Quantcast

Need Math Help- 6" Dia. Honest John Nosecone

The Rocketry Forum

Help Support The Rocketry Forum:

jmmome

Well-Known Member
TRF Supporter
Joined
Jan 11, 2013
Messages
396
Reaction score
104
Location
Maumee (Toledo) OH
May have inhaled too much foam board dust constructing the bottom 14" section of the nosecone. That part was (relatively) easy.

Now I need to construct the top 24" of nosecone, which is a 3:1 ogive. The base of it will be 8" in diameter to match the top of the bottom section I've already created.

So- 24" length for the centerline, and 4" for the radius............how do I calculate my pivot point length & location to create the ogive shape? I throw myself at the mercy of the court.

Thanks!!!!

IMG_5375_1.jpghonest-john-plans.jpg
 

Brent

Well-Known Member
Joined
Jan 19, 2009
Messages
1,623
Reaction score
64
May have inhaled too much foam board dust constructing the bottom 14" section of the nosecone. That part was (relatively) easy.

Now I need to construct the top 24" of nosecone, which is a 3:1 ogive. The base of it will be 8" in diameter to match the top of the bottom section I've already created.

So- 24" length for the centerline, and 4" for the radius............how do I calculate my pivot point length & location to create the ogive shape? I throw myself at the mercy of the court.

Thanks!!!!

View attachment 440923View attachment 440924
 

heada

Well-Known Member
Joined
Jan 18, 2009
Messages
3,725
Reaction score
949
Location
Indianapolis, Indiana
I can't answer your question but just wanted to say when I first read the title of your thread, I thought you needed help with a dishonest john. I had to re-read it and then chuckled to myself a few times.
 

ThirstyBarbarian

Well-Known Member
TRF Supporter
Joined
Feb 11, 2013
Messages
8,547
Reaction score
1,119
I’m not sure how you would do it mathematically. When I did my foam project, I made an Open Rocket file, specified the nosecone type, and then printed out the template at full scale. It took a few sheets of paper that I had to tape together for the full template.
 

boatgeek

Well-Known Member
Joined
Dec 27, 2014
Messages
3,049
Reaction score
1,475
May have inhaled too much foam board dust constructing the bottom 14" section of the nosecone. That part was (relatively) easy.

Now I need to construct the top 24" of nosecone, which is a 3:1 ogive. The base of it will be 8" in diameter to match the top of the bottom section I've already created.

So- 24" length for the centerline, and 4" for the radius............how do I calculate my pivot point length & location to create the ogive shape? I throw myself at the mercy of the court.

Thanks!!!!

View attachment 440923View attachment 440924
I don't know how to do this mathematically, because I always let AutoCAD do the math. 😀 The center of the curve is 5'-10" (70") from the center of the base of the ogive section. In other words, the radius of the ogive section is 74".
 

jmmome

Well-Known Member
TRF Supporter
Joined
Jan 11, 2013
Messages
396
Reaction score
104
Location
Maumee (Toledo) OH
I don't know how to do this mathematically, because I always let AutoCAD do the math. 😀 The center of the curve is 5'-10" (70") from the center of the base of the ogive section. In other words, the radius of the ogive section is 74".
Thanks!
 

K'Tesh

OpenRocket Chuck Norris
TRF Supporter
Joined
Mar 27, 2013
Messages
14,508
Reaction score
1,083
If you're looking for some HoJo nosecones sims I have a couple of them (as part of full sims) over in my OpenRocket Files thread.
1607019716389.png



1607019790076.png


You can scale them up to your necessary diameter, and try something like Thirsty suggested. FWIW, I recently scored a set of original for the BT-55 Hojo (1919). I'm hoping that they'll be sent soon so I can try to scan everything accurately (in a way that I can see easily and clearly) for the 1919 sim. I've been sent decals for other Hojo's but I want to make sure that the 1919's sim matches the kit.
 

kuririn

BARGeezer
TRF Supporter
Joined
Oct 3, 2016
Messages
5,427
Reaction score
3,016
Location
Hawaii
You have the drill/lathe.
You have the dimensional drawing.
Create a template by upscaling the drawing to your pieces.
Cut out the template on cardboard and turn your nose cone.
No math needed (except for simple ratios for the upscale), no equations to deal with.
 

dr wogz

Fly caster
Joined
Feb 5, 2009
Messages
5,874
Reaction score
1,095
Location
Land of Poutine!
+1 on allowing Autocad do the math & print out a template

