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07-14-2012, 09:29 PM
I was thinking that you can compute the angle if you have the two sides. For example if your board is .5 inches and the cut is .7 inches long you would have a 45.58 degree angle, Has any one tried this

07-14-2012, 09:55 PM
I was thinking that you can compute the angle if you have the two sides. For example if your board is .5 inches and the cut is .7 inches long you would have a 45.58 degree angle, Has any one tried thisYou may set the grid size as needed (within Extrude), set the Snap-to-grid function. From there, you need only draw your extrude line (for your 45 degree) from one corner to the opposite corner. Then just continue your extrude layout as desired.

07-14-2012, 10:09 PM
I was trying to find out if anyone had tried this and how the angles turned out.

07-14-2012, 11:48 PM
I was trying to find out if anyone had tried this and how the angles turned out.Your math may be off! You can use designer to see the results by activating the Snap-to-grid function. From there you can draw your connected lines to scale, assuring your right angle. I used 7 and 5 inches. The attached picture shows the figure in designer. The math shows one angle at 54.5 so the smaller angle must be 35.5.
54734

07-15-2012, 12:23 AM
Your math may be off! You can use designer to see the results by activating the Snap-to-grid function. From there you can draw your connected lines to scale, assuring your right angle. I used 7 and 5 inches. The attached picture shows the figure in designer. The math shows one angle at 54.5 so the smaller angle must be 35.5.
54734

Your number is correct, but I used .5 for the height and .7 for the Hipotinoze( Been a long time since geometry class) that would give you the 45.58 degree. We can set the thickness of the board = .5 and the length of the cut .7.

bluecobra
07-15-2012, 08:54 PM
Your number is correct, but I used .5 for the height and .7 for the Hipotinoze( Been a long time since geometry class) that would give you the 45.58 degree. We can set the thickness of the board = .5 and the length of the cut .7.

A right triangle has a base, height & hypotenuse.

If you're looking for 45° angles, both the base & height need to be the same length (also known as an isosceles triangle).

If you want a 45° bevel, make the base .5" long and make the height .5" high or adjust accordingly.

If you need the length of the hypotenuse, you have to trig it out, I think that involves pythagorean theorem (a²+b²=c², where c² is the hypotenuse). So for a .5" triangle it'd be .707.

*i miss my machining days

07-15-2012, 10:00 PM
A right triangle has a base, height & hypotenuse.

If you're looking for 45° angles, both the base & height need to be the same length (also known as an isosceles triangle).

If you want a 45° bevel, make the base .5" long and make the height .5" high or adjust accordingly.

If you need the length of the hypotenuse, you have to trig it out, I think that involves pythagorean theorem (a²+b²=c², where c² is the hypotenuse). So for a .5" triangle it'd be .707.

*i miss my machining days

You are saying the same thing I said except I use the Height and the Hypotenuse. I used the Hypotenuse because its lenght is show by the program, the height is the wood.
If you are looking for the angle needed. You would need to compute the angle of each side (180*(n-2))/ n with n being the number of sides. For example for 6 sides the angle would be 120. or (180*(6-2))/6=120.
Boy this takes me back a couple of years.

bluecobra
07-16-2012, 12:19 PM
You are saying the same thing I said except I use the Height and the Hypotenuse. I used the Hypotenuse because its lenght is show by the program, the height is the wood.
If you are looking for the angle needed. You would need to compute the angle of each side (180*(n-2))/ n with n being the number of sides. For example for 6 sides the angle would be 120. or (180*(6-2))/6=120.
Boy this takes me back a couple of years.

Agreed.
But we didn't have Google back then either to google (geometric shape) calculator.
Triangle calculator: http://ostermiller.org/calc/triangle.html
Regular polygon calculator: http://www.1728.org/polygon.htm

Wonder how good today's kids are with mental math with instant info like this.

07-16-2012, 12:33 PM
Agreed.
But we didn't have Google back then either to google (geometric shape) calculator.
Triangle calculator: http://ostermiller.org/calc/triangle.html
Regular polygon calculator: http://www.1728.org/polygon.htm

Wonder how good today's kids are with mental math with instant info like this.Only as good as we teach them!

Take this problem, and first solve it yourself, then give it to others.

Situation:
You have a 2 mile track, which a car must travel at an average speed of 60 MPH.

Problem:
If the car goes the 1st mile going 30 MPH, how fast must it travel the 2nd mile?
The answer is not 90 MPH!

60 MPH gives you 2 minutes to travel the course. You used 2 minutes traveling 30 MPH the 1st mile. You are out of time! You can not make the goal!

07-16-2012, 12:47 PM
Before my math problem creates a life of its own, re visit post #9. The answer is hidden in "White" text. Just highlight the post and the solution appears!

I don't want to hijack the actual purpose of the thread!

bluecobra
07-16-2012, 01:12 PM
Before my math problem creates a life of its own, re visit post #9. The answer is hidden in "White" text. Just highlight the post and the solution appears!

I don't want to hijack the actual purpose of the thread!

I read your answer and it is incorrect in relation to the original question asked as there is no time element put into place for the 2 mile or 60MPH avg to take place in.

60mph is a unit of speed (velocity), not time.

