What would you do? (Checkride)

Ya'all never seen one of these?
PXL_20210803_183628655.jpg
Lives on my knee board, works for any situation.
 
Fly4Fun! said:
I have my commercial checkride next week. I have been studying my butt off... I finally feel up to the ACS standards on both the ground and flight portion.

However, I have an issue:

The aircraft I am taking my checkride in is an old Piper Arrow. The AFM only offers one chart for takeoff performance, which is max gross. With a 1.5 margin, at max gross, with temperatures that are most likely for an august afternoon, I am not safely clearing a 50 foot obstacles.

With all that said, the situation I have been provided by my DPE doesn't have us departing at max gross. On top of that, I do not nearly need full tanks.

I know about the 30/70 rule for obstacles, and the 50/70 rule for takeoffs without an obstacle, but those are rules of thumbs to back up know performance numbers.

I believe that the correct answer is that in the interest of safety, only the know performance numbers can be utilized, and that a departure with the given conditions (HOT) is unsafe. Departing bellow max gross will have a resultant increase in performance, however how much of an increase is a guess, which is (in my opinion) an unacceptable level of risk.

Am I looking at this wrong?
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Can you do the test from a different airfield nearby?
 
aftCG said:
Ya'all never seen one of these?
View attachment 98899
Lives on my knee board, works for any situation.
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Mine looks a bit different, but same idea. I'm guessing I bought the last one in Aircraft Spruce's stock, because they discontinued the item right after mine arrived.
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Arnold said:
So for the sake of legal nitpicking. What is the fuel reserve requirement for a pilot operating on an IFR flight plan in VFR conditions? I could argue that it is less than the fuel reserve requirement for operating VFR in VFR conditions.

Example. You have 30 minutes fuel in the ship. You want to fly from A to B ins VMC conditions. Flight time is expected to be 15 minutes. Are you legal to fly day VFR from A to B? No.
Are you legal to fly IFR from A to B IFR? Yes.

Is this careless and reckless (91.13(a))? Perhaps.
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@EdFred is correct. Why would you argue otherwise?
 
MauleSkinner said:
@EdFred is correct. Why would you argue otherwise?
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He's a lawyer and "VFR conditions" is not explicit in the definitions, so there's a place for a lawyer to say, well, what he said.

We have a case coming up to the Michigan Supreme Court that could have implications depending on which way they rule. I asked 3 lawyers their opinions and got 3 answers, and the last one, after I said I guess it's true ask 10 lawyers get 10 opinions. He said no, ask 10 lawyers, close the door, draw the blinds and we ask you "what do you want the regulations to say?"
 
EdFred said:
He's a lawyer and "VFR conditions" is not explicit in the definitions, so there's a place for a lawyer to say, well, what he said.

We have a case coming up to the Michigan Supreme Court that could have implications depending on which way they rule. I asked 3 lawyers their opinions and got 3 answers, and the last one, after I said I guess it's true ask 10 lawyers get 10 opinions. He said no, ask 10 lawyers, close the door, draw the blinds and we ask you "what do you want the regulations to say?"
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Exactly! But the Lawyer Pilot Bar Association meeting is coming up and I'll ask around for ten different opinions.

This reminds me of a joke my father used to tell which I (caution lawyer word) herewith abbreviate. A CEO was interviewing accountants and asked only one question. What is 2+2. The first three guys said 4 and the CEO said next. The fourth guy says "what would you like it to be" and got the job.
 
murphey said:
It's a shame that Piper didn't take the same approach to AFM/POH that Cessna did/does. The Cessna chapter on Performance (chp 5) include "fudge factors" for types of runway, altitude, temperature, flaps, and other conditions. Lots of performance data as well as ideas how to address different situations. I'm not saying the C172 manual should be used for the Arrow, but it will give you an idea how to adjust for performance.
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Later POHs (eg mid/late 1970s) can have a lot more information than earlier owners' manuals.
 
Arnold said:
A CEO was interviewing accountants and asked only one question. What is 2+2. The first three guys said 4 and the CEO said next. The fourth guy says "what would you like it to be" and got the job.
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I wish this weren't so close to the truth. :(
 
texasclouds said:
I had a few students that were engineering students. For their pre solo knowledge test they were required to tell me the crosswind component with a given scenario. Well they did some fancy math equations
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Doe multiplying the sine of the wind angle times the wind speed really qualify as fancy math?
 
