So I'm having trouble with an equation in my experiment (https://www.sciencebuddies.org/science- ... p036.shtml). The equation is:
crater depth (km) = shadow length (km) × tan(angle of the sun above the horizon)
d =L tan β
d = crater depth (km)
L = shadow length (km)
β = angle of the sun above the horizon
This is equation 11. I tried to just enter the tan into my calculator, but I get a negative number. Am I actually supposed to calculate the tangent, or am I just entering it wrong? Please help, and thank you!
Help Please with Equation!
Moderators: kgudger, bfinio, Moderators
Re: Help Please with Equation!
Hello. I'm sorry you're having problems. I love this experiment!
The angle of the sun above the horizon should be a value 0-90 in degrees. The tangent value of degrees between 0-90 is always positive. A couple of options of things going wrong. What is your sun angle? Is it in degrees or radians? Is the value greater than 90 degrees?
The angle of the sun above the horizon should be a value 0-90 in degrees. The tangent value of degrees between 0-90 is always positive. A couple of options of things going wrong. What is your sun angle? Is it in degrees or radians? Is the value greater than 90 degrees?
Deana
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- Posts: 13
- Joined: Tue Nov 29, 2011 2:06 pm
- Occupation: Student: 11th grade
- Project Question: Finding frictional force
- Project Due Date: 12/5/15
- Project Status: I am finished with my experiment and analyzing the data
Re: Help Please with Equation!
My angle of the sun above the horizon is 8.2. I found that by using equation 10. It doesn't say whether it's in degrees or radians, but here is how I found that:
Angle of the sun above the horizon = 90 - incidence angle
Angle of the sun above the horizon = 90- 81.8
Angle of the sun above the horizon = 8.2
Then in equation 11 this is how I was originally doing it:
shadow length (km) × tan(angle of sun above horizon) = crater depth (km)
4.13002 × tan(8.2) = -11.4556
I know it can't be a negative number, so am I supposed to do this instead?
4.13002 × tan^-1(8.2) = 5.98623
But I'm still not sure if it's right.
Thanks!
Angle of the sun above the horizon = 90 - incidence angle
Angle of the sun above the horizon = 90- 81.8
Angle of the sun above the horizon = 8.2
Then in equation 11 this is how I was originally doing it:
shadow length (km) × tan(angle of sun above horizon) = crater depth (km)
4.13002 × tan(8.2) = -11.4556
I know it can't be a negative number, so am I supposed to do this instead?
4.13002 × tan^-1(8.2) = 5.98623
But I'm still not sure if it's right.
Thanks!
Re: Help Please with Equation!
Most calculators expect your input in radians, not degrees.
If your calculator is in radians, then tan(8.2) = -2.77. This is where your negative number is coming from.
You have two options. The Microsoft Windows built in calculator allows you to change to degrees before calculating.
Otherwise, you should convert degrees to radians before calculating. Here is the formula:
radians = degrees * pi / 180
8.2 degrees = 0.143 radians
Then, tan(0.143) = 0.144
Here is a good read on the relationship between degrees and radians:
http://math.rice.edu/~pcmi/sphere/drg_txt.html
Hopefully, this will get you back on track. Be sure to write back if you are still having problems. Good luck!
If your calculator is in radians, then tan(8.2) = -2.77. This is where your negative number is coming from.
You have two options. The Microsoft Windows built in calculator allows you to change to degrees before calculating.
Otherwise, you should convert degrees to radians before calculating. Here is the formula:
radians = degrees * pi / 180
8.2 degrees = 0.143 radians
Then, tan(0.143) = 0.144
Here is a good read on the relationship between degrees and radians:
http://math.rice.edu/~pcmi/sphere/drg_txt.html
Hopefully, this will get you back on track. Be sure to write back if you are still having problems. Good luck!
Deana
-
- Posts: 13
- Joined: Tue Nov 29, 2011 2:06 pm
- Occupation: Student: 11th grade
- Project Question: Finding frictional force
- Project Due Date: 12/5/15
- Project Status: I am finished with my experiment and analyzing the data
Re: Help Please with Equation!
