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How to generate an equation for estimating maximum error? 
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#1
Feb1712, 02:08 PM

P: 18

Hi,
I am hoping to generate an equation that will describe a maximum error in estimation of angles. I would like an advice on the type of math/methods/devices (e.g., differential equations?) that I could use to solve this. Here's the problem: Imagine a square with an oblique line inside the square. Lets say the oblique line touches the left bottom corner of the square. Of interest is the error in estimation of angles between the oblique and the left side of the square. I will refer to it as the angle of an oblique. By error I mean the difference between the true angle and the estimated angle. If the length of two parallel sides of a square is misestimated relative to the other two sides, then the angle of an oblique will also be misestimated in a very specific way. Here's the twist: the misestimation of angles will vary depending on the physical angle between an oblique and the left side of a square. If one of the sides of a square is misestimated by less than 50 %, for example, the maximum error will presumably occur for angles that are about 40 degrees to the left side of the square. I would like to generate a function that will show (1) when the maximum error should occur (i.e., at which angle of an oblique?), (2) what is the maximum error (e.g., angle is overestimated by 12 degrees; therefore error = +12 degrees). Maximum error will depend on percent error in misestimation of lengths of a square, and the physical angle of an oblique. I hope this is clear enough. Thank you for your help! 


#2
Feb1712, 03:19 PM

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P: 3,300

For example if the vertices of the figure are given counterclockwise as ABCE with the line passing through A and hitting side BC, how is the estimate of the length BC used to calculate the angle? Does the estimator assume all sides of the square have length equal to his estimate for BC? 


#3
Feb1712, 05:07 PM

P: 18

Thank you for your reply, Stephen. This question is about a square that is perceived as a rectangle. For example, an observer may underestimate sides "AE" and "BC" but accurately estimates sides "AB" and "EC." An observer does not assume that all sides of the square have length equal to his estimate for BC.
A line passing through A and hitting side BC will be at a larger angle from side "AE" in a rectangle above than a square. This is because a component of the line parallel to side "AE" will be underestimated by the same amount (in % error) as the side of the square. As an analogy, consider what happens with a diamond inscribed within a square seen as a rectangle. The diamond will look squashed in the latter. Consequently, the angle between its side and the adjacent side of the square will be overestimated. 


#4
Feb1712, 10:21 PM

P: 4,573

How to generate an equation for estimating maximum error?
Hey PatternSeeker and welcome to the forums.
This might sound anal, but it might help if you give us a simple diagram outlining what you are trying to measure. In the graphic show all the variables and then if you can use an arrow to point to the 'angle' and if necessary work out the relationships between the sides that specify the angle in terms of them. 


#5
Feb1912, 02:55 PM

P: 18

Thanks, chiro. I attached a picture which I hope will clarify things.
The angle of interest is angle alpha. Any suggestions will be much appreciated! Thank you for welcoming me to the forum! I hear good things about it. 


#6
Feb1912, 03:33 PM

Sci Advisor
P: 3,300

PatternSeeker,
You have not clearly described the experiment that collects this data. For example, does the person being tested report "I think angle alpha = 65 degrees"? Or are you deducing the person's opinion about angle alpha from his opinion about other dimensions of the diagram. For example, does the person report "I think AE = 3" or do they report "I think AE = 3/5 of EC?". If the person does not report a perception of the angle, I think calling the 65 degrees a "perceived" angle is misleading. If you are computing the angle, how are you computing it? My speculation is that you know the actual dimensions of AP and RP. When the subject reports his guess g about distance [itex] \overline{AE}[/itex] , you assume all vertical distances are shrunk by the factor [itex] \frac{g}{\overline{AE}} [/itex]. So you compute the "perceived" angle as [itex] \alpha' = \arctan ( \frac{ \frac{g}{\overline{AE}}(\overline{RP})}{\overline{AP}}) [/itex]. Is that what you do? 