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PPC1052 said:
Doe multiplying the sine of the wind angle time the head wind really qualify as fancy math?
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True, and you don't even have to calculate the sine; just remember "a third is a half" (30° gives you half the wind as a crosswind). If you want to get fancier, memorise it in 30 degree increments:

30°: 50%
60°: ~85%
90°: 100%

Since we're counting in round knots, linear interpolation is good enough to fill in rest. And reversing the list gives you cosine for the headwind/tailwind component.

I still find playing with my E6B or CR-6 more fun though, which is why I created https://e6b.org to do practice problems (I'm probably the only regular user).
 
David Megginson said:
True, and you don't even have to calculate the sine; just remember "a third is a half" (30° gives you half the wind as a crosswind). If you want to get fancier, memorise it in 30 degree increments:

30°: 50%
60°: ~85%
90°: 100%

Since we're counting in round knots, linear interpolation is good enough to fill in rest. And reversing the list gives you cosine for the headwind/tailwind component.

I still find playing with my E6B or CR-6 more fun though, which is why I created https://e6b.org to do practice problems (I'm probably the only regular user).
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Beat me to it.:yeahthat: But I use 30/45/90. 45=70% anything over that I treat as a direct crosswind.

The key is to fly safely, exact number don't matter in real life. The real question is: am I comfortable doing this?
 
David Megginson said:
True, and you don't even have to calculate the sine; just remember "a third is a half" (30° gives you half the wind as a crosswind). If you want to get fancier, memorise it in 30 degree increments:

90°: 100%
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Do you really have to memorize that a direct crosswind is a direct crosswind? :goofy:
 
PPC1052 said:
Do you really have to memorize that a direct crosswind is a direct crosswind? :goofy:
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The first level of learning is rote, the correlation and application takes awhile. Especially with the Canadian exchange rate.
 
The “unit circle” is one of the things I remember from high school math…for 30 degrees, sine and cosine are .5 and .866 (60 degrees is the opposite, obviously) and 45 is .7071. Everything else is an estimate in between.
 
Arnold said:
Beat me to it.:yeahthat: But I use 30/45/90. 45=70% anything over that I treat as a direct crosswind.
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Not a bad idea, since when the surface wind is strong enough to be a concern, it's rarely obliging enough to keep coming from exactly the same direction at exactly the same strength.
 
texasclouds said:
Buck, for us feeble minded folk…yes.
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So go back to my "a third is a half" rule. If the wind angle is <30°, then you're getting less than half the wind velocity as a crosswind. If the wind angle is >30°, then you're getting more than half of the wind as a crosswind. If it's >60°, then you're getting basically all of it as a crosswind. That's the same thing as multiplying by the sine, but just with a bit of rounding off. :)

And it's easy enough to do in your head in the couple of seconds between when tower gives you the new winds with your takeoff clearance, and you push the throttle forward to taxi onto the runway. You don't want to be wasting time pulling out an app, a calculator, or an E6B in a situation like that.
 
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David Megginson said:
So go back to my "a third is a half" rule. If the wind angle is <30°, then you're getting less than half the wind velocity as a crosswind. If the wind angle is >30°, then you're getting more than half of the wind as a crosswind. If it's >60°, then you're getting basically all of it as a crosswind. That's the same thing as multiplying by the sine, but just with a bit of rounding off. :)

And it's easy enough to do in your head in the couple of seconds between when tower gives you the new winds with your takeoff clearance, and you push the throttle forward to taxi onto the runway. You don't want to be wasting time pulling out an app, a calculator, or an E6B in a situation like that.
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I've run into pilots that have a hard time multiplying and dividing by 10.
 
murphey said:
Unfortunately, most people have never taken Calculus 101 and don't understand that 1st derivative of acceleration is velocity (speed). Constant acceleration means increasing (linear) velocity. Linear acceleration means the velocity is increasing exponentially.

If you're interested (and I know almost no one here is interested) the second derivative of acceleration is the first derivative of velocity which is distance.
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That doesn't answer his question at all. If an object accelerated linearly it could reach infinite speed.
 