(Come to think of it.. I don't really remember the last time I did "math" [except build & tolerance stacks].. I always draw in ACAD, so it does my trig for me. FEM in Creo.. now that's another story!)
 

jmmome

Well-Known Member
TRF Supporter
Joined
Jan 11, 2013
Messages
396
Reaction score
104
Location
Maumee (Toledo) OH
I don't know how to do this mathematically, because I always let AutoCAD do the math. 😀 The center of the curve is 5'-10" (70") from the center of the base of the ogive section. In other words, the radius of the ogive section is 74".
The 74" dimension worked perfectly for drawing the template- thanks!

Not only will it help me to cut each 1" thick foam piece a little larger, but I'll also transfer the template to 1/2" plywood, cut two of them and glue together, epoxy sandpaper to the curve of the template, and see if it sands down the foam board as well as my first try with a straight 16" sanding board.
 

Nytrunner

Pop lugs, not drugs
TRF Supporter
Joined
Oct 15, 2016
Messages
7,505
Reaction score
3,066
Location
Huntsville AL
Oh c'mon folks, this isn't the Barrowman equations, do the math and keep your brain from turning into mush!

Circles are annoying, do triangles instead!

Example, a circles is such:
$$ r^2=(x-h)^2+(y-k)^2$$

Where [x,y] is the coordinates of your points on the circle and [h,k] are the coordinates of your center.
Since the radius is the same, you could write the circle equation for both points, set them equal to each other and solve for r and k (the radius, and the offset of the center)

If your center is at [0,0] then you end up with
$$x^2+y^2=r^2$$
which is suspiciously similar to the pythagorean theorem (triangle!)

You know 2 points [0,4] and [24,0] both of which are points along the circle that forms your ogive, and thus have the same radius.
Knowing that your circle center is at 0 on the x-axis simplifies things and allows us to skip the circle equation. As with any geometry problem, draw a picture first (exercise left to student)

Treating the cone axis as the x-axis, and cone base as the y-axis, we know your radius extends from circle center along the y-axis to your first point which is 4" above the x-axis [0,4]
If the first point [0,4] is above the x-axis, then your circle center will be below it and k is negative (sign convention is important!)

That means your radius is equal to 4 plus -k (the circle's y-offset)

$$(1): r=4+(-k)$$

If the 2nd point is on the x-axis (y=0) then we can draw a triangle with one leg equal to the center offset (-k) and the other leg equal to it's x position (24")

Applying pythagoras:
$$(2): r^2=24^2+(-k)^2$$
We once again have 2 equations with two variables: vertical circle offset (k) and circle radius (r) which you really want

Draw out -k from equation 1
$$(3): -k=r-4$$
Substitute in equation 2 so the whole thing is in terms of r
$$(4): r^2=24^2+(r-4)^2 $$
Solving for r
$$ 0=576+r^2-8r+16-r^2=>$$
$$8r=592=>$$
$$r=74$$
You have the number you want, but it's best for to finish the calc and check yourself. If the radius is 74 then the circle offset can be found

$$74=-k+4=>$$
$$k=-70$$
That result matches our earlier statement that k would be a negative value reflecting the circle's position below the x-axis
Final check

$$74^2=24^2+(-70)^2=>$$
$$5476=576+4900=>$$
$$5476=5476$$

And if you plug those values into the circle equation at the top for both your coordinates, it'll check out there too
 

boatgeek

Well-Known Member
Joined
Dec 27, 2014
Messages
3,049
Reaction score
1,475
+1 on allowing Autocad do the math & print out a template

(Come to think of it.. I don't really remember the last time I did "math" [except build & tolerance stacks].. I always draw in ACAD, so it does my trig for me. FEM in Creo.. now that's another story!)
I always tell high school students interested in engineering that you need higher math to get an engineering degree, but you don't need higher math to do engineering. I rarely go past a square root, though I sometimes get into trig.

Oh c'mon folks, this isn't the Barrowman equations, do the math and keep your brain from turning into mush!

Circles are annoying, do triangles instead!

Example, a circles is such:
$$ r^2=(x-h)^2+(y-k)^2$$

Where [x,y] is the coordinates of your points on the circle and [h,k] are the coordinates of your center.
Since the radius is the same, you could write the circle equation for both points, set them equal to each other and solve for r and k (the radius, and the offset of the center)

If your center is at [0,0] then you end up with
$$x^2+y^2=r^2$$
which is suspiciously similar to the pythagorean theorem (triangle!)