#7
Feb1912, 04:46 PM

P: 18

Hi Stephen,
Thank you for your reply. I used the term "perceived" for angles that would be estimated by observers. In this example, they would report angle as 65 degrees. Real angles are physical angles. The data I put up on this portal are madeup. I used the formulas such as the one you gave to generate these hypothetical data. % error in perception of length of stimuli is derived from a theory. Just as a reminder, I would like a suggestion about how to develop an equation that will produce an answer to where the maximum error (perceived  real angles) should occur. For example, assume that the % error in misestimation of lengths is  40%. At what real angle between two edges of stimuli should the maximum error in estimated angle occur? I can certainly generate lists of values to figure this out, but I need a formula that will give me a single answer. I provided more details in the attached picture, and in previous posts. Thank you for your attention and suggestions about how to clarify this problem. I hope to hear from you again. Any further suggestions will be appreciated. 


#8
Feb1912, 08:17 PM

Sci Advisor
P: 3,300

Can we agree that the problem is:
Given [itex] \lambda > 0 [/itex] find angle [itex] \alpha [/itex] that maximizes [itex] f(r) =  \arctan(r)  \arctan( \lambda r)  [/itex] where [itex] r = tan(\alpha) [/itex]. 


#9
Feb2012, 01:10 PM

P: 18

Hi Stephen,
That may be so. Can you just tell me what λ stands for? Is it % error? In translation I would state the problem as: given that the % error in estimated lengths is less than zero, find physical angle α that maximizes the change between the estimated angle α' and the physical angle α. I'm not sure if that's exactly what you wrote in the mathematical language :) Please clarify. 


#10
Feb2012, 02:39 PM

Sci Advisor
P: 3,300

For a "40% underestimate", [itex] \lambda [/itex] would be 0.60. (I try to avoid using the terminology "percent" whenever possible. It often introduces confusion.)
Just glancing at the problem, I think the angle of maximum error is [itex] \arctan( \frac{1}{\sqrt{\lambda}}) [/itex], but I must go do something now. I'll look at it closer later. 


#11
Feb2012, 05:22 PM

P: 18

Stephen, I think you got it!
Except that it's not arctan(1√λ), but that formula subtracted from 90 degrees. How did you arrive at that solution? I will try to figure this out on my own, but if you could clarify further, that would be great. Thank you so much 


#12
Feb2012, 07:57 PM

Sci Advisor
P: 3,300

It's one of those calculus problems where you find extrema of a function by setting its derivative equal to zero and solving for values that make that happen.
For [itex] f(r) =  \arctan(r)  \arctan( \lambda r)  [/itex] [itex] f'(r) = \frac{1}{1 + r^2}  \frac{\lambda}{1 + \lambda^2 r^2 } [/itex] if [itex] \lambda < 1 [/itex] If you're looking for a maximum, you must verify that a given solution of [itex] f'(r) = 0 [/itex] produces a maximum instead of a minimum or an inflection point. You also have to check if the endpoints of the range of [itex] r [/itex] are extrema. In this case [itex] r = 0 [/itex] is a minimum. Since angles (in mathematics) are usually measured counterclockwise from a horizontal reference line, I computed the angle that way. 


#13
Feb2112, 11:50 AM

P: 18

Hi Stephen,
I appreciate your work. Can you just explain to mee exactly how did you get f(r) = f(r)=arctan(r)−arctan(λr) where r=tan(α)? If this is for a reported angle minus physical angle, wouldn't the equation read arctan(λr)arctan(r)? Why did you use arctans the way you did? A few more steps would be helpful. 


#14
Feb2112, 01:30 PM

Sci Advisor
P: 3,300

Suppose we have right triangle ABC with hypoteneus AC, let [itex] \alpha = \angle BAC [/itex]. Let [itex] r = \frac{BC}{AB} [/itex]. Then [itex] \alpha = \arctan(r) [/itex]. If [itex] BC [/itex] is multiplied by a factor of [itex] \lambda [/itex] then [itex] r [/itex] is multiplied by a factor of [itex] \lambda [/itex] and [itex] \angle BAC [/itex] becomes [itex] \alpha' = \arctan( \lambda r) [/itex]. So the absolute value of the change in angle is [itex] \arctan(r)  \arctan(\lambda r)  [/itex].



#15
Feb2312, 06:41 PM

P: 18

Thanks, Stephen!



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