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EdFred said:
I didn't say it was required or should be required for the check ride, but it's what *can* be done to determine a takeoff and clearance distance, and doesn't take more than 9th grade math skills. But if someone walked into a checkride with you and dropped that on the table, wouldn't you be like "alright, this person isn't it training to JUST pass the checkride."
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The math may be 9th grade level, but the knowledge and understanding of physics to apply it to the problem in question is not. I would say AP Physics high school level at least.
 
dmspilot said:
The math may be 9th grade level, but the knowledge and understanding of physics to apply it to the problem in question is not. I would say AP Physics high school level at least.
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You don't need to know advanced physics to doing simple aviation arithmetic, any more than you need to know advanced biology to plant a garden. Be careful, or you'll start me on a rant about how pointless it is to confuse student pilots by forcing them to learn about Bernoulli ... ;)
 
David Megginson said:
You don't need to know advanced physics to doing simple aviation arithmetic, any more than you need to know advanced biology to plant a garden. Be careful, or you'll start me on a rant about how pointless it is to confuse student pilots by forcing them to learn about Bernoulli ... ;)
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I said high school, that's not what I would call "advanced physics".
 
dmspilot said:
I said high school, that's not what I would call "advanced physics".
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That's more advanced than any physics I ever studied. :)

I took one science course in all of high school (and the minimum mandatory math), and focused on languages, social sciences, humanities, and music.
 
dmspilot said:
The math may be 9th grade level, but the knowledge and understanding of physics to apply it to the problem in question is not. I would say AP Physics high school level at least.
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Not sure what school system you grew up in, but this is not even close to AP level, and I had the introduction to this in the 8th grade - and that was in the Upper Peninsula where they were a year behind the other schools I went to down state.
 
EdFred said:
I've run into pilots that have a hard time multiplying and dividing by 10.
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I throw 100 pounds of baggage in on W&B problems for an oral just to see if the applicant uses a calculator to compute the MOMENT/100 that the airplane paperwork uses.
 
EdFred said:
Not sure what school system you grew up in, but this is not even close to AP level, and I had the introduction to this in the 8th grade - and that was in the Upper Peninsula where they were a year behind the other schools I went to down state.
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Plugging numbers into a formula, which could be done by 8th graders, is not equivalent to solving the conceptual problem that OP posted. If it was so easy why would this thread even be here? Does OP only have a 7th grade education?
 
dmspilot said:
Plugging numbers into a formula, which could be done by 8th graders, is not equivalent to solving the conceptual problem that OP posted. If it was so easy why would this thread even be here? Does OP only have a 7th grade education?
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I was familiar with F = ma before high school. Even the concept is understood by little kids (though they don't have the math to go with it at the time) that it's easier if two of them pull a wagon than if only one does, or that it's harder to pull something heavy than it is something that is light.

I don' t know the OP's education level as I am not him, nor did I ask him.
 
Woah now, y’all leave Daniel Bernoulli out of this! You can’t lean to fly until you’ve heard the three theories of lift.

1 Bernoulli’s Venturi
2 Newton’s 3rd Law
3 PFM
 
A dog can recognize human faces, understand human speech, follow commands, and perform simple tasks. Ergo it must be easy for a dog to assemble a jigsaw puzzle of a person's face.
 
dmspilot said:
A dog can recognize human faces, understand human speech, follow commands, and perform simple tasks. Ergo it must be easy for a dog to assemble a jigsaw puzzle of a person's face.
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If few enough pieces and you took the time to train it to do so, yes.

But I'm fairly certain that most pilots have an intelligence level higher than that of a dog. Not sure if present company is excluded or not.
 
EdFred said:
But I'm fairly certain that most pilots have an intelligence level higher than that of a dog.
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A dog doesn't know what an analogy is. Clearly, spiders don't either.
 
EdFred said:
Sure, beyond the FAA learning. You know the takeoff distance and rotation speed for full gross. Based on that, you can determine the time it took to get to that speed. With that information you can determine the average acceleration for the take off run. With that information you can determine the force being applied over the take off run. Now you can plug your less than gross weight number back into the formulas used to determine that. You now have a new acceleration number. But it gets slightly better. Your V speeds also decreases based on the square root of the ratio of current weight to gross weight. So you can also calculate your new rotation speed. Coupled with the new acceleration rate, you can calculate the new take off distance. You can also determine with the new Vx based on current weight the horizontal distance needed to clear the 50' object.

Is it a lot of work? Yeah, maybe, but it shows to the DPE that you actually put some thought into the process.
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Can also use the Koch chart.
 
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