You know 2 points [0,4] and [24,0] both of which are points along the circle that forms your ogive, and thus have the same radius.
Knowing that your circle center is at 0 on the x-axis simplifies things and allows us to skip the circle equation. As with any geometry problem, draw a picture first (exercise left to student)

Treating the cone axis as the x-axis, and cone base as the y-axis, we know your radius extends from circle center along the y-axis to your first point which is 4" above the x-axis [0,4]
If the first point [0,4] is above the x-axis, then your circle center will be below it and k is negative (sign convention is important!)

That means your radius is equal to 4 plus -k (the circle's y-offset)

$$(1): r=4+(-k)$$

If the 2nd point is on the x-axis (y=0) then we can draw a triangle with one leg equal to the center offset (-k) and the other leg equal to it's x position (24")

Applying pythagoras:
$$(2): r^2=24^2+(-k)^2$$
We once again have 2 equations with two variables: vertical circle offset (k) and circle radius (r) which you really want

Draw out -k from equation 1
$$(3): -k=r-4$$
Substitute in equation 2 so the whole thing is in terms of r
$$(4): r^2=24^2+(r-4)^2 $$
Solving for r
$$ 0=576+r^2-8r+16-r^2=>$$
$$8r=592=>$$
$$r=74$$
You have the number you want, but it's best for to finish the calc and check yourself. If the radius is 74 then the circle offset can be found

$$74=-k+4=>$$
$$k=-70$$
That result matches our earlier statement that k would be a negative value reflecting the circle's position below the x-axis
Final check

$$74^2=24^2+(-70)^2=>$$
$$5476=576+4900=>$$
$$5476=5476$$

And if you plug those values into the circle equation at the top for both your coordinates, it'll check out there too
On the other hand, it took me about 45 seconds to do the layout in CAD... :headspinning:
 

jmmome

Well-Known Member
TRF Supporter
Joined
Jan 11, 2013
Messages
396
Reaction score
104
Location
Maumee (Toledo) OH
Oh c'mon folks, this isn't the Barrowman equations, do the math and keep your brain from turning into mush!

Circles are annoying, do triangles instead!

Example, a circles is such:
$$ r^2=(x-h)^2+(y-k)^2$$

Where [x,y] is the coordinates of your points on the circle and [h,k] are the coordinates of your center.
Since the radius is the same, you could write the circle equation for both points, set them equal to each other and solve for r and k (the radius, and the offset of the center)

If your center is at [0,0] then you end up with
$$x^2+y^2=r^2$$
which is suspiciously similar to the pythagorean theorem (triangle!)

You know 2 points [0,4] and [24,0] both of which are points along the circle that forms your ogive, and thus have the same radius.
Knowing that your circle center is at 0 on the x-axis simplifies things and allows us to skip the circle equation. As with any geometry problem, draw a picture first (exercise left to student)

Treating the cone axis as the x-axis, and cone base as the y-axis, we know your radius extends from circle center along the y-axis to your first point which is 4" above the x-axis [0,4]
If the first point [0,4] is above the x-axis, then your circle center will be below it and k is negative (sign convention is important!)

That means your radius is equal to 4 plus -k (the circle's y-offset)

$$(1): r=4+(-k)$$

If the 2nd point is on the x-axis (y=0) then we can draw a triangle with one leg equal to the center offset (-k) and the other leg equal to it's x position (24")

Applying pythagoras:
$$(2): r^2=24^2+(-k)^2$$
We once again have 2 equations with two variables: vertical circle offset (k) and circle radius (r) which you really want

Draw out -k from equation 1
$$(3): -k=r-4$$
Substitute in equation 2 so the whole thing is in terms of r
$$(4): r^2=24^2+(r-4)^2 $$
Solving for r
$$ 0=576+r^2-8r+16-r^2=>$$
$$8r=592=>$$
$$r=74$$
You have the number you want, but it's best for to finish the calc and check yourself. If the radius is 74 then the circle offset can be found

$$74=-k+4=>$$
$$k=-70$$
That result matches our earlier statement that k would be a negative value reflecting the circle's position below the x-axis
Final check

$$74^2=24^2+(-70)^2=>$$
$$5476=576+4900=>$$
$$5476=5476$$

And if you plug those values into the circle equation at the top for both your coordinates, it'll check out there too
And to think i used to teach that stuff in high school in 1976 & 1977............your equations brought back memories of blank faces on kids in my class.
 