For example, there is no limitation to somebody speeding up to 120mph to make up "time" before the 2 mile marker comes up.

cnsranch
07-16-2012, 01:42 PM
Forget all that, Mark - here's the REAL answer.....

Recruit Superman to make the last mile - he can travel faster than the speed of sound, making time not only stand still, but go backward!!!!

I gotta get a life.

07-16-2012, 01:54 PM
Only as good as we teach them!

Take this problem, and first solve it yourself, then give it to others.

Situation:
You have a 2 mile track, which a car must travel at an average speed of 60 MPH.

Problem:
If the goes the 1st mile going 30 MPH, how fast must it travel the 2nd mile?
The answer is not 90 MPH!

60 MPH gives you 2 minutes to travel the course. You used 2 minutes traveling 30 MPH the 1st mile. You are out of time! You can not make the goal!

We will never know because it is a electric car and it can't go faster than 30mph

bergerud
07-16-2012, 03:01 PM
I read your answer and it is incorrect in relation to the original question asked as there is no time element put into place for the 2 mile or 60MPH avg to take place in.

60mph is a unit of speed (velocity), not time.

For example, there is no limitation to somebody speeding up to 120mph to make up "time" before the 2 mile marker comes up.

Bud is correct. It is a matter of definition. The average of 30 and 90 is 60 as numbers but not as speeds. The average speed is defined as the total distance divided by the total time. It is not a simple number but a rate which involves time. To calculate average speed, you need a weighted average. For example, if I go at 30 mph for 1 hour and then at 60 mph for 2 hours, the average speed is

((1)(30)+(2)(60))/(1+2)=50.

Note that this is just total distance divided by total time.

bluecobra
07-16-2012, 05:05 PM
Thought you guys would like to know that I've removed all traces of my foot from my mouth. :mrgreen:

I totally disregarded the basic fundamentals of physics, even reaffirming Bud's answer with my incorrect answer.

Anybody care to point me in the direction of the line for crow pie? ;)

...superman & electric cars huh?

TerryT
07-16-2012, 08:12 PM
Bud is correct. It is a matter of definition. The average of 30 and 90 is 60 as numbers but not as speeds. The average speed is defined as the total distance divided by the total time. It is not a simple number but a rate which involves time. To calculate average speed, you need a weighted average. For example, if I go at 30 mph for 1 hour and then at 60 mph for 2 hours, the average speed is

((1)(30)+(2)(60))/(1+2)=50.

Note that this is just total distance divided by total time.

Seems to me that as the question was posed, to average 60 mph over two miles with the first mile at 30mph, the correct answer would indeed be 90 mph for the 2nd mile.
If the question had been posed as averaging 60 mph over a two mile course with a two minute time limit THEN it would be impossible if you only did 30mph for the first mile.

30mph for 1 mile and 90 mph for the 2nd mile would average 60 mph over the two mile course. It would just take 2 minutes and roughly 40 seconds to do. Similarly in a side by side drag race with both cars leaving the starting line at the same time and traveling the same distance, 1/4 mile, the slower car (speed) may win if it covers the distance faster (time).
Since everyone agrees with Buds answer I must be making a mistake somewhere but I'm not seeing it.

mathman
07-16-2012, 10:34 PM
Seems to me that as the question was posed, to average 60 mph over two miles with the first mile at 30mph, the correct answer would indeed be 90 mph for the 2nd mile.
If the question had been posed as averaging 60 mph over a two mile course with a two minute time limit THEN it would be impossible if you only did 30mph for the first mile.

2 miles at 60mph takes two minutes, so if you average 60mph over the two miles you HAVE to traverse the two miles in two minutes. The first mile at 30mph takes two minutes - anything more is going to keep the average speed under 60mph. 1 mile at 30mph followed by 1 mile at 90mph results in an average speed of 45mph. Two miles at exactly 45mph means each mile takes 1-1/3 minutes, or 2-2/3 minutes for the two miles. 2/3 of a minute is 40 seconds so total time is 2 minutes 40 seconds, which you have already shown is how long it will take to cover the two miles with the first mile at 30mph and the second at 90mph.

bergerud
07-16-2012, 11:07 PM
Terry: average speed is total distance divided by total time. It is not the average of the speeds! If I were to travel all day at 10 mph and then rip at 100 mph for 10 seconds would the average speed for the day be 55 mph?

TerryT
07-16-2012, 11:24 PM
Thanks guys, I hate to seem stupid, I'm trying to understand, maybe it's just because of no sleep last night but I'm just not getting it yet LOL.

TerryT
07-16-2012, 11:30 PM
I GOT IT!!! It just took awhile. Of course 2 miles in 160 seconds can't be an average of 60 mph. Geez

mathman
07-17-2012, 12:14 AM
I GOT IT!!! It just took awhile. Of course 2 miles in 160 seconds can't be an average of 60 mph. Geez

The way the question is worded is designed to misdirect the reader - it pushes you to concentrate on getting an average of 60 out of 30 and [something_to_be_determined] rather than what the real point is, which, in this case, is realizing that there is a 2-minute time limit imposed by the nature of the problem even though it was never explicitly stated. Fun problems. ;)

TerryT
07-17-2012, 09:15 AM
It was fun, I like that kind of stuff, but I usually do a little better than that.