Hi again,
I understand all of that now, but now I'm having trouble with the ratios. My first one is 0.595143/413002, but I don't know how to reduce that. I was hoping for something more simple like 1 to 3, or 2 to 3. I converted it to a fraction on my calculator, and it says 119029/826004, so should I just leave it at that, or find something close to it but more simple?
Thanks again!
p.s this is part of equation 12
I understand all of that now, but now I'm having trouble with the ratios. My first one is 0.595143/413002, but I don't know how to reduce that. I was hoping for something more simple like 1 to 3, or 2 to 3. I converted it to a fraction on my calculator, and it says 119029/826004, so should I just leave it at that, or find something close to it but more simple?
Thanks again!
p.s this is part of equation 12
Re: Help Please with Equation!
So, your crater is only 0.5 km deep but over 413,000 km wide? That is how I interpret equation 12. Is that right?
Deana
-
- Posts: 13
- Joined: Tue Nov 29, 2011 2:06 pm
- Occupation: Student: 11th grade
- Project Question: Finding frictional force
- Project Due Date: 12/5/15
- Project Status: I am finished with my experiment and analyzing the data
Re: Help Please with Equation!
Oh I'm sorry, it's actually 4.13002, I forgot the decimal point!
Re: Help Please with Equation!
Lol. That sounds more feasible! I like your idea of coming up with a ratio. It makes concepts easier to understand sometimes.
When dealing with fractions, our brain is trained to reduce. In this case, since your numerator isn't a whole number, think in the opposite direction. Come up with a multiplying factor to get your numerator to a whole number. It doesn't have to be exact. Your numerator is close enough to 0.5. Work with 0.5 to come up with a multiplier. Don't forget to apply the multiplier to the denominator, as well. Then, you will have a good ratio.
I'm being vague on purpose. You're doing great so far. I want you to try to figure this out first. . Write back with what you think.
When dealing with fractions, our brain is trained to reduce. In this case, since your numerator isn't a whole number, think in the opposite direction. Come up with a multiplying factor to get your numerator to a whole number. It doesn't have to be exact. Your numerator is close enough to 0.5. Work with 0.5 to come up with a multiplier. Don't forget to apply the multiplier to the denominator, as well. Then, you will have a good ratio.
I'm being vague on purpose. You're doing great so far. I want you to try to figure this out first. . Write back with what you think.
Deana
-
- Posts: 13
- Joined: Tue Nov 29, 2011 2:06 pm
- Occupation: Student: 11th grade
- Project Question: Finding frictional force
- Project Due Date: 12/5/15
- Project Status: I am finished with my experiment and analyzing the data
Re: Help Please with Equation!
I think I figured it out. To make the ratio as simplified as possible, the first number should most likely be one. So, in order to find the factor that I need to make that happen, I divided 1 by the crater depth for each crater. Then I would multiply that answer to both the crater depth and crater diameter (the numerator and the denominator) to get something close to a normal ratio. Here is an example:
Crater 1 Image 1
Crater Depth = 0.595145
1/0.595145 = 1.68026
0.595145 × 1.68026 = 1 (the numerator)
Crater Diameter = 6.00497
6.00497 × 1.68026 = 10.0899 (the denominator)
Final ratio (rounded) = 1:10
Does this process makes sense, and is it correct?
Thanks for the help!
Crater 1 Image 1
Crater Depth = 0.595145
1/0.595145 = 1.68026
0.595145 × 1.68026 = 1 (the numerator)
Crater Diameter = 6.00497
6.00497 × 1.68026 = 10.0899 (the denominator)
Final ratio (rounded) = 1:10
Does this process makes sense, and is it correct?
Thanks for the help!
Re: Help Please with Equation!
Before, your diameter was 4.13. Now, you are saying it is 6. Is this for a different image? Other than that, your calculations are correct. Good job!
Deana