RocketRev

Lifetime Supporter
TRF Lifetime Supporter
Joined
Jul 25, 2012
Messages
260
Reaction score
131
Uh....... blank faces..........on the kids..............yes..............that's what I was going to say too..........

Sorry, I'm laughing to hard!
 

itsmeGriff

Member
Joined
Nov 3, 2020
Messages
8
Reaction score
7
Location
Central IL
I can't answer your question but just wanted to say when I first read the title of your thread, I thought you needed help with a dishonest john. I had to re-read it and then chuckled to myself a few times.
Baahahaa, some brightness to my crappy day!
 

Ez2cDave

Well-Known Member
TRF Supporter
Joined
Jan 18, 2009
Messages
4,111
Reaction score
1,167
I don't know how to do this mathematically, because I always let AutoCAD do the math. 😀 The center of the curve is 5'-10" (70") from the center of the base of the ogive section. In other words, the radius of the ogive section is 74".
Time to go "Old School" . . . LOL !

See PDF's below.

Dave F.
 

Attachments

jmmome

Well-Known Member
TRF Supporter
Joined
Jan 11, 2013
Messages
396
Reaction score
104
Location
Maumee (Toledo) OH
So far- so good. First 14" of 6" body tube dia. to 8" fattest part of the nosecone done, including adding the 6" long shoulder which will go into the body tube.

IMG_5377_1.jpg

Next 8" of disks of the final 24" of the ogive portion are cut. I'll cut a 24" long ogive template, attach sandpaper to it's edge, and see if I can hinge it to the tip of the nosecone end of the sanding lathe. I'll start sanding from the 8" diameter, and work my way down to the tip by adding consecutively smaller foam board disks. I could "freehand" the ogive sanding, but I'd like to try this first.
 

Cape Byron

The BAR formerly known as Skippy-2
TRF Supporter
Joined
Aug 4, 2019
Messages
950
Reaction score
845
Location
Northern Rivers, Australia
Rocketeers that don't like math? Scary! :oops:
I like maths fine, but it doesn't like me. I had to cram maths with a tutor to get into Uni as a mature age student studying Physics and Chemistry (and some other stuff). I have three university degrees, but maths I still struggle with. Use it or lose it, as they say.
 

Ez2cDave

Well-Known Member
TRF Supporter
Joined
Jan 18, 2009
Messages
4,111
Reaction score
1,167

Nytrunner

Pop lugs, not drugs
TRF Supporter
Joined
Oct 15, 2016
Messages
7,505
Reaction score
3,066
Location
Huntsville AL
Rocketeers have become dependent on "plugging everything into" a program or an app . . .
I wholly support the use of those tools (and use them 90% of the time). Once you know how to do the math, the tools let you do it more efficiently and in better time (and better understand the results and limitations).
 

Ez2cDave

Well-Known Member
TRF Supporter
Joined
Jan 18, 2009
Messages
4,111
Reaction score
1,167
I wholly support the use of those tools (and use them 90% of the time). Once you know how to do the math, the tools let you do it more efficiently and in better time (and better understand the results and limitations).
The "downfall of Modern Man" began with the invention of the "pocket calculator" and has progressed from there.

Dave F.
 

caveduck

semi old rocketeer
Joined
Jun 6, 2011
Messages
1,478
Reaction score
177
LOL I totally get the math but I bet you could look at the picture, turn that curve by hand and no one would be able to tell without a caliper :)
 

jmmome

Well-Known Member
TRF Supporter
Joined
Jan 11, 2013
Messages
396
Reaction score
104
Location
Maumee (Toledo) OH
Wow- thanks to all!

Finished the nosecone- looks pretty close to what I was hoping for.

38" long. Fits a 6" Blue Tube. 8" (maybe a tad more) at its fattest diameter.

Still need to add the rest of the 3/8" threaded rod central spine, add the BB/epoxy nose weight between inches 6 to 12, and epoxy the four individual sections together. First photo shows some of the guides/stops I made to help keep the diameters consistent between the top of the previous section to the bottom of the next section. The ogive-shaped block & the straight block are what I attached 60 grit sandpaper to.

Next..........the dreaded fiberglassing.

IMG_5379_1.jpgIMG_5443_1.jpg
 
Top