boils down to x velocity. At 10 degrees most of the velocity is in y, very little in x, so very little air resistance contributes to deceleration of x velocity. As x velocity increases at 20 degrees y velocity decreases so more air resistance to decelerate x, less for y.

Thank you for your response. But isn't it the other way round; at 10 degrees (i.e. 10 degrees upwards from the horizontal) most of the velocity is in the horizontal (x), rather than the vertical (y), right?

 

[A.T.]

The fact that the impact is zero at 0 and 90 degrees only indicates that the impact increases with increasing angle

The fact that impact is zero at 0° and 90° and non-zero in between, implies that it must increase with increasing angle, for some angles in that range.

but doesn't explain it per say.

Showing mathematically that something must be true is a form of explanation to me.

 

[jerromyjon]

I think you are looking at it backwards, 0 is straight up and 90 is horizontal in your baseball example chart. 90 degrees isn't included in the chart because "grounders?" don't fly very far...

 

[eddywalrus]

The fact that impact is zero at 0° and 90° and non-zero in between, implies that it must increase with increasing angle, for some angles in that range.Showing mathematically that something must be true is a form of explanation to me.

Thank you for your contribution, but it doesn't really satisfy my curiosity :(

I think you are looking at it backwards, 0 is straight up and 90 is horizontal in your baseball example chart. 90 degrees isn't included in the chart because "grounders?" don't fly very far...

In the document it refers to angles above the horizontal:

Thank you for your help though!

 

[jerromyjon]

It appears to contradict actual physics then, because at 10 degrees just above horizontal at 60 mph the baseball would travel 95 percent of the distance in air than it would in vacuum according to the chart.